Question

Help me! Giải nhanh giúp mình với :

Answer 1

\(a,A=\dfrac{x^2-x}{x-1}.ĐKXĐ:x\ne1.\\ B=\dfrac{x^2-4}{x^2+2x}.ĐKXĐ:x\ne0;-2.\)

\(b,A=\dfrac{x^2-x}{x-1}.\\ A=\dfrac{x\left(x-1\right)}{x-1}=x.\\ B=\dfrac{x^2-4}{x^2+2x}.\\ B=\dfrac{\left(x-2\right)\left(x+2\right)}{x\left(x+2\right)}.\\ B=\dfrac{x-2}{x}.\)

\(c,A-B=x-\dfrac{x-2}{x}.\\ A-B=\dfrac{x^2-x+2}{x}.\\ A-B=\dfrac{x^2-2x+1+x+1}{x}.\\ A-B=\dfrac{\left(x-1\right)^2+\left(x+1\right)}{x}.\)

Ta có: 

\(x>0.\\ \left(x-1\right)^2\ge0\forall x\in R.\\ x+1>0\left(x>0\right).\\ \Rightarrow A-B>0.\\ \Rightarrow A>B.\)

Answer 2