a, \(\left(x-y+1\right)\left(x+y+1\right)=x^2+xy+x-xy-y^2-y+x+y+1\)

\(=x^2+2x-y^2+1\)

b, \(\left(2x+3\right)^2+\left(2x-3\right)^2-2\left(4x^2-9\right)=4x^2+12x+9+4x^2-12x+9-8x^2+18\)

\(=36\)

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

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

c) \(xy+z+y+xz=y\left(x+1\right)+z\left(x+1\right)=\left(x+1\right)\left(y+z\right)\) đề vậy còn ra

tks bn

a) (2x-5)y+2y-10=0  <=>  2xy-3y = 10  <=>  y(2x-3)=10  <=>  y=\(\frac{10}{2x-3}\) với y là số nguyên 

=> 2x-3 là ước của 10 

ta có bảng sau

b)

3xy + 21x-y-11=0  <=>  y(3x-1)=-(21x-11) <=>  -y=\(\frac{21x-11}{3x-1}\) =\(\frac{7\left(3x-1\right)-4}{3x-1}\)=7-\(\frac{4}{3x-1}\)với -y nguyên nên 3x-1 là ước của 4

a) ( 2x - 5 )y + 2y - 10 = 0

<=> 2xy - 5y + 2y - 10 = 0

<=> 2xy - 3y - 10 = 0

<=> y( 2x - 3 ) - 10 = 0

<=> y( 2x - 3 ) = 10

Ta có bảng sau :

Vì x , y nguyên nên các cặp ( x ; y ) = { ( 2 ; 10 ) , ( 1 ; -10 ) , ( 4 ; 2 ) , ( -1 ; -2 ) }

b) 3xy + 21x - y - 11 = 0

<=> 3x( y + 7 ) - 1( y + 7 ) - 4 = 0

<=> ( 3x - 1 )( y + 7 ) - 4 = 0

<=> ( 3x - 1 )( y + 7 ) = 4

Ta có bảng sau :

Vì x, y nguyên nên các cặp ( x ; y ) = { ( 0 ; -11 ) , ( 1 ; -5 ) , ( -1 ; -8 ) }

A=10

B=5

C=2

a) \(\left(x+2y\right)^2=x^2+4xy+4y^2\)

b) \(\left(3x-\frac{1}{8}y\right)^2=9x^2-\frac{3}{4}xy+\frac{1}{64}y^2\)

c) \(\left(-6x-\frac{2}{5}\right)^2=36x^2+\frac{24}{5}x+\frac{4}{25}\)

d) \(\left(xy^2+1\right)\left(xy^2-1\right)=x^2y^4-1\)

e) \(\left(x-y\right)^2\left(x+y\right)^2=\left(x^2-y^2\right)^2=x^4-2x^2y^2+y^4\)

f) \(\left(\frac{1}{2}x-\frac{1}{3}y-1\right)^2=\frac{1}{4}x^2+\frac{1}{9}y^2+1-\frac{1}{3}xy-x+\frac{2}{3}y\)

a, \(\left(x+2y\right)^2=x^2+4xy+4y^2\)

b, \(\left(3x-\frac{1}{8}y\right)^2=9x^2-\frac{3}{4}xy+\frac{1}{64}y^2\)

e, \(\left(x-y\right)^2\left(x+y\right)^2=x^4-2x^2y^2+y^4\)

Bài làm:

Ta có: \(3\left(3-2x^2\right)+3x\left(2x-1\right)=9\)

\(\Leftrightarrow9-6x^2+6x^2-3x=9\)

\(\Leftrightarrow3x=0\)

\(\Rightarrow x=0\)

\(3\left(2-2x^2\right)+3x\left(2x-1\right)=9\)

\(\Leftrightarrow3\left(2-2x^2+2x^2-x\right)=9\)

\(\Leftrightarrow2-x=3\)

\(\Leftrightarrow x=-1\)

\(A=5-8x+x^2=-8x+x^2+6-11\)

\(=\left(x-4\right)^2-11\)

Vì \(\left(x-4\right)^2\ge0\forall x\)\(\Rightarrow\left(x-4\right)^2-11\ge-11\)

Dấu "=" xảy ra \(\Leftrightarrow\left(x-4\right)^2=0\Leftrightarrow x-4=0\Leftrightarrow x=4\)

Vậy Amin = - 11 <=> x = 4

\(B=\left(2-x\right)\left(x+4\right)=-x^2-2x+8\)

\(=-\left(x^2+2x+1\right)+9=-\left(x+1\right)^2+9\)

Vì \(\left(x+1\right)^2\ge0\forall x\)\(\Rightarrow-\left(x+1\right)^2+9\le9\)

Dấu "=" xảy ra \(\Leftrightarrow-\left(x+1\right)^2=0\Leftrightarrow x+1=0\Leftrightarrow x=-1\)

Vậy Bmax = 9 <=> x = - 1

a) \(A=x\left(x+2\right)+y\left(y-2\right)-2xy+37\)

\(=x^2+2x+y^2-2y-2xy+37\)

\(=\left(x^2-2xy+y^2\right)+\left(2x-2y\right)+37\)

\(=\left(x-y\right)^2+2\left(x-y\right)+37\)

Thay \(x-y=7\)vào biểu thức ta được: 

\(A=7^2+2.7+37=49+14+37=100\)

b) Ta có: \(x+y=3\)\(\Rightarrow\left(x+y\right)^2=9\)\(\Rightarrow x^2+y^2+2xy=9\)

mà \(x^2+y^2=5\)\(\Rightarrow5+2xy=9\)

\(\Rightarrow2xy=4\)\(\Rightarrow xy=2\)

Vậy \(xy=2\)

a) A = x( x + 2 ) + y( y - 2 ) - 2xy + 37

= x² + 2x + y² - 2y - 2xy + 37

= ( x² - 2xy + y² ) + ( 2x - 2y ) + 37

= ( x - y )² + 2( x - y ) + 37

Thế x - y = 7 vào A ta được :

A = 7² + 2.7 + 37 = 49 + 14 + 37 = 100

Vậy A = 100 khi x - y = 7

b) x + y = 3 => ( x + y )² = 9

=> x² + 2xy + y² = 9

=> 5 + 2xy = 9 ( sử dụng gt x² + y² = 5 )

=> 2xy = 4

=> xy = 2 

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

\(\Leftrightarrow2\left(a^2+b^2\right)-\left(a+b\right)^2=0\)

\(\Leftrightarrow\left(2a^2+2b^2\right)-\left(a^2+2ab+b^2\right)=0\)

\(\Leftrightarrow2a^2+2b^2-a^2-2ab-b^2=0\)

\(\Leftrightarrow a^2-2ab+b^2=0\)

\(\Leftrightarrow\left(a-b\right)^2=0\)

\(\Leftrightarrow a-b=0\)

\(\Leftrightarrow a=b\)( đpcm )

2( a² + b² ) = ( a + b )²

<=> 2a² + 2b² = a² + 2ab + b²

<=> 2a² + 2b² - a² - 2ab - b² = 0

<=> a² - 2ab + b² = 0

<=> ( a - b )² = 0 (*)

( a - b )² ≥ 0 ∀ a, b

Đẳng thức xảy ra ( tức (*) ) <=> a - b = 0 => a = b ( đpcm )