Bài 1:
a) Ta có:
\(VT=\dfrac{3x+6}{x+2}=3\) (ĐK: \(x\ne-2\))
\(=\dfrac{3\left(x+2\right)}{x+2}\)
\(=\dfrac{3}{1}\)
\(=3=VP\left(dpcm\right)\)
b) Ta có:
\(VT=\dfrac{x^2+2x}{3x+6}\) (ĐK: \(x\ne-2\))
\(=\dfrac{x\left(x+2\right)}{3\left(x+2\right)}\)
\(=\dfrac{x}{3}=VP\left(dpcm\right)\)
c) Ta có:
\(VT=\dfrac{x-1}{x^2-1}\) (ĐK: \(x\ne\pm1\))
\(=\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\)
\(=\dfrac{1}{x+1}=VP\left(dpcm\right)\)
d) Ta có:
\(VT=\dfrac{x^2+3x-4}{x-1}\) (ĐK: \(x\ne1\))
\(=\dfrac{x^2+4x-x-4}{x-1}\)
\(=\dfrac{x\left(x+4\right)-\left(x+4\right)}{x-1}\)
\(=\dfrac{\left(x-1\right)\left(x+4\right)}{x-1}\)
\(=x+4=VP\left(dpcm\right)\)
Bài 2:
Với \(x\ne0\)
\(\dfrac{x^2-2x+1}{x\left(x-1\right)}=\dfrac{\left(x-1\right)\left(x-1\right)}{x\left(x-1\right)}=\dfrac{x-1}{x}\)
\(\dfrac{2x-2}{2x}=\dfrac{2\left(x-1\right)}{2x}=\dfrac{x-1}{x}\)
Vậy cả 3 phân thức đều bằng nhau.
