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

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

d: \(=9x^2+6x+1-9x^2+6x-1=12x\)

a: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1+4-2\left(4x^2-12x+9\right)\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

e: \(\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)=8x^3+27y^3\)

A = (3x + y)^2 - 3y . ( 2x - 1/3y )

=2y²+9x²

B = ( x - 2 )^2 + ( x + 2 )^2 - 2. ( x - 2 ) ( x + 2)

=2⁴

C = ( x - y ) ( x^2 + xy + y^2 ) + 2y^3

=y³+x³

D = ( x -5 ) ( x+ 5 ) -(x - 8 ) (x + 4)

=4x+7

E = (3x + 1 )^2 - 2 . ( 9x^2 - 1 ) + ( 3x - 1 ) ^2

=4

F = ( x - 3 ) ( x + 3 ) - ( x - 3 )^2

=6x-18

Lời giải:

a. $=(x-y)(x+y)=[(-1)-(-3)][(-1)+(-3)]=2(-4)=-8$
b. $=3x^4-2xy^3+x^3y^2+3x^2y+12xy+15y-12xy-12$

$=3x^4-2xy^3+x^3y^2+3x^2y+15y-12$
=3-2.1(-2)^3+1^3.(-2)^2+3.1^2(-2)+15(-2)-12$
$=-25$
c.

$=2x^4+3x^3y-4x^3y-12xy+12xy=2x^4-x^3y$

$=x^3(2x-y)=(-1)^3[2(-1)-2]=-1.(-4)=4$

d. 

$=2x^2y+4x^2-5xy^2-10x+3xy^2-3x^2y$

$=(2x^2y-3x^2y)+4x^2+(-5xy^2+3xy^2)-10x$

$=-x^2y+4x^2-2xy^2-10x$

$=-3^2.(-2)+4.3^2-2.3(-2)^2-10.3=0$

a: Ta có: \(\left(2x-1\right)^2-2\left(2x-3\right)^2+4\)

\(=4x^2-4x+1-2\left(4x^2-12x+9\right)+4\)

\(=4x^2-4x+5-8x^2+24x-18\)

\(=-4x^2+20x-13\)

b: \(\left(3x+2\right)^2+2\left(3x+2\right)\left(1-2y\right)+\left(1-2y\right)^2\)

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

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

1 a

2c

3b

4d

5c

6c

\(a.\left(3x-1\right)^2+\left(x+3\right)\left(2x-1\right)\)

\(=9x^2-6x+1-2x^2+x-6x+3\)

\(=7x^2-11x+4\)

Bài 1:

a: ĐKXĐ: \(x+4\ne0\)

=>\(x\ne-4\)

b: ĐKXĐ: \(2x-1\ne0\)

=>\(2x\ne1\)

=>\(x\ne\dfrac{1}{2}\)

c: ĐKXĐ: \(x\left(y-3\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne0\\y-3\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne0\\y\ne3\end{matrix}\right.\)

d: ĐKXĐ: \(x^2-4y^2\ne0\)

=>\(\left(x-2y\right)\left(x+2y\right)\ne0\)

=>\(x\ne\pm2y\)

e: ĐKXĐ: \(\left(5-x\right)\left(y+2\right)\ne0\)

=>\(\left\{{}\begin{matrix}x\ne5\\y\ne-2\end{matrix}\right.\)

 Bài 2:

a: \(\dfrac{-12x^3y^2}{-20x^2y^2}=\dfrac{12x^3y^2}{20x^2y^2}=\dfrac{12x^3y^2:4x^2y^2}{20x^2y^2:4x^2y^2}=\dfrac{3x}{5}\)

b: \(\dfrac{x^2+xy-x-y}{x^2-xy-x+y}\)

\(=\dfrac{\left(x^2+xy\right)-\left(x+y\right)}{\left(x^2-xy\right)-\left(x-y\right)}\)

\(=\dfrac{x\left(x+y\right)-\left(x+y\right)}{x\left(x-y\right)-\left(x-y\right)}=\dfrac{\left(x+y\right)\left(x-1\right)}{\left(x-y\right)\left(x-1\right)}\)

\(=\dfrac{x+y}{x-y}\)

c: \(\dfrac{7x^2-7xy}{y^2-x^2}\)

\(=\dfrac{7x\left(x-y\right)}{\left(y-x\right)\left(y+x\right)}\)

\(=\dfrac{-7x\left(x-y\right)}{\left(x-y\right)\left(x+y\right)}=\dfrac{-7x}{x+y}\)
d: \(\dfrac{7x^2+14x+7}{3x^2+3x}\)

\(=\dfrac{7\left(x^2+2x+1\right)}{3x\left(x+1\right)}\)

\(=\dfrac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{7\left(x+1\right)}{3x}\)

e: \(\dfrac{3y-2-3xy+2x}{1-3x-x^3+3x^2}\)

\(=\dfrac{3y-2-x\left(3y-2\right)}{1-3x+3x^2-x^3}\)

\(=\dfrac{\left(3y-2\right)\left(1-x\right)}{\left(1-x\right)^3}=\dfrac{3y-2}{\left(1-x\right)^2}\)

g: \(\dfrac{x^2+7x+12}{x^2+5x+6}\)

\(=\dfrac{\left(x+3\right)\left(x+4\right)}{\left(x+3\right)\left(x+2\right)}\)

\(=\dfrac{x+4}{x+2}\)

a, (\(x\) + y).(\(x\) + y)² - 3\(xy\).(\(x\) + y) 

= (\(x+y\))³ - 3\(x^2\)y - 3\(xy^2\)

= \(x^3\) + 3\(x^2\).y + 3\(xy^2\) + y³ - 3\(x^2\).y  - 3\(xy^2\)

= \(x^3\) + y³ 

b, (\(x-y\)).(\(x-y\))² - 3\(xy\).(\(x-y\)) 

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

= \(x^3\) - 3\(x^2\)y + 3\(xy^2\) - y³ - 3\(x^2\)y + 3\(xy^2\)

= \(x^3\) - 6\(x^2\)y + 6\(xy^2\) - y³

c, (\(x\) - 2y)² + 4y²

=  \(x^2\) - 4\(xy\) + 4y² + 4y²

=  \(x^2\) - 4\(xy\) + 8y²