Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn<\/p>\n\n\n\n
Trong tam gi\u00e1c vu\u00f4ng, n\u1ebfu bi\u1ebft \u0111\u1ed9 d\u00e0i c\u1ee7a 2 c\u1ea1nh g\u00f3c vu\u00f4ng, li\u1ec7u r\u1eb1ng c\u00f3 th\u1ec3 t\u00ednh \u0111\u01b0\u1ee3c chi\u1ec1u d\u00e0i c\u1ee7a c\u1ea1nh huy\u1ec1n? C\u00e1c b\u00e9 s\u1ebd c\u00f3 \u0111\u01b0\u1ee3c c\u00e2u tr\u1ea3 l\u1eddi khi theo d\u00f5i b\u00e0i h\u1ecdc sau \u0111\u00e2y: \u0110\u1ecbnh l\u00fd Pytago trong tam gi\u00e1c vu\u00f4ng. \u0110\u01b0\u1ee3c xem nh\u01b0 m\u1ed9t trong nh\u1eefng \u0111\u1ecbnh l\u00fd kinhh \u0111i\u1ec3n c\u1ee7a to\u00e1n h\u1ecdc. Pytago \u0111\u00e3 gi\u00fap h\u00ecnh h\u1ecdc ti\u1ebfn th\u00eam m\u1ed9t b\u01b0\u1edbc d\u00e0i trong h\u00e0nh tr\u00ecnh ph\u00e1t tri\u1ec3n. C\u00f9ng Vietlearn kh\u00e1m ph\u00e1 n\u1ed9i dung ki\u1ebfn th\u1ee9c v\u1ec1 \u0111\u1ecbnh l\u00fd n\u00e0y ngay n\u00e0o.<\/p>\n\n\n\n
V\u00ed d\u1ee5<\/p>\n\n\n\n
V\u1ebd tam gi\u00e1c vu\u00f4ng c\u00f3 hai c\u1ea1nh g\u00f3c vu\u00f4ng l\u1ea7n l\u01b0\u1ee3t b\u1eb1ng 3 v\u00e0 4,<\/p>\n\n\n\n
Nh\u1eadn x\u00e9t t\u1ed5ng b\u00ecnh ph\u01b0\u01a1ng 2 c\u1ea1nh g\u00f3c vu\u00f4ng so v\u1edbi c\u1ea1nh huy\u1ec1n<\/p>\n\n\n\n
=> Ta th\u1ea5y b\u00ecnh ph\u01b0\u01a1ng 2 c\u1ea1nh g\u00f3c vu\u00f4ng b\u1eb1ng b\u00ecnh ph\u01b0\u01a1ng c\u1ea1nh huy\u1ec1n.<\/p>\n\n\n\n
Ta c\u00f3 \u0111\u1ecbnh l\u00fd:<\/p>\n\n\n\n
\u0110\u1ecbnh l\u00fd Pytago<\/p>\n\n\n\n
Ch\u00fa \u00fd: N\u1ed9i dung \u0111\u1ecbnh l\u00fd Pytago \u0111\u01b0\u1ee3c th\u1eeba nh\u1eadn m\u00e0 kh\u00f4ng c\u1ea7n ph\u1ea3i ch\u1ee9ng minh<\/p>\n\n\n\n
V\u00ed d\u1ee5:<\/p>\n\n\n\n
V\u1ebd tam gi\u00e1c MNO c\u00f3 \u0111\u1ed9 d\u00e0i c\u00e1c c\u1ea1nh MN, NO, MO l\u1ea7n l\u01b0\u1ee3t l\u00e0 3 , 4 v\u00e0 5 cm. D\u00f9ng th\u01b0\u1edbc \u0111o \u0111\u1ed9 \u0111\u1ec3 \u0111o g\u00f3c N<\/p>\n\n\n\n
=> Ta c\u00f3 g\u00f3c N = 90<\/p>\n\n\n\n
D\u1ef1a tr\u00ean \u0111\u1ecbnh l\u00fd Pytago, ta c\u00f3<\/p>\n\n\n\n
X\u00e9t tam gi\u00e1c ABC:<\/p>\n\n\n\n
Ta c\u00f3 BC2 = AB2 + AC2<\/p>\n\n\n\n
=> G\u00f3c BAC = 90<\/p>\n\n\n\n
Ng\u01b0\u1ee3c l\u1ea1i v\u1edbi \u0111\u1ecbnh l\u00fd Pytago, \u0111\u1ecbnh l\u00fd Pytago \u0111\u1ea3o \u0111\u01b0\u1ee3c s\u1eed d\u1ee5ng \u0111\u1ec3 ch\u1ee9ng minh tam gi\u00e1c vu\u00f4ng khi bi\u1ebft chi\u1ec1u d\u00e0i c\u00e1c c\u1ea1nh c\u1ee7a tam gi\u00e1c \u0111\u00f3.<\/p>\n\n\n\n
+Trong tam gi\u00e1c vu\u00f4ng, b\u00ecnh ph\u01b0\u01a1ng c\u1ea1nh huy\u1ec1n b\u1eb1ng b\u00ecnh ph\u01b0\u01a1ng c\u00e1c c\u1ea1nh g\u00f3c vu\u00f4ng<\/p>\n\n\n\n
+Ng\u01b0\u1ee3c l\u1ea1i, n\u1ebfu 1 tam gi\u00e1c c\u00f3 m\u1ed9t c\u1ea1nh b\u1eb1ng b\u00ecnh ph\u01b0\u01a1ng 2 c\u1ea1nh c\u00f2n l\u1ea1i th\u00ec \u0111\u00f3 l\u00e0 tam gi\u00e1c vu\u00f4ng, c\u1ea1nh \u0111\u00f3 \u0111\u01b0\u1ee3c g\u1ecdi l\u00e0 c\u1ea1nh huy\u1ec1n.<\/p>\n\n\n\n
M\u1ed1i li\u00ean h\u1ec7 gi\u1eefa c\u00e1c c\u1ea1nh trong tam gi\u00e1c vu\u00f4ng \u0111\u00e3 \u0111\u01b0\u1ee3c con ng\u01b0\u1eddi ph\u00e1t hi\u1ec7n t\u1eeb th\u1eddi c\u1ed5 \u0111\u00e2\u1ecb, tr\u01b0\u1edbc c\u1ea3 Pytago, t\u1eeb v\u0103n minh Ai C\u1eadp t\u1edbi v\u00f9ng L\u01b0\u1ee1ng H\u00e0, v\u0103n minh \u1ea4n H\u1eb1ng t\u1edbi v\u0103n minh Trung Hoa c\u1ed5 \u0111\u1ea1i. Tuy nhi\u00ean, ph\u1ea3i t\u1edbi th\u1eddi Hy L\u1ea1p c\u1ed5 \u0111\u1ea1i, \u0111\u1ecbnh l\u00fd n\u00e0y m\u1edbi \u0111\u01b0\u1ee3c ch\u1ee9ng minh b\u1edfi Pyatago \u2013 nh\u00e0 to\u00e1n h\u1ecdc n\u1ed5i ti\u1ebfng Hy L\u1ea1p th\u1eddi b\u1ea5y gi\u1edd. Kh\u00f4ng ch\u1ec9 \u0111\u01b0\u1ee3c \u1ee9ng d\u1ee5ng trong h\u00ecnh h\u1ecdc \u0111\u01a1n gi\u1ea3n, Pytago \u0111\u01b0\u1ee3c \u1ee9ng d\u1ee5ng ph\u1ed5 bi\u1ebfn trong c\u00e1c l\u0129nh v\u1ef1c to\u00e1n h\u1ecdc nh\u01b0 vi ph\u00e2n, t\u00edch ph\u00e2n, h\u00ecnh h\u1ecdc kh\u00f4ng gian,\u2026 V\u00ec v\u1eady, n\u00f3 \u0111\u01b0\u1ee3c xem nh\u01b0 th\u00e0nh t\u1ef1u th\u00fac \u0111\u1ea9y s\u1ef1 ph\u00e1t tri\u1ec3n c\u1ee7a c\u1ea3 n\u1ec1n to\u00e1n h\u1ecdc.<\/p>\n\n\n\n
B\u00e0i t\u1eadp 1:<\/p>\n\n\n\n
X\u00e9t tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A, cho b\u1ea3ng sau, t\u00ednh chi\u1ec1u d\u00e0i c\u1ea1nh huy\u1ec1n BC.<\/p>\n\n\n\n
AB 3 5 11 9 18 6 7<\/p>\n\n\n\n
AC 4 7 6 17 6 12 4<\/p>\n\n\n\n
BC ? ? ? ? ? ? ?<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/p>\n\n\n\n
V\u00ec tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A, theo \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<\/p>\n\n\n\n
BC2 = AC2 + AB2<\/p>\n\n\n\n
=> BC = \u221a(AC2 + BC2)<\/p>\n\n\n\n
AB 3 5 11 9 18 6 7<\/p>\n\n\n\n
AC 4 7 6 17 6 12 4<\/p>\n\n\n\n
BC 5 8,6 12,5 19,2 19 13 8,1<\/p>\n\n\n\n
B\u00e0i t\u1eadp 2:<\/p>\n\n\n\n
X\u00e9t tam gi\u00e1c ABC vu\u00f4ng t\u1ea1i A:<\/p>\n\n\n\n
Bi\u1ebft chi\u1ec1u d\u00e0i c\u1ea1nh AB = 4 cm, chi\u1ec1u d\u00e0i c\u1ea1nh BC = 6 cm, t\u00ednh chi\u1ec1u d\u00e0i c\u1ea1nh AC<\/p>\n\n\n\n
Bi\u1ebft chi\u1ec1u d\u00e0i c\u1ea1nh AC = 2 cm, chi\u1ec1u d\u00e0i c\u1ea1nh BC = 7 cm, t\u00ednh chi\u1ec1u d\u00e0i c\u1ea1nh AB<\/p>\n\n\n\n
Bi\u1ebft chi\u1ec1u d\u00e0i c\u1ea1nh AB = 3 cm, chi\u1ec1u d\u00e0i c\u1ea1nh AC = 5 cm, t\u00ednh chi\u1ec1u d\u00e0i c\u1ea1nh BC<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i<\/p>\n\n\n\n
=> AC\u00b2 = BC\u00b2 \u2013 AB\u00b2<\/p>\n\n\n\n
=> AC\u00b2 = 6\u00b2 \u2013 4\u00b2<\/p>\n\n\n\n
=> AC = \u221a20<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ee7a c\u1ea1nh AC l\u00e0 \u221a20 cm<\/p>\n\n\n\n
=> AB\u00b2 = BC\u00b2 \u2013 AC\u00b2<\/p>\n\n\n\n
=> AB\u00b2 = 7\u00b2 \u2013 2 \u00b2<\/p>\n\n\n\n
=> AB = \u221a45<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ea1nh AB = \u221a45 cm<\/p>\n\n\n\n
=> BC\u00b2 = 3\u00b2 + 5\u00b2<\/p>\n\n\n\n
=> BC = \u221a34<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ea1nh BC l\u00e0\u221a34<\/p>\n\n\n\n
B\u00e0i t\u1eadp 3:<\/p>\n\n\n\n
T\u00ednh chi\u1ec1u d\u00e0i c\u1ea1nh huy\u1ec1n c\u1ee7a c\u00e1c tam gi\u00e1c sau, bi\u1ebft:<\/p>\n\n\n\n
a. Tam gi\u00e1c MNO vu\u00f4ng t\u1ea1i M c\u00f3 c\u1ea1nh MO = 4 cm, c\u1ea1nh MN = 5 cm<\/p>\n\n\n\n
b. Tam gi\u00e1c PQR vu\u00f4ng t\u1ea1i P c\u00f3 c\u1ea1nh PQ = 7 cm, c\u1ea1nh PR = 6 cm<\/p>\n\n\n\n
c. Tam gi\u00e1c BCD vu\u00f4ng t\u1ea1i B c\u00f3 c\u1ea1nh BC = 8 cm, c\u1ea1nh BD = 2 cm<\/p>\n\n\n\n
d. Tam gi\u00e1c IKL vu\u00f4ng t\u1ea1i I c\u00f3 c\u1ea1nh IL = 4,5 cm, c\u1ea1nh IK = 8 cm<\/p>\n\n\n\n
L\u1eddi gi\u1ea3i:<\/p>\n\n\n\n
a. V\u00ec tam gi\u00e1c MNO vu\u00f4ng t\u1ea1i M, NO l\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng, do \u0111\u00f3, ta \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago trong tam gi\u00e1c vu\u00f4ng:<\/p>\n\n\n\n
NO2 = MN2 + MO2<\/p>\n\n\n\n
=> NO2 = 42 + 52<\/p>\n\n\n\n
=> NO2 = 41<\/p>\n\n\n\n
=> NO = \u221a41<\/p>\n\n\n\n
=> NO = 6,4<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ea1nh NO c\u1ee7a tam gi\u00e1c MNO l\u00e0 6,4 cm<\/p>\n\n\n\n
b. V\u00ec tam gi\u00e1c PQR vu\u00f4ng t\u1ea1i P, QR l\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng, do \u0111\u00f3, ta \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago trong tam gi\u00e1c vu\u00f4ng:<\/p>\n\n\n\n
QR2 = PQ2 + PR2<\/p>\n\n\n\n
=> QR2 = 72 + 62<\/p>\n\n\n\n
=> QR2 = 85<\/p>\n\n\n\n
=> QR = \u221a85<\/p>\n\n\n\n
=> QR = 9,2<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ea1nh QR c\u1ee7a tam gi\u00e1c PQR l\u00e0 9,2 cm<\/p>\n\n\n\n
c. V\u00ec tam gi\u00e1c BCD vu\u00f4ng t\u1ea1i B, CD l\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng, do \u0111\u00f3, ta \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago trong tam gi\u00e1c vu\u00f4ng:<\/p>\n\n\n\n
CD2 = BC2 + BD2<\/p>\n\n\n\n
=> CD2 = 82 + 22<\/p>\n\n\n\n
=> CD2 = 70<\/p>\n\n\n\n
=> CD = \u221a70<\/p>\n\n\n\n
=> CD = 8,4<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ea1nh CD c\u1ee7a tam gi\u00e1c BCD l\u00e0 8,4 cm<\/p>\n\n\n\n
c. V\u00ec tam gi\u00e1c IKL vu\u00f4ng t\u1ea1i I, KL l\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng, do \u0111\u00f3, ta \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago trong tam gi\u00e1c vu\u00f4ng:<\/p>\n\n\n\n
KL2 = IL2 + IK2<\/p>\n\n\n\n
=> KL2 = 4,52 + 82<\/p>\n\n\n\n
=> KL2 = 84,25<\/p>\n\n\n\n
=>KL = \u221a84,25<\/p>\n\n\n\n
=> KL = 9,2<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ea1nh CD c\u1ee7a tam gi\u00e1c BCD l\u00e0 9,2 cm<\/p>\n\n\n\n
B\u00e0i 53 s\u00e1ch gi\u00e1o khoa:<\/p>\n\n\n\n
a. V\u00ec x l\u00e0 c\u1ea1nh huy\u1ec1n c\u1ee7a tam gi\u00e1c, \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3<\/p>\n\n\n\n
x\u00b2 = 12\u00b2 + 5\u00b2<\/p>\n\n\n\n
=> x\u00b2 = 169<\/p>\n\n\n\n
=> x = 13<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ee7a x l\u00e0 13<\/p>\n\n\n\n
b. V\u00ec x l\u00e0 c\u1ea1nh huy\u1ec1n c\u1ee7a tam gi\u00e1c, \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd pytago ta c\u00f3<\/p>\n\n\n\n
x\u00b2 = 1\u00b2 + 2 \u00b2<\/p>\n\n\n\n
=> x\u00b2 = 5<\/p>\n\n\n\n
=> x = \u221a5 = 2,34<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ee7a x l\u00e0 2,34<\/p>\n\n\n\n
c. V\u00ec x l\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng, \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3<\/p>\n\n\n\n
29\u00b2 = x\u00b2 + 21\u00b2<\/p>\n\n\n\n
=> x\u00b2 = 29\u00b2 \u2013 21\u00b2<\/p>\n\n\n\n
=> x\u00b2 = 841 \u2013 441<\/p>\n\n\n\n
=> x\u00b2 = 400<\/p>\n\n\n\n
=> x = 20<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ee7a x l\u00e0 20<\/p>\n\n\n\n
d. V\u00ec x l\u00e0 c\u1ea1nh g\u00f3c vu\u00f4ng, \u00e1p d\u1ee5ng \u0111\u1ecbnh l\u00fd Pytago ta c\u00f3:<\/p>\n\n\n\n
=> x\u00b2 = \u221a7\u00b2 + 3\u00b2<\/p>\n\n\n\n
=> x\u00b2 = 7 + 9<\/p>\n\n\n\n
=> x = 4<\/p>\n\n\n\n
V\u1eady chi\u1ec1u d\u00e0i c\u1ee7a x l\u00e0 4<\/p>\n\n\n\n
L\u1eddi k\u1ebft: Hy v\u1ecdng v\u1edbi n\u1ed9i dung b\u00e0i h\u1ecdc tr\u00ean Vietlearn \u0111\u00e3 gi\u00fap c\u00e1c b\u00e9 n\u1eafm v\u1ee9ng ki\u1ebfn th\u1ee9c v\u1ec1 \u0111\u1ecbnh l\u00fd Pytago. \u0110\u1eb7c bi\u1ec7t, \u0111\u1ec3 ti\u1ebfp thu ki\u1ebfn th\u1ee9c b\u00e0i h\u1ecdc m\u1ed9t c\u00e1ch hi\u1ec7u qu\u1ea3, c\u00e1c b\u1ea1n h\u1ecdc sinh<\/a> n\u00ean \u00f4n luy\u1ec7n v\u00e0 gi\u1ea3i c\u00e1c b\u00e0i t\u1eadp v\u1ec1 tam gi\u00e1c vu\u00f4ng \u0111\u1ec3 c\u1ee7ng c\u1ed1 ki\u1ebfn th\u1ee9c. Ho\u1eb7c c\u00e1c b\u1ea1n c\u0169ng c\u00f3 th\u1ec3 tham kh\u1ea3o nh\u1eefng b\u00e0i to\u00e0n n\u00e2ng cao \u0111\u1ec3 l\u00e0m quen v\u1edbi d\u1ea1ng c\u00e2u h\u1ecfi v\u1eadn d\u1ee5ng v\u00e0 gi\u00e0nh \u0111i\u1ec3m cao trong c\u00e1c \u0111\u1ee3t ki\u1ec3m tra. Theo d\u00f5i Vietlearn th\u01b0\u1eddng xuy\u00ean \u0111\u1ec3 c\u1eadp nh\u1eadt nh\u1eefng b\u00e0i h\u1ecdc b\u1ed5 \u00edch.<\/p>\n\n\n\n V\u1ec1 Vietlearn<\/p>\n\n\n\n H\u1ecdc tr\u1ef1c tuy\u1ebfn t\u1ea1i Vietlearn<\/p>\n\n\n\n Vietlearn l\u00e0 c\u00f4ng ty Edtech v\u1ec1 gi\u00e1o d\u1ee5c tr\u1ef1c tuy\u1ebfn, cung c\u1ea5p tr\u1ea3i nghi\u1ec7m h\u1ecdc t\u1eadp<\/a> c\u00e1 nh\u00e2n cho h\u00e0ng tr\u0103m ngh\u00ecn h\u1ecdc sinh<\/a>, sinh vi\u00ean v\u00e0 nh\u00e0 tr\u01b0\u1eddng \u0111\u1ec3 gi\u1ea3i \u0111\u00e1p nh\u1eefng y\u00eau c\u1ea7u trong vi\u1ec7c h\u1ecdc t\u1eadp<\/a> Anh ng\u1eef th\u00f4ng qua m\u1ea1ng l\u01b0\u1edbi c\u00e1c chuy\u00ean gia<\/a> v\u00e0 gi\u00e1o vi\u00ean<\/a> kh\u1eafp to\u00e0n c\u1ea7u m\u00e0 ch\u00fang t\u00f4i g\u1ecdi l\u00e0 c\u00e1c gia s\u01b0<\/a> h\u1ecdc thu\u1eadt qu\u1ed1c t\u1ebf.<\/p>\n\n\n\n Vietlearn mong mu\u1ed1n tr\u1edf th\u00e0nh h\u1ec7 th\u1ed1ng h\u1ecdc t\u1eadp th\u00edch \u1ee9ng s\u1eed d\u1ee5ng c\u00f4ng ngh\u1ec7 tr\u00ed tu\u1ec7 nh\u00e2n t\u1ea1o (AI) v\u00e0 d\u1eef li\u1ec7u l\u1edbn h\u00e0ng \u0111\u1ea7u \u0110\u00f4ng Nam \u00c1. S\u1ee9 m\u1ec7nh c\u1ee7a Vietlearn l\u00e0 truy\u1ec1n c\u1ea3m h\u1ee9ng, truy\u1ec1n l\u1eeda, v\u00e0 b\u1ed3i d\u01b0\u1ee1ng th\u1ebf h\u1ec7 tr\u1ebb. Vietlearn mong mu\u1ed1n t\u1ea1o ra s\u1ef1 thay \u0111\u1ed5i v\u1ec1 tr\u00ed tu\u1ec7, nh\u1eadn th\u1ee9c x\u00e3 h\u1ed9i truy\u1ec1n c\u1ea3m h\u1ee9ng , gi\u00fap c\u00e1c em ph\u00e1t huy h\u1ebft ti\u1ec1m n\u0103ng trong vi\u1ec7c h\u1ecdc c\u0169ng nh\u01b0 \u0111i\u1ec3m m\u1ea1nh c\u1ee7a m\u00ecnh.<\/p>\n\n\n\n \u0110\u0103ng k\u00fd kh\u00f3a h\u1ecdc cho con ngay h\u00f4m nay!<\/p>\n\n\n\n <\/p>\n\n\n\n Hai g\u00f3c \u0111\u1ed1i \u0111\u1ec9nh v\u00e0 ki\u1ebfn th\u1ee9c c\u01a1 b\u1ea3n \u2013 To\u00e1n l\u1edbp 7 l\u00e0 chuy\u1ec7n nh\u1ecf<\/p>\n\n\n\n Ho\u0323c Toa\u0301n 7 cu\u0300ng Vietlearn: 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 h\u01b0\u0303u ti\u0309<\/p>\n\n\n\n T\u1ed5ng ba g\u00f3c c\u1ee7a m\u1ed9t tam gi\u00e1c \u2013 To\u00e1n l\u1edbp 7 l\u00e0 chuy\u1ec7n nh\u1ecf<\/p>\n\n\n\n\n\n\n","protected":false},"excerpt":{"rendered":" Ghi ch\u00fa: T\u00e0i li\u1ec7u tr\u00edch d\u1eabn Trong tam gi\u00e1c vu\u00f4ng, n\u1ebfu bi\u1ebft \u0111\u1ed9 d\u00e0i c\u1ee7a 2 c\u1ea1nh g\u00f3c vu\u00f4ng, li\u1ec7u r\u1eb1ng c\u00f3 th\u1ec3 t\u00ednh \u0111\u01b0\u1ee3c chi\u1ec1u d\u00e0i c\u1ee7a c\u1ea1nh huy\u1ec1n? C\u00e1c b\u00e9 s\u1ebd c\u00f3 \u0111\u01b0\u1ee3c c\u00e2u tr\u1ea3 l\u1eddi khi theo d\u00f5i b\u00e0i h\u1ecdc sau \u0111\u00e2y: \u0110\u1ecbnh l\u00fd Pytago trong tam gi\u00e1c vu\u00f4ng. \u0110\u01b0\u1ee3c xem nh\u01b0 […]<\/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-2539","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\/2539","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=2539"}],"version-history":[{"count":0,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/2539\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/media?parent=2539"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/categories?post=2539"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/tags?post=2539"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}