Bài 1:
\(a.x^2-y^2-5x+5y\\ =\left(x+y\right)\left(x-y\right)-5\left(x-y\right)\\ =\left(x-y\right)\left(x+y-5\right)\\ b.2x^2-5x-7\\ =\left(2x^2+2x\right)+\left(-7x-7\right)\\ =2x\left(x+1\right)-7\left(x+1\right)\\ =\left(x+1\right)\left(2x-7\right)\)
Bài 2:
\(\dfrac{4x^2-16}{x^2+2x}=\dfrac{A}{x}\\ =>\dfrac{A}{x}=\dfrac{4\left(x^2-4\right)}{x\left(x+2\right)}=\dfrac{4\left(x+2\right)\left(x-2\right)}{x\left(x+2\right)}=\dfrac{4\left(x-2\right)}{x}\\ =>A=4\left(x-2\right)=4x-8\)
Bài 1:
a: \(x^2-y^2-5x+5y\)
=(x-y)(x+y)-5(x-y)
=(x-y)(x+y-5)
b: \(2x^2-5x-7\)
\(=2x^2-7x+2x-7\)
=x(2x-7)+(2x-7)
=(2x-7)(x+1)
Bài 2:
\(\dfrac{A}{x}=\dfrac{4x^2-16}{x^2+2x}\)
=>\(\dfrac{A}{x}=\dfrac{4\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}\)
=>\(\dfrac{A}{x}=\dfrac{4\left(x-2\right)}{x}\)
=>A=4(x-2)=4x-8
