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

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

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

\(=2^2=4\)

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

\(=4x^2-12xy+9y^2+4x^2+12xy+9y^2+4x^2-9y^2\)

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

\(=12x^2+9y^2\)

a ) ( 12x - 5 )( 4x - 1 ) + 3x( 1 - 16x )

= ( 12x - 5 )( 4x - 1 ) - 3x( 16x - 1 )

= ( 12x - 5 )( 4x - 1 ) - 3x( 4x + 1 )( 4x - 1 )

= ( 4x - 1 ) [ 12x - 5 - 3x( 4x + 1 ) ]

= ( 4x - 1 ) ( 12x - 5 - 12x² - 3x )

= ( 4x - 1 )( 12x² - 9x + 5 )

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

\(=4x^4-4x^2\)

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

\(=4x^4+4x^2y-2x^2+2y^2\)

c. \(\left(2x+1\right)\left(2x-1\right)-4x^2=\left(2x\right)^2-1^2-4x^2=4x^2-1-4x^2=-1\)

d. \(\left(2x^{3y}+y\right)^2-\left(y-2x^{3y}\right)^2\)

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

\(=2y.2.2x^{3y}=4y.2x^{3y}\)

Câu 4:B

Câu 7: A

Câu 8: B

Câu 11: C

\(a,=\dfrac{x^2+4x+3-2x^2+2x+x^2-4x+3}{\left(x-3\right)\left(x+3\right)}=\dfrac{2\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}=\dfrac{2}{x-3}\\ b,=\dfrac{1-2x+3+2y+2x-4}{6x^3y}=\dfrac{2y}{6x^3y}=\dfrac{1}{x^2}\\ c,=\dfrac{75y^2+18xy+10x^2}{30x^2y^3}\\ d,=\dfrac{5x+8-x}{4x\left(x+2\right)}=\dfrac{4\left(x+2\right)}{4x\left(x+2\right)}=\dfrac{1}{x}\\ c,=\dfrac{x^2+2+2x-2-x^2-x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{x-1}{\left(x-1\right)\left(x^2+x+1\right)}=\dfrac{1}{x^2+x+1}\)

1/2x=2/3y=3/4

=> 2x=3y/2=4/3

chia cho 6 => x/3=y/4= x-y/3-4= 15/-1=-15

=> x= -45;y=-60  

ko có z