Th\u1ebf n\u00e0o l\u00e0 ph\u01b0\u01a1ng ph\u00e1p thay th\u1ebf d\u1ea7n? Tr\u01b0\u1edbc h\u1ebft ta gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh x2= 2. Th\u1ebf ch\u1eb3ng ph\u1ea3i nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \u221a2 sao?<\/p>\n\n\n\n
Kh\u00f4ng sai. Th\u1ebf nh\u01b0ng \u221a2 b\u1eb1ng bao nhi\u00eau, c\u00f3 th\u1ec3 bi\u1ec3u di\u1ec5n b\u1eb1ng m\u1ed9t s\u1ed1 l\u1ebb th\u1eadp ph\u00e2n kh\u00f4ng? \u221a2 l\u00e0 m\u1ed9t s\u1ed1 v\u00f4 t\u1ec9. Ch\u00fang ta hi v\u1ecdng kh\u00f4ng c\u1ea7n ph\u00e9p t\u00ednh khai c\u0103n c\u0169ng c\u00f3 th\u1ec3 t\u00ecm \u0111\u01b0\u1ee3c gi\u00e1 tr\u1ecb g\u1ea7n \u0111\u00fang c\u1ee7a \u221a2 d\u01b0\u1edbi d\u1ea1ng m\u1ed9t s\u1ed1 l\u1ebb th\u1eadp ph\u00e2n c\u00f3 \u0111\u1ed9 ch\u00ednh x\u00e1c bao nhi\u00eau c\u0169ng \u0111\u01b0\u1ee3c. Ta bi\u1ebfn \u0111\u1ed5i x2 = 2 th\u00e0nh x = 1\/2 (x + 2\/x). Ta l\u00e0m ph\u00e9p t\u00ednh \u01b0\u1edbc l\u01b0\u1ee3ng ch\u1ecdn x1 = 1. \u0110\u01b0a v\u00e0o bi\u1ec3u th\u1ee9c v\u00e0 t\u00ednh x2 = 1\/2(1 + 2\/1). L\u1ea1i \u0111\u01b0a x2 thay v\u00e0o bi\u1ec3u th\u1ee9c v\u00e0 t\u00ednh x3 = 1\/2(1,5 + 2\/1,5) \u2248 1,41666<\/p>\n\n\n\n
Ta c\u00f3 th\u1ec3 thay d\u1ea7n c\u00e1c gi\u00e1 tr\u1ecb xn v\u00e0o bi\u1ec3u th\u1ee9c v\u00e0 t\u00ednh xn+1 = 1\/2(xn + 2\/xn) ; n = 1,2,3,\u2026,n t\u1ea1o th\u00e0nh m\u1ed9t d\u00e3y s\u1ed1 l\u00e0 c\u00e1c gi\u00e1 tr\u1ecb g\u1ea7n \u0111\u00fang c\u1ee7a \u221a2. T\u1eeb t\u00ednh to\u00e1n c\u00f3 th\u1ec3 t\u00ecm \u0111\u01b0\u1ee3c x4 = 1,414215686, x5 = 1,414213562. Gi\u00e1 tr\u1ecb c\u1ee7a x5 \u0111\u1ea1t \u0111\u1ebfn \u0111\u1ed9 ch\u00ednh x\u00e1c c\u1ee7a s\u1ed1 l\u1ebb th\u1ee9 9, so v\u1edbi ph\u01b0\u01a1ng ph\u00e1p khai c\u0103n \u1edf \u0111\u00e2y c\u00e1c t\u00ednh to\u00e1n th\u1ef1c hi\u1ec7n d\u1ec5 d\u00e0ng v\u00e0 nhanh ch\u00f3ng h\u01a1n. Nh\u01b0 v\u1eady \u1edf \u0111\u00e2y \u0111\u00e3 \u0111\u01b0a s\u1ed1 xn v\u00e0o c\u00f4ng th\u1ee9c \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb s\u1ed1 xn+1, l\u1ea1i \u0111\u01b0a s\u1ed1 xn+2 thay v\u00e0o c\u00f4ng th\u1ee9c \u0111\u1ec3 t\u00ednh xn+2\u2026v\u00e0 c\u00f3 th\u1ec3 ti\u1ebfp t\u1ee5c c\u00e1c gi\u00e1 tr\u1ecb xn ti\u1ebfp sau, t\u1eeb \u0111\u00f3 nh\u1eadn \u0111\u01b0\u1ee3c d\u00e3y s\u1ed1 x1, x2, ..,x3. \u1ede \u0111\u00e2y \u0111\u1ec3 t\u00ednh gi\u00e1 tr\u1ecb xn ta \u0111\u00e3 \u0111\u01b0a gi\u00e1 tr\u1ecb x = x1 v\u00e0o h\u00e0m f(x) thay th\u1ebf d\u1ea7n v\u00e0o f(x) v\u00e0 xn+1 = f(x) n =1,2,3\u2026n \u0111\u1ebfn khi \u0111\u1ea1t \u0111\u01b0\u1ee3c gi\u00e1 tr\u1ecb c\u00f3 \u0111\u1ed9 ch\u00ednh x\u00e1c c\u1ea7n thi\u1ebft. \u0110\u1ea5y ch\u00ednh l\u00e0 ph\u01b0\u01a1ng ph\u00e1p thay th\u1ebf d\u1ea7n.<\/p>\n\n\n\n
Xem ra th\u00ec ph\u01b0\u01a1ng ph\u00e1p thay th\u1ebf d\u1ea7n c\u00f3 th\u1ec3 c\u00f3 ch\u00fat \u00edt phi\u1ec1n ph\u1ee9c khi t\u00ednh to\u00e1n. Thay th\u1ebf, r\u1ed3i l\u1ea1i t\u00ednh, c\u1ee9 ti\u1ebfp t\u1ee5c li\u1ec7u c\u00f3 th\u1ec3 t\u1ea1o \u0111i\u1ec1u ki\u1ec7n xu\u1ea5t hi\u1ec7n t\u00ednh sai ch\u0103ng? Th\u1ef1c ra ng\u00e0y nay v\u1edbi s\u1ef1 xu\u1ea5t hi\u1ec7n m\u00e1y t\u00ednh c\u00f3 t\u1ed1c \u0111\u1ed9 t\u00ednh nhanh v\u00e0 ch\u00ednh x\u00e1c n\u00ean c\u00f3 th\u1ec3 \u00e1p d\u1ee5ng \u0111\u1ec3 gi\u1ea3i c\u00e1c ph\u01b0\u01a1ng tr\u00ecnh ph\u1ee9c t\u1ea1p h\u01a1n ph\u01b0\u01a1ng tr\u00ecnh x2= 2 nhi\u1ec1u m\u00e0 kh\u00f4ng h\u1ec1 g\u1eb7p kh\u00f3 kh\u0103n g\u00ec, v\u00ec v\u1eady ph\u01b0\u01a1ng ph\u00e1p thay th\u1ebf d\u1ea7n c\u00f3 nhi\u1ec1u l\u1ee3i \u00edch \u0111\u1ec3 gi\u1ea3i quy\u1ebft nhi\u1ec1u v\u1ea5n \u0111\u1ec1 th\u1ef1c t\u1ebf.<\/p>\n\n\n\n
T\u1eeb kho\u00e1: Ph\u01b0\u01a1ng ph\u00e1p thay th\u1ebf d\u1ea7n.<\/p>\n\n\n\n
<\/p>\n\n\n\n\n\n\n","protected":false},"excerpt":{"rendered":"
Th\u1ebf n\u00e0o l\u00e0 ph\u01b0\u01a1ng ph\u00e1p thay th\u1ebf d\u1ea7n? Tr\u01b0\u1edbc h\u1ebft ta gi\u1ea3i ph\u01b0\u01a1ng tr\u00ecnh x2= 2. Th\u1ebf ch\u1eb3ng ph\u1ea3i nghi\u1ec7m c\u1ee7a ph\u01b0\u01a1ng tr\u00ecnh l\u00e0 \u221a2 sao? Kh\u00f4ng sai. Th\u1ebf nh\u01b0ng \u221a2 b\u1eb1ng bao nhi\u00eau, c\u00f3 th\u1ec3 bi\u1ec3u di\u1ec5n b\u1eb1ng m\u1ed9t s\u1ed1 l\u1ebb th\u1eadp ph\u00e2n kh\u00f4ng? \u221a2 l\u00e0 m\u1ed9t s\u1ed1 v\u00f4 t\u1ec9. Ch\u00fang ta hi v\u1ecdng […]<\/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":[15,9],"tags":[],"class_list":["post-7706","post","type-post","status-publish","format-standard","hentry","category-hoi-dap-khoa-hoc","category-toan-hoc"],"_links":{"self":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/7706","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=7706"}],"version-history":[{"count":0,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/posts\/7706\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/media?parent=7706"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/categories?post=7706"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.vietlearn.org\/kien-thuc\/wp-json\/wp\/v2\/tags?post=7706"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}