a) Ta có:
\(VT=\dfrac{\left(x-2\right)^3}{x^2-2x}\)
\(=\dfrac{\left(x-2\right)^3}{x\left(x-2\right)}\)
\(=\dfrac{\left(x-2\right)^2}{x}=VP\left(dpcm\right)\)
b) Ta có:
\(VT=\dfrac{x^2-4x+3}{x-3}\)
\(=\dfrac{x^2-x-3x+3}{x-3}=\dfrac{x\left(x-1\right)-3\left(x-1\right)}{x-3}\)
\(=\dfrac{\left(x-3\right)\left(x-1\right)}{x-3}=x-1\)
\(VP=\dfrac{x^2-5x+4}{x-4}\)
\(=\dfrac{x^2-x-4x+4}{x-4}=\dfrac{x\left(x-1\right)-4\left(x-1\right)}{x-4}\)
\(=\dfrac{\left(x-1\right)\left(x-4\right)}{x-4}=x-1\)
\(\Rightarrow VT=VP=x-1\left(dpcm\right)\)
c) Ta có:
\(VT=\dfrac{1-x}{-5x+1}\)
\(=\dfrac{-x+1}{-\left(5x-1\right)}\)
\(=\dfrac{-\left(x-1\right)}{-\left(5x-1\right)}\)
\(=\dfrac{x-1}{5x-1}=VP\left(dpcm\right)\)
d) Ta có:
\(VT=\dfrac{x^3+27}{x^2-3x+9}\)
\(=\dfrac{x^3+3^3}{x^2-3x+9}\)
\(=\dfrac{\left(x+3\right)\left(x^2-3x+9\right)}{x^2-3x+9}\)
\(=x+3=VP\left(dpcm\right)\)
