Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn<\/p>\n\n\n\n
\u01a0\u0309 ba\u0300i tr\u01b0\u01a1\u0301c Vietlearn \u0111a\u0303 cu\u0300ng ba\u0323n ti\u0300m hi\u00ea\u0309u ca\u0301c ki\u00ea\u0301n th\u01b0\u0301c v\u00ea\u0300 s\u00f4\u0301 h\u01b0\u0303u ti\u0309, phe\u0301p c\u00f4\u0323ng, tr\u01b0\u0300, nh\u00e2n, chia. V\u01a1\u0301i ba\u0300i vi\u00ea\u0301t na\u0300y, Vietlearn se\u0303 cu\u0300ng ba\u0323n ti\u0300m hi\u00ea\u0309u lu\u0303y th\u01b0\u0300a cu\u0309a m\u00f4\u0323t s\u00f4\u0301 h\u01b0\u0303u ti\u0309. Lu\u0303y th\u01b0\u0300a cu\u0309a m\u00f4\u0323t s\u00f4\u0301 h\u01b0\u0303u ti\u0309 la\u0300 gi\u0300? Ti\u0301ch cu\u0309a lu\u0303y th\u01b0\u0300a, th\u01b0\u01a1ng cu\u0309a lu\u0303y ti\u0301nh nh\u01b0 th\u00ea\u0301 na\u0300o? Lu\u0303y th\u01b0\u0300a cu\u0309a lu\u0303y th\u01b0\u0300a co\u0301 kho\u0301 kh\u00f4ng? T\u00e2\u0301t ca\u0309 se\u0303 co\u0301 trong ba\u0300i vi\u00ea\u0301t sau \u0111\u00e2y:<\/p>\n\n\n\n
Ki\u00ea\u0301n th\u01b0\u0301c c\u01a1 ba\u0309n<\/p>\n\n\n\n
Ki\u00ea\u0301n th\u01b0\u0301c c\u01a1 ba\u0309n<\/p>\n\n\n\n
Lu\u0303y th\u01b0\u0300a cu\u0309a s\u00f4\u0301 mu\u0303 t\u01b0\u0323 nhi\u00ean<\/p>\n\n\n\n
Ta co\u0301 s\u00f4\u0301 h\u01b0\u0303u ti\u0309 x, lu\u0303y th\u01b0\u0300a b\u00e2\u0323c n. Ki\u0301 hi\u00ea\u0323u x^n (n la\u0300 s\u00f4\u0301 t\u01b0\u0323 nhi\u00ean, n > 1).<\/p>\n\n\n\n
Lu\u0303y th\u01b0\u0300a b\u00e2\u0323c n cu\u0309a s\u00f4\u0301 h\u01b0\u0303u ti\u0309 x t\u01b0\u0301c la\u0300 ti\u0301ch cu\u0309a n th\u01b0\u0300a s\u00f4\u0301 x.<\/p>\n\n\n\n
x^n = x\u2026x ( x thu\u00f4\u0323c Q, n thu\u00f4\u0323c N, n > 1)<\/p>\n\n\n\n
Vi\u0301 du\u0323: 5^3 = 5 x 5 x 5 ; 2^8 = 2 x 2 x 2 x 2 x 2 x 2 x 2 x 2<\/p>\n\n\n\n
Ta co\u0301 s\u00f4\u0301 h\u01b0\u0303u ti\u0309 x = a\/b thi\u0300 lu\u0303y th\u01b0\u0300a b\u00e2\u0323c n cu\u0309a x la\u0300 (a\/b)^n = a^n \/ b^n<\/p>\n\n\n\n
Vi\u0301 du\u0323: (1\/2)^2 = 1^2 \/ 2^2 ; (3\/4)^4 = 3^4\/ 4^4<\/p>\n\n\n\n
Xem ngay: Ho\u0323c Toa\u0301n 7 cu\u0300ng Vietlearn: Phe\u0301p nh\u00e2n, chia s\u00f4\u0301 h\u01b0\u0303u ti\u0309<\/p>\n\n\n\n
Ti\u0301ch cu\u0309a hai lu\u0303y th\u01b0\u0300a cu\u0300ng c\u01a1 s\u00f4\u0301<\/p>\n\n\n\n
Ta co\u0301 lu\u0303y th\u01b0\u0300a cu\u0309a hai s\u00f4\u0301 h\u01b0\u0303u ti\u0309 la\u0300 x^m va\u0300 x^n (trong \u0111o\u0301 x thu\u00f4\u0323c Q, m va\u0300 n thu\u00f4\u0323c N)<\/p>\n\n\n\n
Ti\u0301ch cu\u0309a 2 s\u00f4\u0301 h\u01b0\u0303u ti\u0309 tr\u00ean la\u0300 x^m . x^n = x^(m + n)<\/p>\n\n\n\n
K\u00ea\u0301t lu\u00e2\u0323n: Ti\u0301ch cu\u0309a lu\u0303y th\u01b0\u0300a cu\u0300ng c\u01a1 s\u00f4\u0301 cu\u0309a s\u00f4\u0301 h\u01b0\u0303u ti\u0309 b\u0103\u0300ng t\u00f4\u0309ng cu\u0309a ca\u0301c lu\u0303y th\u01b0\u0300a.<\/p>\n\n\n\n
Vi\u0301 du\u0323: 2^3 x 2^4 = 2^(3+4) = 2^7<\/p>\n\n\n\n
5^3 x 5^12 = 5^(3 +12) = 5^15<\/p>\n\n\n\n
Th\u01b0\u01a1ng cu\u0309a hai lu\u0303y th\u01b0\u0300a cu\u0300ng c\u01a1 s\u00f4\u0301<\/p>\n\n\n\n
Ta co\u0301 lu\u0303y th\u01b0\u0300a cu\u0309a hai s\u00f4\u0301 h\u01b0\u0303u ti\u0309 la\u0300 x^m va\u0300 x^n (trong \u0111o\u0301 x thu\u00f4\u0323c Q, m va\u0300 n thu\u00f4\u0323c N)<\/p>\n\n\n\n
Th\u01b0\u01a1ng cu\u0309a 2 s\u00f4\u0301 h\u01b0\u0303u ti\u0309 tr\u00ean la\u0300 x^m . x^n = x^(m \u2013 n) ( x kha\u0301c 0, m > ho\u0103\u0323c = n)<\/p>\n\n\n\n
K\u00ea\u0301t lu\u00e2\u0323n: Th\u01b0\u01a1ng cu\u0309a lu\u0303y th\u01b0\u0300a cu\u0300ng c\u01a1 s\u00f4\u0301 cu\u0309a s\u00f4\u0301 h\u01b0\u0303u ti\u0309 b\u0103\u0300ng hi\u00ea\u0323u cu\u0309a ca\u0301c lu\u0303y th\u01b0\u0300a.<\/p>\n\n\n\n
Vi\u0301 du\u0323: 6^3 : 6^2 = 6^(3-2) = 6^1<\/p>\n\n\n\n
8^8 : 8^5 = 8^(8 \u2013 5) = 8^3<\/p>\n\n\n\n
Lu\u0303y th\u01b0\u0300a cu\u0309a lu\u0303y th\u01b0\u0300a<\/p>\n\n\n\n
Lu\u0303y th\u01b0\u0300a cu\u0309a lu\u0303y th\u01b0\u0300a la\u0300 (x^m)^n = x^(m x n)<\/p>\n\n\n\n
K\u00ea\u0301t lu\u00e2\u0323n: Khi xu\u00e2\u0301t hi\u00ea\u0323n lu\u0303y th\u01b0\u0300a cu\u0309a lu\u0303y th\u01b0\u0300a cu\u0309a s\u00f4\u0301 h\u01b0\u0303u ti\u0309 ta th\u01b0\u0323c hi\u00ea\u0323n nh\u00e2n hai lu\u0303y th\u01b0\u0300a cu\u0309a s\u00f4\u0301 \u0111o\u0301.<\/p>\n\n\n\n
Vi\u0301 du\u0323: (2^3)^2 = 2^(3 x 2) = 2^6<\/p>\n\n\n\n
(8^7)^6 = 8^(7 x 6) = 8^42<\/p>\n\n\n\n
Ba\u0300i t\u00e2\u0323p v\u00e2\u0323n du\u0323ng<\/p>\n\n\n\n
Tr\u0103\u0301c nghi\u00ea\u0323m<\/p>\n\n\n\n
Ba\u0300i 1: K\u00ea\u0301t qua\u0309 \u0111u\u0301ng cu\u0309a (2\/3)^3 b\u0103\u0300ng:<\/p>\n\n\n\n
A. 8\/9<\/p>\n\n\n\n
B. 8\/27<\/p>\n\n\n\n
C. 4\/9<\/p>\n\n\n\n
D. 4\/27<\/p>\n\n\n\n
L\u01a1\u0300i gia\u0309i: (2\/3)^3 = 2^3 \/ 3^3 = 8\/27<\/p>\n\n\n\n
Ba\u0300i 2: K\u00ea\u0301t qua\u0309 \u0111u\u0301ng cu\u0309a phe\u0301p ti\u0301nh (1\/7)^2 x 7^2 b\u0103\u0300ng:<\/p>\n\n\n\n
A. 7<\/p>\n\n\n\n
B. 1\/7<\/p>\n\n\n\n
C. 1\/49<\/p>\n\n\n\n
D. 1<\/p>\n\n\n\n
L\u01a1\u0300i gia\u0309i: D<\/p>\n\n\n\n
Ba\u0300i 3: S\u1ed1 x^12 (v\u1edbi x \u22600) kh\u00f4ng b\u1eb1ng s\u1ed1 n\u00e0o sau \u0111\u00e2y?<\/p>\n\n\n\n
A. x^18 : x^16<\/p>\n\n\n\n
B. x^4 . x^8<\/p>\n\n\n\n
C. x^2 . x^6<\/p>\n\n\n\n
D. (x^3)^4<\/p>\n\n\n\n
L\u01a1\u0300i gia\u0309i: Ta c\u00f3: x^18 : x^16 = x^ (18-16) = x^2 (x \u22600) n\u00ean A \u0111\u00fang<\/p>\n\n\n\n
x^8 . x^4 = x^ (8+4) = x^12 n\u00ean B \u0111\u00fang<\/p>\n\n\n\n
(x^3)^4 = x^ (3.4) = x^12 n\u00ean D \u0111\u00fang<\/p>\n\n\n\n
=> \u0110\u00e1p \u00e1n C sai<\/p>\n\n\n\n
B\u00e0i 4: S\u1ed1 224 vi\u1ebft d\u01b0\u1edbi d\u1ea1ng l\u0169y th\u1eeba v\u1edbi s\u1ed1 m\u0169 b\u1eb1ng 8 l\u00e0<\/p>\n\n\n\n
A. 8^8<\/p>\n\n\n\n
B. 9^8<\/p>\n\n\n\n
C. 6^8<\/p>\n\n\n\n
D. \u0110\u00e1p a\u0301n kh\u00e1c<\/p>\n\n\n\n
L\u01a1\u0300i gia\u0309i:<\/p>\n\n\n\n
Ta c\u00f3: 2^24 = 2^ (3.8) = (2^3)^8 = 8^8<\/p>\n\n\n\n
Ch\u1ecdn \u0111\u00e1p \u00e1n A.<\/p>\n\n\n\n
Ba\u0300i 5: \u0110a\u0301p a\u0301n na\u0300o kh\u00f4ng \u0111u\u0301ng:<\/p>\n\n\n\n
A. (-2019)^0 = 1<\/p>\n\n\n\n
B. (0.5) x (0.5)^2 = 1\/4<\/p>\n\n\n\n
C. 4^6 : 4^4 = 16<\/p>\n\n\n\n
D. (-3)^3 x (-3)^2 = (-3)^5<\/p>\n\n\n\n
L\u01a1\u0300i gia\u0309i:<\/p>\n\n\n\n
4^6 : 4^4 = 4^ (6 \u2013 4) = 4^2 = 16 n\u00ean C \u0111\u00fang<\/p>\n\n\n\n
(-3)^3 . (-3)^2 = (-3)^(3+2) = (-3)^5 n\u00ean D \u0111\u00fang<\/p>\n\n\n\n
=> \u0110\u00e1p \u00e1n B sai.<\/p>\n\n\n\n
T\u01b0\u0323 lu\u00e2\u0323n<\/p>\n\n\n\n
Ba\u0300i 1: Ti\u0301nh gia\u0301 tri\u0323 cu\u0309a ca\u0301c bi\u00ea\u0309u th\u01b0\u0301c sau:<\/p>\n\n\n\n
L\u01a1\u0300i gia\u0309i:<\/p>\n\n\n\n
Ba\u0300i 2: Co\u0301 c\u00e1c s\u1ed1 t\u1ef1 nhi\u00ean n sao cho: 2.32 \u2265 2n > 8<\/p>\n\n\n\n
L\u01a1\u0300i gia\u0309i:<\/p>\n\n\n\n
\u0110i\u00ea\u0300u c\u00e2\u0300n nh\u01a1\u0301 khi la\u0300m ba\u0300i toa\u0301n lu\u0303y th\u01b0\u0300a<\/p>\n\n\n\n
Lu\u0303y th\u01b0\u0300a la\u0300 ph\u00e2\u0300n ki\u00ea\u0301n th\u01b0\u0301c m\u01a1\u0301i. Ca\u0301c da\u0323ng ba\u0300i v\u00ea\u0300 lu\u0303y th\u01b0\u0300a r\u00e2\u0301t \u0111a da\u0323ng. \u0110\u00ea\u0309 la\u0300m t\u00f4\u0301t ca\u0301c da\u0323ng ba\u0300i v\u00ea\u0300 lu\u0303y th\u01b0\u0300a cu\u0309a m\u00f4\u0323t s\u00f4\u0301 h\u01b0\u0303u ti\u0309 ca\u0301c ba\u0323n ho\u0323c sinh c\u00e2\u0300n nh\u01a1\u0301 nh\u01b0\u0303ng chu\u0301 y\u0301 sau:<\/p>\n\n\n\n
Chu\u0301 y\u0301 \u0111i\u00ea\u0300u ki\u00ea\u0323n cu\u0309a ba\u0300i toa\u0301n cho s\u0103\u0303n nh\u01b0: l\u01a1\u0301n h\u01a1n, nho\u0309 h\u01a1n, trong khoa\u0309ng,\u2026 \u0110a\u0301p a\u0301n cu\u00f4\u0301i cu\u0300ng c\u00e2\u0300n tho\u0309a ma\u0303n ca\u0301c \u0111i\u00ea\u0300u ki\u00ea\u0323n na\u0300y.<\/p>\n\n\n\n
Chu\u0301 y\u0301 \u0111\u00ea\u0301n ca\u0301c \u0111i\u00ea\u0300u ki\u00ea\u0323n b\u0103\u0301t bu\u00f4\u0323c cu\u0309a ph\u00e2n s\u00f4\u0301: m\u00e2\u0303u s\u00f4\u0301 pha\u0309i kha\u0301c 0, t\u01b0\u0309 s\u00f4\u0301 thu\u00f4\u0323c Z,\u2026 Kh\u00f4ng tho\u0309a ma\u0303n ca\u0301c \u0111i\u00ea\u0300u ki\u00ea\u0323n na\u0300y t\u01b0\u0301c ph\u00e2n s\u00f4\u0301 kh\u00f4ng t\u00f4\u0300n ta\u0323i.<\/p>\n\n\n\n
Khi m\u01a1\u0301i la\u0300m quen v\u01a1\u0301i lu\u0303y th\u01b0\u0300a r\u00e2\u0301t d\u00ea\u0303 nh\u00e2\u0300m l\u00e2\u0303n c\u00f4ng th\u01b0\u0301c c\u00f4\u0323ng, tr\u01b0\u0300, nh\u00e2n, chia \u01a1\u0309 s\u00f4\u0301 mu\u0303. Ca\u0301c ba\u0323n c\u00e2\u0300n n\u0103\u0301m ch\u0103\u0301c ca\u0301c c\u00f4ng th\u01b0\u0301c \u0111\u00ea\u0309 tra\u0301nh sai k\u00ea\u0301t qua\u0309.<\/p>\n\n\n\n
Ha\u0303y luy\u00ea\u0323n t\u00e2\u0323p t\u01b0 duy ti\u0301nh nhanh b\u0103\u0300ng ca\u0301ch ta\u0301ch ca\u0301c s\u00f4\u0301 ti\u0300m th\u01b0\u0300a s\u00f4\u0301 chung.<\/p>\n\n\n\n
H\u1ecdc t\u1ed1t l\u1edbp 7<\/p>\n\n\n\n
Bi\u0301 m\u00e2\u0323t Toa\u0301n ho\u0323c<\/p>\n\n\n\n
T\u0103ng thu\u00ea bao \u0111i\u00ea\u0323n thoa\u0323i khi t\u0103ng s\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i t\u01b0\u0300 7 con s\u00f4\u0301 l\u00ean 8 con s\u00f4\u0301<\/p>\n\n\n\n
Theo ba\u0323n, khi t\u0103ng s\u00f4\u0301 ca\u0301c s\u00f4\u0301 trong s\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i co\u0301 khi\u00ea\u0301n s\u00f4\u0301 ng\u01b0\u01a1\u0300i du\u0300ng, s\u00f4\u0301 thu\u00ea bao \u0111i\u00ea\u0323n thoai t\u0103ng l\u00ean hay kh\u00f4ng? C\u00e2u tra\u0309 l\u01a1\u0300i la\u0300 co\u0301. Ba\u0323n co\u0301 bi\u00ea\u0301t vi\u0300 sao kh\u00f4ng? N\u00ea\u0301u kh\u00f4ng bi\u00ea\u0301t ha\u0303y cu\u0300ng Vietlearn ly\u0301 gia\u0309i ngay sau \u0111\u00e2y:<\/p>\n\n\n\n
Ly\u0301 gia\u0309i<\/p>\n\n\n\n
S\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i bao g\u00f4\u0300m 8 ch\u01b0\u0303 s\u00f4\u0301 co\u0301 th\u00ea\u0309 la\u0300: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.<\/p>\n\n\n\n
S\u00f4\u0301 \u0111\u00e2\u0300u ti\u00ean cu\u0309a s\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i se\u0303 kh\u00f4ng th\u00ea\u0309 la\u0300 con s\u00f4\u0301 0. Vi\u0300 v\u00e2\u0323y, vi\u0323 tri\u0301 th\u01b0\u0301 nh\u00e2\u0301t chu\u0301ng ta co\u0301 9 s\u01b0\u0323 l\u01b0\u0323a cho\u0323n kha\u0301c nhau t\u01b0\u0300 1 \u0111\u00ea\u0301n 9. T\u01b0\u0300 vi\u0323 tri\u0301 th\u01b0\u0301 2 \u0111\u00ea\u0301n vi\u0323 tri\u0301 th\u01b0\u0301 8, chu\u0301ng ta se\u0303 co\u0301 10 s\u01b0\u0323 l\u01b0\u0323a cho\u0323n kha\u0301c nhau ma\u0300 kh\u00f4ng c\u00e2\u0300n quan t\u00e2m co\u0301 pha\u0309i la\u0300 s\u00f4\u0301 0 hay kh\u00f4ng. 7 vi\u0323 tri\u0301 na\u0300y co\u0301 th\u00ea\u0309 \u0111i\u00ea\u0300n ca\u0301c s\u00f4\u0301 t\u01b0\u0300 0 \u0111\u00ea\u0301n 9 co\u0301 th\u00ea\u0309 l\u0103\u0323p la\u0323i. Chu\u0301ng ta se\u0303 \u0111i ti\u0301nh xem 8 vi\u0323 tri\u0301 chu\u0301ng ta co\u0301 bao nhi\u00eau l\u01b0\u0323a cho\u0323n.<\/p>\n\n\n\n
Vi\u0323 tri\u0301 th\u01b0\u0301 nh\u00e2\u0301t co\u0301 9 l\u01b0\u0323a cho\u0323n. Vi\u0323 tri\u0301 th\u01b0\u0301 2 \u0111\u00ea\u0301n th\u01b0\u0301 8 co\u0301 10 l\u01b0\u0323a cho\u0323n. Ta co\u0301 phe\u0301p toa\u0301n:<\/p>\n\n\n\n
9 x 10 x 10 x 10 x 10 x 10 x 10 x 10 = 9 x 10^7 s\u00f4\u0301 thu\u00ea bao<\/p>\n\n\n\n
T\u01b0\u01a1ng t\u01b0\u0323 nh\u01b0 tr\u00ean, v\u01a1\u0301i s\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i co\u0301 7 s\u00f4\u0301 ta co\u0301 9 x 10^6 s\u00f4\u0301 thu\u00ea bao.<\/p>\n\n\n\n
Nh\u01b0 v\u00e2\u0323y, khi t\u0103ng s\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i t\u01b0\u0300 7 ch\u01b0\u0303 s\u00f4\u0301 l\u00ean 8 ch\u01b0\u0303 s\u00f4\u0301, ta se\u0303 t\u0103ng \u0111\u01b0\u01a1\u0323c 9 x 10^7 \u2013 9 x 10^6 = 81.000.000 s\u00f4\u0301 thu\u00ea bao.<\/p>\n\n\n\n
T\u01b0\u01a1ng \u01b0\u0301ng v\u01a1\u0301i 81 tri\u00ea\u0323u thu\u00ea bao se\u0303 la\u0300 81 kha\u0301ch ha\u0300ng.<\/p>\n\n\n\n
Th\u01b0\u0323c t\u00ea\u0301 co\u0301 th\u01b0\u0323c s\u01b0\u0323 t\u0103ng l\u00ean 81 tri\u00ea\u0323u thu\u00ea bao?<\/p>\n\n\n\n
Tr\u00ean c\u01a1 s\u01a1\u0309 ti\u0301nh toa\u0301n thi\u0300 \u0111u\u0301ng la\u0300 nh\u01b0 v\u00e2\u0323y, nh\u01b0ng th\u01b0\u0323c t\u00ea\u0301 \u0111i\u00ea\u0300u na\u0300y kh\u00f4ng h\u0103\u0309n chi\u0301nh xa\u0301c. S\u00f4\u0301 thu\u00ea bao co\u0301 t\u0103ng nh\u01b0\u0303ng kh\u00f4ng t\u01a1\u0301i 81 tri\u00ea\u0323u vi\u0300: co\u0301 nh\u01b0\u0303ng s\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i \u0111\u0103\u0323c bi\u00ea\u0323t du\u0300ng cho mu\u0323c \u0111i\u0301ch \u0111\u0103\u0323c bi\u00ea\u0323t nh\u01b0: s\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i cu\u0309a c\u00f4ng an 113 , s\u00f4\u0301 \u0111i\u00ea\u0323n thoa\u0323i c\u01b0\u0301u ho\u0309a 114,\u2026<\/p>\n\n\n\n
Nh\u1eefng s\u1ed1 \u0111i\u1ec7n tho\u1ea1i ki\u1ec3u n\u00e0y kh\u00f4ng nhi\u1ec1u. M\u0103\u0323c du\u0300 ch\u1ec9 d\u00f9ng c\u00f3 3 con s\u1ed1 nh\u01b0ng nh\u1eefng s\u1ed1 \u0111i\u1ec7n tho\u1ea1i kh\u00e1c b\u1eaft \u0111\u1ea7u b\u1eb1ng nh\u1eefng con s\u1ed1 n\u00e0y c\u0169ng kh\u00f4ng s\u1eed d\u1ee5ng \u0111\u01b0\u1ee3c. L\u1ea5y v\u00ed d\u1ee5 s\u1ed1 110.<\/p>\n\n\n\n
Khi s\u1ed1 \u0111i\u1ec7n tho\u1ea1i l\u00e0 7 ch\u1eef s\u1ed1, nh\u1eefng s\u1ed1 \u0111i\u1ec7n tho\u1ea1i b\u1eaft \u0111\u1ea7u b\u1eb1ng 110 s\u1ebd c\u00f3 10 x 10 x 10 x 10 = 10.000 v\u00ec con s\u1ed1 110 \u0111\u1eb7c bi\u1ec7t n\u00e0y m\u00e0 kh\u00f4ng s\u1eed d\u1ee5ng \u0111\u01b0\u1ee3c. Khi s\u1ed1 \u0111i\u1ec7n tho\u1ea1i t\u0103ng l\u00ean l\u00e0m 8 con s\u1ed1 th\u00ec s\u1ebd c\u00f3 10 x 10 x 10 x 10 x 10=100.000 s\u1ed1 \u0111i\u1ec7n tho\u1ea1i b\u1eaft \u0111\u1ea7u b\u1eb1ng s\u1ed1 110 kh\u00f4ng s\u1eed d\u1ee5ng \u0111\u01b0\u1ee3c v\u00ec c\u00f9ng m\u1ed9t nguy\u00ean nh\u00e2n nh\u01b0 tr\u00ean.<\/p>\n\n\n\n
L\u01a1\u0300i k\u00ea\u0301t<\/p>\n\n\n\n
Tr\u00ean la\u0300 t\u00f4\u0309ng h\u01a1\u0323p ki\u00ea\u0301n th\u01b0\u0301c v\u00ea\u0300 lu\u0303y th\u01b0\u0300a cu\u0309a m\u00f4\u0323t s\u00f4\u0301 ma\u0300 Vietlearn mu\u00f4\u0301n g\u01b0\u0309i \u0111\u00ea\u0301n ca\u0301c ba\u0323n ho\u0323c sinh. Vietlearn mong r\u0103\u0300ng nh\u01b0\u0303ng th\u00f4ng tin na\u0300y h\u01b0\u0303u i\u0301ch va\u0300 h\u00f4\u0303 tr\u01a1\u0323 cho qua\u0301 tri\u0300nh ho\u0323c t\u00e2\u0323p cu\u0309a ca\u0301c ba\u0323n. Mo\u0323i th\u1eafc m\u0103\u0301c c\u00e2\u0300n gia\u0309i \u0111a\u0301p ha\u0303y li\u00ean h\u00ea\u0323 ngay v\u01a1\u0301i Vietlearn \u0111\u00ea\u0309 \u0111\u01b0\u01a1\u0323c t\u01b0 v\u00e2\u0301n. N\u00ea\u0301u ba\u0323n mu\u00f4\u0301n trau d\u00f4\u0300i th\u00eam ki\u00ea\u0301n th\u01b0\u0301c v\u00ea\u0300 Toa\u0301n ho\u0323c, Ho\u0301a ho\u0323c, Anh ng\u01b0\u0303, V\u00e2\u0323t Ly\u0301 tham kha\u0309o ngay kho\u0301a ho\u0323c tr\u01b0\u0323c tuy\u00ea\u0301n K12 cu\u0309a Vietlearn. Kho\u0301a ho\u0323c da\u0300nh cho ca\u0301c ba\u0323n ho\u0323c sinh t\u01b0\u0300 lo\u0301p 1 \u0111\u00ea\u0301n l\u01a1\u0301p 12. Vietlearn gia s\u01b0<\/a> tr\u01b0\u0323c tuy\u00ea\u0301n ha\u0300ng \u0111\u00e2\u0300u Vi\u00ea\u0323t Nam. Vietlearn t\u01b0\u0323 tin giu\u0301p ba\u0323n ho\u0323c t\u00e2\u0323p hi\u00ea\u0323u qua\u0309 h\u01a1n, n\u00e2ng cao \u0111i\u00ea\u0309m s\u00f4\u0301 trong th\u01a1\u0300i gian ng\u0103\u0301n.<\/p>\n\n\n\n <\/p>\n\n\n\n Ki\u00ea\u0301n th\u01b0\u0301c v\u00ea\u0300 T\u00e2\u0323p h\u01a1\u0323p Q ca\u0301c s\u00f4\u0301 h\u01b0\u0303u ti\u0309 \u2013 Gi\u1ecfi to\u00e1n l\u1edbp 7<\/p>\n\n\n\n Ly\u0301 thuy\u00ea\u0301t v\u00e0 Ba\u0300i t\u00e2\u0323p c\u00f4\u0323ng tr\u01b0\u0300 s\u00f4\u0301 h\u01b0\u0303u ti\u0309 \u2013 Gi\u1ecfi to\u00e1n<\/p>\n\n\n\n\n\n\n","protected":false},"excerpt":{"rendered":" Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn \u01a0\u0309 ba\u0300i tr\u01b0\u01a1\u0301c Vietlearn \u0111a\u0303 cu\u0300ng ba\u0323n ti\u0300m hi\u00ea\u0309u ca\u0301c ki\u00ea\u0301n th\u01b0\u0301c v\u00ea\u0300 s\u00f4\u0301 h\u01b0\u0303u ti\u0309, phe\u0301p c\u00f4\u0323ng, tr\u01b0\u0300, nh\u00e2n, chia. V\u01a1\u0301i ba\u0300i vi\u00ea\u0301t na\u0300y, Vietlearn se\u0303 cu\u0300ng ba\u0323n ti\u0300m hi\u00ea\u0309u lu\u0303y th\u01b0\u0300a cu\u0309a m\u00f4\u0323t s\u00f4\u0301 h\u01b0\u0303u ti\u0309. Lu\u0303y th\u01b0\u0300a cu\u0309a m\u00f4\u0323t s\u00f4\u0301 h\u01b0\u0303u ti\u0309 la\u0300 gi\u0300? Ti\u0301ch cu\u0309a […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":"","_disable_autopaging":false},"categories":[9,4],"tags":[],"class_list":["post-2663","post","type-post","status-publish","format-standard","hentry","category-toan-hoc","category-trung-hoc"],"_links":{"self":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/2663","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/comments?post=2663"}],"version-history":[{"count":0,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/2663\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/media?parent=2663"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/categories?post=2663"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/tags?post=2663"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}