TA có: \(A=x^3-y^3-36xy=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy=\left(x-y\right)\left[\left(x-y\right)^2+3xy\right]-36xy\)

\(=12.12^2+3.12xy-36xy=12^3\)

Ta có: \(2x^2+\frac{y^2}{4}+\frac{1}{x^2}=4\)

=> \(\left(x^2+\frac{y^2}{4}\right)+\left(x^2+\frac{1}{x^2}\right)=4\)

Lại có: \(x^2+\frac{y^2}{4}\ge2.x.\frac{y}{2}=xy\) Và \(x^2+\frac{1}{x^2}\ge2.x.\frac{1}{x}=2\)

=> \(4\ge xy+2\)=> \(2\ge xy\)

=> \(A=2016+xy\le2016+2=2018\)

=> A_min=2018

\(\sqrt[]{\sqrt{ }\frac{ }{ }\sqrt[]{}3\hept{\begin{cases}\\\\\end{cases}}3\frac{ }{ }\sqrt{ }\cos\hept{\begin{cases}\\\\\end{cases}}\Omega3\cong}\)

\(\left(\frac{x^2-1}{x^4-x^2+1}-\frac{1}{x^2+1}\right)\times\left(x^4+\frac{1-x^4}{1+x^2}\right)\)

ĐK : ...

\(=\left(\frac{\left(x^2-1\right)\left(x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}-\frac{x^4-x^2+1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right)\times\left(x^4+\frac{\left(1-x^2\right)\left(1+x^2\right)}{\left(1+x^2\right)}\right)\)

\(=\left(\frac{x^4-1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}-\frac{x^4-x^2+1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right)\times\left(x^4+1-x^2\right)\)

\(=\left(\frac{x^4-1-x^4+x^2-1}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\right)\times\left(x^4-x^2+1\right)\)

\(=\frac{x^2-2\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\)

\(=\frac{x^2-2}{x^2+1}\)

Mình sửa dòng 5 một chút nhé 

\(=\frac{\left(x^2-2\right)\left(x^4-x^2+1\right)}{\left(x^4-x^2+1\right)\left(x^2+1\right)}\)( như kia dễ bị nhầm )

kokokokokokoko biết

a/b+c + b/c+a + c/a+b = 1 

=> (a+b+c)(a/b+c + b/c+a + c/a+b = (a+b+c).1

=> a²/ b+c + a + b²/c+a + b + c²/a+b + c = a+b+c

=> a²/b+c + b²/c+a + c²/a+b = (a+b+c)-(a+b+c) = 0

 đúng thì k cho mình nka

ĐK:\(a+b+c\ne0\)

Khi đó:\(\frac{a}{a+b}+\frac{b}{c+a}+\frac{c}{a+b}=1\)

\(\Leftrightarrow\left(a+b+c\right)\left(\frac{a}{c+b}+\frac{b}{a+c}+\frac{c}{a+b}\right)=a+b+c\)

\(\Leftrightarrow\frac{\left(a+b+c\right)a}{c+b}+\frac{\left(a+b+c\right)b}{a+c}+\frac{\left(a+b+c\right)c}{a+b}=a+b+c\)

\(\Leftrightarrow\frac{a^2}{b+c}+a+\frac{b^2}{c+a}+b+\frac{c^2}{a+b}+c=a+b+c\)

\(\Leftrightarrow\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}=0\)

a) 2x( x - 7 ) - ( x + 3 )( x - 2 ) - ( x + 4 )( x - 4 )

= 2x² - 14x - ( x² + x - 6 ) - ( x² - 16 )

= 2x² - 14x - x² - x + 6 - x² + 16

= 22 - 15x

b) ( 2x + 5 )( x - 2 ) - 3( x + 2 )² + ( x + 1 )²

= 2x² + x - 10 - 3( x² + 4x + 4 ) + x² + 2x + 1

= 3x² + 3x - 9 - 3x² - 12x - 12

= -9x - 21

c) ( x + 3 )( x - 3 ) - ( x + 5 )( x - 1 ) - ( x - 4 )²

= x² - 9 - ( x² + 4x - 5 ) - ( x² - 8x + 16 )

= x² - 9 - x² - 4x + 5 - x² + 8x - 16

= -x² + 4x - 20

d) 2x( x + 1 )² - ( x - 1 )³ - ( x - 2 )( x² + 2x + 4 ) 

= 2x( x² + 2x + 1 ) - ( x³ - 3x² + 3x - 1 ) - ( x³ - 8 )

= 2x³ + 4x² + 2x - x³ + 3x² - 3x + 1 - x³ + 8

= 7x² - x + 9

e) ( x + 5 )( x - 5 )( x + 2 ) - ( x + 2 )³

= ( x² - 25 )( x + 2 ) - ( x³ + 6x² + 12x + 8 )

= x³ + 2x² - 25x - 50 - x³ - 6x² - 12x - 8

= -4x² - 37x - 58

Bài 1 : 

Ta có : \(VP=\left(a+b\right)^4=\left(a+b\right)\left(a+b\right)^3\)

\(=\left(a+b\right)\left(a^3+3a^2b+3ab^2+b^3\right)=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

=> HĐT ko đc CM 

Bài 2 : 

a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)

\(=x^3+2x^2+4x-2x^2-4x-8-x+1+7=x^3-x=x\left(x^2-1\right)\)

Sửa đề : b, \(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x+1\right)\)

\(=8\left(x^3-1\right)-8x^3+1=8x^3-8-8x^3+1=-7\)

Xin phép chủ nahf cho mjnh sửa đề:D

\(\left(a+b\right)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

a,\(\left(a+b\right)^4\)

\(=\left[\left(a+b\right)^2\right]^2\)

\(=\left(a^2+2ab+b^2\right)^2\)

\(=\left[\left(a^2+2ab\right)+b^2\right]^2\)

\(=\left(a^2+2ab\right)^2+2\left(a^2+2ab\right)b^2+b^4\)

\(=a^4+4a^3b+4a^2b^2+2a^2b^2+4ab^3+b^4\)

\(=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)

Bài 2:

a,\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)

\(=\left(x^3-8\right)-\left(x-1\right)+7\)

b,\(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x-1\right)\)

\(=8\left(x^3-1\right)-\left(8x^3-1\right)\)

\(=8x^3-8-8x^3+1\)

\(=-7\)

b, \(x^3+2x^2+2x+1=\left(x^2+x+1\right)\left(x+1\right)\)

c, \(x^3-4x^2+12x-27=\left(x^2-x+9\right)\left(x-3\right)\)

d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)

e, sai đề 

a, \(\left(ab-1\right)^2+\left(a+b\right)^2=\left(a^2+1\right)\left(b^2+1\right)\)

b, \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2+x+1\right)\)

c, \(x^3-4x^2+12x-27=\left(x-3\right)\left(x^2-x+9\right)\)

d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)

e, cho mình sửa đề xíu

\(x^4+2x^3+2x^2+2x+1=\left(x+1\right)^2\left(x^2+1\right)\)

\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)

\(=\left[a.\left(a+b+c\right)+bc\right]\left[b.\left(a+b+c\right)+ac\right]\left[c.\left(a+b+c\right)+ab\right]\)

\(=\left(a^2+ab+ac+bc\right)\left(ba+b^2+bc+ac\right)\left(ca+cb+c^2+ab\right)\)

\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ba+b^2\right)+\left(bc+ac\right)\right]\left[\left(ca+c^2\right)\left(cb+ab\right)\right]\)

\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(b+a\right)\right]\left[c\left(a+c\right)b\left(b+b\right)\right]\)

\(=\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)\)

\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)

\(\Rightarrowđpcm\)

\(\left(am+bc\right)\left(bm+ac\right)\left(cm+ab\right)\)

\(=\left[a\left(a+b+c\right)+bc\right]\left[b\left(a+b+c\right)+ac\right]\left[c\left(a+b+c\right)+ab\right]\)

\(=\left(a^2+ab+ac+bc\right)\left(ab+b^2+bc+ac\right)\left(ac+bc+c^2+ab\right)\)

\(=\left[\left(a^2+ab\right)+\left(ac+bc\right)\right]\left[\left(ab+b^2\right)+\left(bc+ac\right)\right]\left[\left(ac+c^2\right)+\left(bc+ab\right)\right]\)

\(=\left[a\left(a+b\right)+c\left(a+b\right)\right]\left[b\left(a+b\right)+c\left(a+b\right)\right]\left[c\left(a+c\right)+b\left(a+c\right)\right]\)

\(=\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\)

\(=\left(a+b\right)^2\left(a+c\right)^2\left(b+c\right)^2\)

\(\Rightarrowđpcm\)