giúp mk vs ạ đang cần gấp ạ
Bài 26:
a: ĐKXĐ: x<>-2
\(\dfrac{\left(x+2\right)^2}{2x+4}=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}=\dfrac{x+2}{2}\)
b: ĐKXĐ: x<>-2
\(\dfrac{x^2+4x+4}{2x+4}\)
\(=\dfrac{x^2+2\cdot x\cdot2+2^2}{2\left(x+2\right)}\)
\(=\dfrac{\left(x+2\right)^2}{2\left(x+2\right)}=\dfrac{x+2}{2}\)
c: ĐKXĐ: x<>-2
\(\dfrac{\left(1-x\right)\left(-x-2\right)}{x+2}\)
\(=\dfrac{\left(x-1\right)\left(x+2\right)}{x+2}\)
=x-1
d: \(\dfrac{x^2-y^2}{x+y}=\dfrac{\left(x-y\right)\left(x+y\right)}{x+y}=x-y\)
Bài 27:
a: ĐKXĐ: x<>2
\(\dfrac{3x-6}{x^3-6x^2+12x-8}\)
\(=\dfrac{3\left(x-2\right)}{x^3-3\cdot x^2\cdot2+3\cdot x\cdot2^2-2^3}\)
\(=\dfrac{3\left(x-2\right)}{\left(x-2\right)^3}=\dfrac{3}{\left(x-2\right)^2}\)
b: ĐKXĐ: x<>-2
\(\dfrac{x^3+2x^2}{x^3+6x^2+12x+8}\)
\(=\dfrac{x^2\left(x+2\right)}{x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3}\)
\(=\dfrac{x^2\left(x+2\right)}{\left(x+2\right)^3}=\dfrac{x^2}{\left(x+2\right)^2}\)