7 \u00a0h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u00e1ng nh\u1edb l\u1edbp 8
\n<\/strong><\/h2>\n

1)\u00a0\u00a0\u00a0\u00a0\u00a0 (A + B)^{2<\/sup>\u00a0= A2<\/sup>\u00a0+ 2AB + B2<\/sup><\/strong><\/span><\/p>\n}

2)\u00a0\u00a0\u00a0\u00a0\u00a0 (A \u2013 B)^{2<\/sup>\u00a0= A2<\/sup>\u00a0\u2013 2AB + B2<\/sup><\/strong><\/p>\n}

3)\u00a0\u00a0\u00a0\u00a0\u00a0 A^{2<\/sup>\u00a0\u2013 B2<\/sup>\u00a0= (A \u2013 B)(A + B)<\/strong><\/p>\n}

4)\u00a0\u00a0\u00a0\u00a0\u00a0 (A + B)^{3<\/sup>\u00a0= A3<\/sup>\u00a0+ 3A2<\/sup>B + 3AB2<\/sup>\u00a0+ B3<\/sup><\/strong><\/p>\n}

5)\u00a0\u00a0\u00a0\u00a0\u00a0 (A \u2013 B)^{3<\/sup>\u00a0= A3<\/sup>\u00a0\u2013 3A2<\/sup>B + 3AB2<\/sup>\u00a0\u2013 B3<\/sup><\/strong><\/p>\n}

6)\u00a0\u00a0\u00a0\u00a0\u00a0 A^{3<\/sup>\u00a0+ B3<\/sup>\u00a0= (A + B)(A2<\/sup>\u00a0\u2013 AB + B2<\/sup>)<\/strong><\/p>\n}

7)\u00a0\u00a0\u00a0\u00a0\u00a0 A^{3<\/sup>\u00a0\u2013 B3<\/sup>\u00a0= (A \u2013 B)(A2<\/sup>\u00a0+ AB + B2<\/sup>)<\/strong><\/p>\n}

B\u00e0i t\u1eadp \u00e1p d\u1ee5ng<\/h2>\n

D\u1ea1ng 1 : t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c.<\/strong><\/p>\n

B\u00e0i 1 :t\u00ednh gi\u00e1 tr\u1ecb c\u1ee7a bi\u1ec3u th\u1ee9c : A = x^{2<\/sup>\u00a0\u2013 4x + 4 t\u1ea1i x = -1<\/p>\n}

Ta c\u00f3 : A = x^{2<\/sup>\u00a0\u2013 4x + 4 = A = x2<\/sup>\u00a0\u2013 2.x.2 + 22<\/sup>\u00a0= (x \u2013 2)2<\/sup><\/p>\n}

T\u1ea1i x = -1 : A = ((-1) \u2013 2)^{2<\/sup>=(-3)2<\/sup>= 9<\/p>\n}

V\u1eady : A(-1) = 9<\/p>\n

D\u1ea1ng 2 : Ch\u1ee9ng minh bi\u1ec3u th\u1ee9c A kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn :<\/strong><\/p>\n

B = (x \u2013 1)^{2<\/sup>\u00a0+ (x + 1)(3 \u2013 x)<\/p>\n}

GI\u1ea2I.<\/p>\n

B =(x \u2013 1)^{2<\/sup>\u00a0+ (x + 1)(3 \u2013 x)<\/p>\n}

= x^{2<\/sup>\u00a0\u2013 2x + 1 \u2013 x2<\/sup>\u00a0+ 3x + 3 \u2013 x<\/p>\n}

= 4\u00a0: h\u1eb1ng s\u1ed1 kh\u00f4ng ph\u1ee5 thu\u1ed9c v\u00e0o bi\u1ebfn x.<\/p>\n

D\u1ea1ng 3 : T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c :<\/strong><\/p>\n

C = x^{2<\/sup>\u00a0\u2013 2x + 5<\/p>\n}

GI\u1ea2I.<\/p>\n

Ta c\u00f3 : C = x^{2<\/sup>\u00a0\u2013 2x + 5 = (x2<\/sup>\u00a0\u2013 2x + 1) + 4 = (x \u2013 1)2\u00a0<\/sup>+ 4<\/p>\n}

M\u00e0\u00a0: (x \u2013 1)^{2\u00a0<\/sup>\u2265 0 v\u1edbi m\u1ecdi x.<\/p>\n}

Suy ra\u00a0: (x \u2013 1)^{2\u00a0<\/sup>+ 4 \u2265 4 hay C \u2265 4<\/p>\n}

D\u1ea5u \u201c=\u201d\u00a0x\u1ea3y ra khi\u00a0: x \u2013 1 = 0 hay x = 1<\/p>\n

N\u00ean\u00a0: C_{min\u00a0<\/sub>= 4 khi x = 1<\/p>\n}

D\u1ea1ng 4 : T\u00ecm gi\u00e1 tr\u1ecb l\u1edbn nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c :<\/strong><\/p>\n

D = 4x \u2013 x^{2<\/sup><\/p>\n}

GI\u1ea2I.<\/p>\n

Ta c\u00f3 : D = 4x \u2013 x^{2\u00a0<\/sup>= 4 \u2013 4 + 4x \u2013 x2\u00a0<\/sup>= 4 \u2013 (4 + x2\u00a0<\/sup>\u2013 4x) = 4 \u2013 (x \u2013 2)2<\/sup><\/p>\n}

M\u00e0\u00a0: -(x \u2013 2)^{2\u00a0<\/sup>\u2264 0 v\u1edbi m\u1ecdi x.<\/p>\n}

Suy ra\u00a0: 4 \u2013 (x \u2013 2)^{2\u00a0<\/sup>\u2264 4 hay D \u2264 4<\/p>\n}

D\u1ea5u \u201c=\u201d\u00a0x\u1ea3y ra khi\u00a0: x \u2013 2 = 0 hay x = 2<\/p>\n

N\u00ean\u00a0: D_{max\u00a0<\/sub>= 4 khi x = 2.<\/p>\n}

D\u1ea1ng 5 :Ch\u1ee9ng minh \u0111\u1eb3ng th\u1ee9c :\u00a0\u00a0\u00a0<\/strong><\/p>\n

(a + b)^{3<\/sup>\u00a0\u2013 (a \u2013 b)3<\/sup>\u00a0= 2b(3a2<\/sup>\u00a0+ b2<\/sup>)<\/p>\n}

GI\u1ea2I.<\/p>\n

VT = (a + b)^{3<\/sup>\u00a0\u2013 (a \u2013 b)3<\/sup><\/p>\n}

=\u00a0(<\/strong>a^{3<\/sup>\u00a0+ 3a2<\/sup>b + 3ab2<\/sup>\u00a0+ b3<\/sup>) \u2013 (a3<\/sup>\u00a0\u2013 3a2<\/sup>b + 3ab2<\/sup>\u00a0\u2013 b3<\/sup>)<\/strong><\/p>\n}

= a^{3<\/sup>\u00a0+ 3a2<\/sup>b + 3ab2<\/sup>\u00a0+ b3<\/sup>\u00a0\u2013 a3<\/sup>\u00a0+ 3a2<\/sup>b \u2013 3ab2<\/sup>\u00a0+ b3<\/sup><\/p>\n}

= 6a^{2<\/sup>b + 2b3<\/sup><\/p>\n}

= 2b(3a^{2<\/sup>\u00a0+ b2<\/sup>) ->\u0111pcm.<\/p>\n}

V\u1eady : (a + b)^{3<\/sup>\u00a0\u2013 (a \u2013 b)3<\/sup>\u00a0= 2b(3a2<\/sup>\u00a0+ b2<\/sup>)<\/p>\n}

D\u1ea1ng 6\u00a0 : Ch\u1ee9ng minh b\u1ea5t \u0111\u1eb3ng th\u1ee9c :\u00a0\u00a0\u00a0<\/strong><\/p>\n

B\u00e0i t\u1eadp \u00e1p d\u1ee5ng 7 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u00e1ng nh\u1edb n\u00e2ng cao\u00a0<\/strong><\/h3>\n

B\u00e0i 1.<\/strong>\u00a0Cho \u0111a th\u1ee9c 2x\u00b2 \u2013 5x + 3 . Vi\u1ebft \u0111a th\u1ee9c tr\u00ean d\u01b0\u1edbi d\u1ea1ng 1 \u0111a th\u1ee9c c\u1ee7a bi\u1ebfn y trong \u0111\u00f3 y = x + 1.<\/p>\n

L\u1eddi Gi\u1ea3i<\/strong><\/p>\n

Theo \u0111\u1ec1 b\u00e0i ta c\u00f3:\u00a0 y = x + 1 => x = y \u2013 1.<\/p>\n

A = 2x\u00b2 \u2013 5x + 3<\/p>\n

= 2(y \u2013 1)\u00b2 \u2013 5(y \u2013 1) + 3 = 2(y\u00b2 \u2013 2y + 1) \u2013 5y + 5 + 3 = 2y\u00b2 \u2013 9y + 10<\/p>\n

B\u00e0i 2.<\/strong>\u00a0T\u00ednh nhanh k\u1ebft qu\u1ea3 c\u00e1c bi\u1ec3u th\u1ee9c sau:<\/p>\n

a) 127\u00b2 + 146.127 + 73\u00b2<\/p>\n

b) 9^{8<\/sup>\u00a0.28<\/sup>\u00a0\u2013 (184<\/sup>\u00a0\u2013 1)(184<\/sup>\u00a0+ 1)<\/p>\n}

c) 100\u00b2 \u2013 99\u00b2 + 98\u00b2 \u2013 97\u00b2 + \u2026+ 2\u00b2 \u2013 1\u00b2<\/p>\n

d) (20\u00b2 + 18\u00b2 + 16\u00b2 +\u2026+ 4\u00b2 + 2\u00b2) \u2013 ( 19\u00b2 + 17\u00b2 + 15\u00b2 +\u2026+ 3\u00b2 + 1\u00b2)<\/p>\n

L\u1eddi Gi\u1ea3i<\/strong><\/p>\n

a)\u00a0A = 127\u00b2 + 146.127 + 73\u00b2<\/strong>\u00a0= 127\u00b2 + 2.73.127 + 73\u00b2\u00a0 = (127 + 73)\u00b2 = 200\u00b2 = 40000 .<\/p>\n

b)\u00a0B = 9^{8<\/sup>\u00a0.28<\/sup>\u00a0\u2013 (184<\/sup>\u00a0\u2013 1)(184<\/sup>\u00a0+ 1)<\/strong>\u00a0= 188<\/sup>\u00a0 \u2013 (188<\/sup>\u00a0 \u2013 1) = 1<\/p>\n}

c)\u00a0C = 100\u00b2 \u2013 99\u00b2 + 98\u00b2 \u2013 97\u00b2 + \u2026+ 2\u00b2 \u2013 1\u00b2<\/strong><\/p>\n

= (100 + 99)(100 \u2013 99) + (98 + 97)(98 \u2013 97) +\u2026+ (2 + 1)(2 \u2013 1)<\/p>\n

= 100 + 99 + 98 + 97 +\u2026+\u00a0 2 + 1 = 5050.<\/p>\n

d)\u00a0D = (20\u00b2 + 18\u00b2 + 16\u00b2 +\u2026+ 4\u00b2 + 2\u00b2) \u2013 ( 19\u00b2 + 17\u00b2 + 15\u00b2 +\u2026+ 3\u00b2 + 1\u00b2)<\/strong><\/p>\n

= (20\u00b2 \u2013 19\u00b2) + (18\u00b2 \u2013 17\u00b2) + (16\u00b2 \u2013 15\u00b2)+ \u2026+ (4\u00b2 \u2013 3\u00b2) + (2\u00b2 \u2013 1\u00b2)<\/p>\n

= (20 + 19)(20 \u2013 19) + (18 + 17)(18 \u2013 17) + ( 16 +15)(16 \u2013 15)+ \u2026+ (4 + 3)(4 \u2013 3) + (2 + 1)(2 \u2013 1)<\/p>\n

= 20 + 19 + 18 + 17 + 16 +15 + \u2026+ 4 + 3 + 2 + 1 = 210<\/p>\n

B\u00e0i 3.<\/strong>\u00a0So s\u00e1nh hai s\u1ed1 sau, s\u1ed1 n\u00e0o l\u1edbn h\u01a1n?<\/p>\n

a) A = (2 + 1)(2^{2<\/sup>\u00a0+ 1)(24<\/sup>\u00a0+ 1)(28<\/sup>\u00a0+ 1)(216<\/sup>\u00a0+ 1) v\u00e0 B = 232<\/sup><\/p>\n}

b) A = 1989.1991 v\u00e0 B = 1990^{2<\/sup><\/p>\n}

L\u1eddi Gi\u1ea3i<\/strong><\/p>\n

a) Ta nh\u00e2n 2 v\u1ebf c\u1ee7a A v\u1edbi 2 \u2013 1, ta \u0111\u01b0\u1ee3c:<\/p>\n

A = (2 \u2013 1)(2 + 1)(2^{2<\/sup>\u00a0+ 1)(24<\/sup>\u00a0+ 1)(28<\/sup>\u00a0+ 1)(216<\/sup>\u00a0+ 1)<\/p>\n}

Ta \u00e1p d\u1ee5ng \u0111\u1eb3ng th\u1ee9c ( a- b)(a + b) = a\u00b2 \u2013 b\u00b2 nhi\u1ec1u l\u1ea7n, ta \u0111\u01b0\u1ee3c:<\/p>\n

A = 2^{32<\/sup>\u00a0\u2013 1.<\/p>\n}

=> V\u1eady A < B.<\/p>\n

b) Ta \u0111\u1eb7t 1990 = x => B = x\u00b2<\/p>\n

V\u1eady A = (x \u2013 1)(x + 1) = x\u00b2 \u2013 1<\/p>\n

=> B > A l\u00e0 1.<\/p>\n

B\u00e0i 4.<\/strong>\u00a0Ch\u1ee9ng minh r\u1eb1ng:<\/p>\n

a) a(a \u2013 6) + 10 > 0.<\/p>\n

b) (x \u2013 3)(x \u2013 5) + 4 > 0.<\/p>\n

c) a\u00b2 + a + 1 > 0.<\/p>\n

L\u1eddi Gi\u1ea3i<\/strong><\/p>\n

a) VT = a\u00b2 \u2013 6a + 10 = (a \u2013 3)\u00b2 + 1 \u2265 1<\/p>\n

=> VT > 0<\/p>\n

b) VT = x\u00b2 \u2013 8x + 19 = (x \u2013 4)\u00b2 + 3 \u2265 3<\/p>\n

=> VT > 0<\/p>\n

c) a\u00b2 + a + 1 = a\u00b2 + 2.a.\u00bd + \u00bc + \u00be = (a + \u00bd )\u00b2 + \u00be \u2265 \u00be >0.<\/p>\n

B\u00e0i 5.<\/strong>\u00a0T\u00ecm gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a c\u00e1c bi\u1ec3u th\u1ee9c sau:<\/p>\n

a) A = x\u00b2 \u2013 4x + 1<\/p>\n

b) B = 4x\u00b2 + 4x + 11<\/p>\n

c) C = 3x\u00b2 \u2013 6x \u2013 1<\/p>\n

L\u1eddi Gi\u1ea3i<\/strong><\/p>\n

a) Ta s\u1ebd bi\u1ebfn \u0111\u1ed5i A= x\u00b2 \u2013 4x + 1 = x\u00b2 \u2013 4x + 4 \u2013 3 = ( x- 2)\u00b2 \u2013 3<\/p>\n

Do ( x- 2)\u00b2 > 0 n\u00ean => ( x- 2)\u00b2 \u2013 3 \u2265 -3<\/p>\n

V\u1eady gi\u00e1 tr\u1ecb nh\u1ecf nh\u1ea5t c\u1ee7a bi\u1ec3u th\u1ee9c A(A_{min) = -3\u00a0 khi v\u00e0 ch\u1ec9 khi x = 2.<\/sub><\/p>\n}

b) B = 4x\u00b2 + 4x + 11 = (2x + 1)\u00b2 + 10<\/p>\n

V\u1eady B_{min<\/sub>\u00a0 = 10 khi v\u00e0 ch\u1ec9 khi x = -\u00bd.<\/p>\n}

c) C = 3x\u00b2 \u2013 6x \u2013 1 = 3(x \u2013 1)\u00b2 \u2013 4<\/p>\n

V\u1eady C_{min<\/sub>\u00a0 = -4 khi v\u00e0 ch\u1ec9 khi x = 1.<\/p>\n}

B\u00e0i 6.<\/strong>\u00a0Cho a + b + c = 2p. Ch\u1ee9ng minh r\u1eb1ng: 2bc + b\u00b2 + c\u00b2 \u2013 a\u00b2 = 4p(p \u2013 a)<\/p>\n

L\u1eddi Gi\u1ea3i<\/strong><\/p>\n

Ta s\u1ebd \u0111i bi\u1ebfn \u0111\u1ed5i VP.<\/p>\n

VP = 2p(2p \u2013 2a) = (a + b + c)( a + b \u2013 c) = ( b + c )\u00b2 \u2013 a\u00b2 = b\u00b2 + 2bc + c\u00b2 \u2013 a\u00b2 = VT (\u0111ccm)<\/p>\n

B\u00e0i 7.<\/strong>\u00a0Hi\u1ec7u c\u00e1c b\u00ecnh ph\u01b0\u01a1ng c\u1ee7a 2 s\u1ed1 t\u1ef1 nhi\u00ean ch\u1eb5n li\u00ean ti\u1ebfp b\u1eb1ng 36. T\u00ecm hai s\u1ed1 \u1ea5y.<\/p>\n

L\u1eddi Gi\u1ea3i<\/strong><\/p>\n

G\u1ecdi 2 s\u1ed1 ch\u1eb5n li\u00ean ti\u1ebfp l\u00e0 x v\u00e0 x + 2 (x ch\u1eb5n). Ta c\u00f3:<\/p>\n

(x + 2)\u00b2 \u2013 x\u00b2 = 36<\/p>\n

<=> x\u00b2 + 4x + 4 \u2013 x\u00b2 = 36<\/p>\n

<=> 4x = 32<\/p>\n

<=> x = 8<\/p>\n

=> s\u1ed1 th\u1ee9 2 l\u00e0 8+2 = 10<\/p>\n

\u0110\u00e1p s\u1ed1: 8 v\u00e0 10<\/p>\n

B\u00e0i 8.<\/strong>\u00a0T\u00ecm 3 s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp bi\u1ebft r\u1eb1ng t\u1ed5ng c\u00e1c t\u00edch c\u1ee7a t\u1eebng c\u1eb7p 2 s\u1ed1 trong 3 s\u1ed1 \u1ea5y b\u1eb1ng 74<\/p>\n

L\u1eddi Gi\u1ea3i<\/strong><\/p>\n

G\u1ecdi 3 s\u1ed1 t\u1ef1 nhi\u00ean li\u00ean ti\u1ebfp l\u00e0: x \u2013 1, x, x + 1 ( \u0111k: x>0)<\/p>\n

V\u1eady ta c\u00f3:\u00a0 x(x \u2013 1) + (x \u2013 1)(x + 1) + x(x + 1)= 74<\/p>\n

Ta nh\u00e2n v\u00e0o v\u00e0 r\u00fat g\u1ecdn \u0111i ta c\u00f3:<\/p>\n

x\u00b2 = 25 <=> x = -5 , x = 5<\/p>\n

So s\u00e1nh v\u1edbi \u0110k: x>o => x = 5 (t\/m).<\/p>\n

V\u1eady \u0111\u00e1p s\u1ed1: 4, 5, 6.<\/p>\n

 <\/p>\n

Th\u00f4ng qua l\u00fd thuy\u1ebft c\u1ee7a 7 h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u00e1ng nh\u1edb v\u00e0 b\u00e0i t\u1eadp \u00e1p d\u1ee5ng hy v\u1ecdng b\u1ea1n s\u1ebd n\u1eafm v\u1eefng ki\u1ebfn th\u1ee9c m\u00f4n to\u00e1n l\u1edbp 8. Ch\u00fac c\u00e1c b\u1ea1n h\u1ecdc t\u1ed1t<\/p>\n","protected":false},"excerpt":{"rendered":"

7 \u00a0h\u1eb1ng \u0111\u1eb3ng th\u1ee9c \u0111\u00e1ng nh\u1edb l\u1edbp 8 1)\u00a0\u00a0\u00a0\u00a0\u00a0 (A + B)2\u00a0= A2\u00a0+ 2AB + B2 2)\u00a0\u00a0\u00a0\u00a0\u00a0 (A \u2013 B)2\u00a0= A2\u00a0\u2013 2AB + B2 3)\u00a0\u00a0\u00a0\u00a0\u00a0 A2\u00a0\u2013 B2\u00a0= (A \u2013 B)(A… <\/p>\n","protected":false},"author":9,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[416,429,407],"tags":[],"amp_enabled":true,"_links":{"self":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts\/8499"}],"collection":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/users\/9"}],"replies":[{"embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/comments?post=8499"}],"version-history":[{"count":0,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/posts\/8499\/revisions"}],"wp:attachment":[{"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/media?parent=8499"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/categories?post=8499"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/onthitot.com\/wp-json\/wp\/v2\/tags?post=8499"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}