Bài 1:

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

b) Sửa đề: \(\left(x^2+y^2\right)^2-\left(2xy\right)^2=\left(x^2-2xy+y^2\right)\left(x^2+2xy+y^2\right)\)

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

c) \(x\left(x-3y\right)^2+y\left(y-3x\right)^2=x\left(x^2-6xy+9y^2\right)+y\left(y^2-6xy+9x^2\right)\)

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

\(=x^3+3x^2y+3xy^2+y^3=\left(x+y\right)^3\)

Bài 2:

a) \(\left(a+b\right)^3+\left(a-b\right)^3=\left(a+b+a-b\right)\left[\left(a+b\right)^2-\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

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

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

b) \(\left(a+b\right)^3-\left(a-b\right)^3=\left(a+b-a+b\right)\left[\left(a+b\right)^2+\left(a+b\right)\left(a-b\right)+\left(a-b\right)^2\right]\)

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

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

Ta có : x⁴ + x² + 1

= x⁴ + x² + x² + 1 - x²

= (x² + 1)² - x²

= (x² + 1 - x)(x² + 1 + x)

x⁴ + x² + 1

= x⁴ + 2x² + 1 - x²

= ( x² + 1 )² - x²

= ( x² - x + 1 )( x² + x + 1 )

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

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

\(\Leftrightarrow3x=0\)

\(\Rightarrow x=0\)

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

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

\(\Leftrightarrow2-x=3\)

\(\Leftrightarrow x=-1\)