Question

Đặt \(\left\{{}\begin{matrix}a=\sqrt{2}\\b=\sqrt[3]{2}\end{matrix}\right.\). Chứng minh: \(\dfrac{1}{a-b}-\dfrac{1}{b}=a+b+\dfrac{a}{b}+\dfrac{b}{a}+1\)

Answer 1

https://i.imgur.com/z3wJcoK.gif

Answer 2

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

\(\Leftrightarrow1=\left(a-b\right)\left(a+b+\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{1}{b}+1\right)\circledast\)

VP của \(\circledast\Leftrightarrow a^2+ab+\dfrac{a^2}{b}+b+\dfrac{a}{b}+a-ab-b^2-a-\dfrac{b^2}{a}-1-b=a^2-b^2+\dfrac{a^2}{b}-\dfrac{b^2}{a}+\dfrac{a}{b}-1\)

Do : \(a^2=2;b^3=2;\dfrac{a^2}{b}=\dfrac{2}{b}=b^2;\dfrac{a}{b}=\dfrac{b^2}{a}\)

\(\Rightarrow2-\dfrac{2}{b}+\dfrac{2}{b}-\dfrac{a}{b}+\dfrac{a}{b}-1=1=VT\)

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