Question

Bài 1: 

a) (x+4) (x+3)-7x

b) ( x+4)² + x - 16

c) \(\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}\)

Bài 2:

a) 7a-7b

b) x² - 8x + 16

c) ax - ay + 3x - 3y

d) x² + 6x +9 - y²

 

Answer 1

\(\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}\)

\(=\dfrac{4\left(x-3\right)+x+7}{x^2-9}\)

\(=\dfrac{4x-12+x+7}{x^2-9}\)

\(=\dfrac{5x-5}{x^2-9}\)

Answer 2

Bài 1:
\(a,\left(x+4\right)\left(x+3\right)-7x=x^2+4x+3x+12-7x=x^2+12\\ b,\left(x+4\right)^2+x-16=x^2+8x+16+x-16=x^2+9x\\ c,\dfrac{4}{x+3}+\dfrac{x+7}{x^2-9}=\dfrac{4\left(x-3\right)}{\left(x+3\right)\left(x-3\right)}+\dfrac{x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{4x-12+x+7}{\left(x+3\right)\left(x-3\right)}=\dfrac{5x-5}{\left(x+3\right)\left(x-3\right)}\)

Bài 2:
\(7a-7b=7\left(a-b\right)\\ b,x^2-8x+16=\left(x-4\right)^2\\ c,ax-ay+3x-3y=a\left(x-y\right)+3\left(x-y\right)=\left(a+3\right)\left(x-y\right)\\ d,x^2+6x+9-y^2=\left(x+3\right)^2-y^2=\left(x-y+3\right)\left(x+y+3\right)\)