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