Question

Đặt \(a=\sqrt{2};b=\sqrt[3]{2}\) . CM: \(\dfrac{1}{a-b}-\dfrac{1}{b}=a+b+\dfrac{a}{b}+\dfrac{b}{a}+1\) 

Answer 1

Ta cần CM:

\(a+b+\dfrac{a}{b}+\dfrac{b}{a}+1+\dfrac{1}{b}-\dfrac{1}{a-b}=0\)

Vậy ta xét: \(a+b+\dfrac{a}{b}+\dfrac{b}{a}+1+\dfrac{1}{b}-\dfrac{1}{a-b}\) 

\(=\dfrac{ab\left(a^2-b^2\right)}{ab\left(a-b\right)}+\dfrac{a\left(a+1\right)\left(a-b\right)}{ab\left(a-b\right)}+\dfrac{b^2\left(a-b\right)}{ab\left(a-b\right)}+\dfrac{ab\left(a-b-1\right)}{ab\left(a-b\right)}\) 

\(=\dfrac{a^3b-ab^3+a^3-a^2b+a^2-ab+ab^2-b^3+a^2b-ab^2-ab}{ab\left(a-b\right)}=0\) 

\(\Rightarrow ab\left(a^2-2\right)+a\left(a^2-b^3\right)+\left(a^2-b^3\right)=0\) (Vì \(ab\left(a-b\right)\ne0\)) 

Đúng vì khi thay \(a=\sqrt{2};b=\sqrt[3]{2}\) , ta đc \(VT=0\) . Vậy ta có điều phải CM.