điều kiện: x + y khác 0, z khác 0, chắc vậy :v

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

= \(\frac{\left(x+y-z\right)\left(x+y+z\right)}{\left(x+y-z\right)\left(x+y\right)}\)

= \(\frac{x+y+z}{x+y}\)

rút gọn phân thức giùm mk nha

a) \(\left(5x-4\right)^2-49x^2\)

\(=\left(5x-4\right)^2-\left(7x\right)^2\)

\(=\left(12x-4\right)\left(-2x-4\right)\)

\(=-6\left(3x-1\right)\left(x+2\right)\)

c) \(x^2-y^2-x+y\)

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

\(=\left(x+y-1\right)\left(x-y\right)\)

d)\(4x^2-9y^2+4x-6y\)

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

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

e) \(-x^2+5x+2xy-5y-y^2\)

\(=-\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\)

\(=-\left(x-y\right)^2+5\left(x-y\right)\)

\(=\left(x-y\right)\left(y-x+5\right)\)

f) \(y^2\left(x^2+y\right)-zx^2-zy\)

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

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

Ta có : y²( x² + y ) - zx² - zy

= y²( x² + y ) - z( x² + y )

= ( x² + y )( y² - z )

\(y^2\left(x^2+y\right)-zx^2-zy\)

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

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

TL :

\(y^2\left(x^2+y\right)-zx^2-zy\)

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

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

Hok tốt

Ta có: \(y^2\left(x^2+y\right)-zx^2-zy\)

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

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

đề có thiếu k bạn

a)đề sai

b)4x(x-2y)+8y(2y-x)

=4x²-8xy+16y²-8xy

=16y²-16xy+4x²

=4(4y²-4xy-x²)

=4(2y-x)²

c)3x(x+1)^2-5x^2(x+1)+7(x+1)

=(3x²+3x)(x+1)-(x+1)(5x²+7)

=(x+1)(3x²+3x-5x²+7)

=(x+1)(-2x²+3x+7)

a)\(A=3x^2+6xy+3y^2-3z^2=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y-z\right)\left(x+y+z\right)\)b) \(A=\left(x+y\right)^2-2\left(x+y\right)+1=\left(x+y-1\right)^2\)

c) \(A=x^2+y^2+2xy+yz+zx=\left(x+y\right)^2+z\left(x+y\right)=\left(x+y\right)\left(x+y+z\right)\)