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3 tháng 8 2020

Ta có : \(\frac{\frac{\left(a-b\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}-b\sqrt{b}+2a\sqrt{a}}{a\sqrt{a}-b\sqrt{b}}+\frac{3a+3\sqrt{ab}}{b-a}\)

\(=\frac{\frac{\left(\sqrt{a}-\sqrt{b}\right)^3\left(\sqrt{a}+\sqrt{b}\right)^3}{\left(\sqrt{a}-\sqrt{b}\right)^3}+2a\sqrt{a}-b\sqrt{b}}{\sqrt{a}^3-\sqrt{b}^3}+\frac{3\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{-\left(a-b\right)}\)

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

\(=\frac{a\sqrt{a}+3a\sqrt{b}+3b\sqrt{a}+b\sqrt{b}+2a\sqrt{a}-b\sqrt{b}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=\frac{3a\sqrt{b}+3\sqrt{a}b+3a\sqrt{a}}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)\(=\frac{3\sqrt{a}\left(\sqrt{ab}+b+a\right)}{\left(\sqrt{a}-\sqrt{b}\right)\left(a+\sqrt{ab}+b\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}\)

\(=-\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}+\frac{3\sqrt{a}}{-\left(\sqrt{a}-\sqrt{b}\right)}=0\)

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