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NT
15 tháng 11 2023 lúc 21:09

7:

\(A=\sqrt{6x}+\sqrt{\dfrac{6}{x}}-\sqrt{\dfrac{2x}{3}}\)

\(=\sqrt{6x}-\sqrt{\dfrac{6x}{9}}+\sqrt{\dfrac{6x}{x^2}}\)

\(=\sqrt{6x}-\dfrac{1}{3}\sqrt{6x}+\dfrac{1}{x}\cdot\sqrt{6x}\)

\(=\sqrt{6x}\left(1-\dfrac{1}{3}+\dfrac{1}{x}\right)=\sqrt{6x}\left(\dfrac{2}{3}+\dfrac{1}{x}\right)\)

8: \(5\sqrt{a}-\sqrt{64a}+2\sqrt{9a}\)

\(=5\sqrt{a}-8\sqrt{a}+2\cdot3\sqrt{a}\)

\(=-3\sqrt{a}+6\sqrt{a}=3\sqrt{a}\)

9: \(\sqrt{\dfrac{a}{b}}+\sqrt{ab}+\dfrac{a}{b}\sqrt{\dfrac{b}{a}}\)

\(=\sqrt{\dfrac{ab}{b^2}}+\sqrt{ab}+\dfrac{a}{\sqrt{a}}\cdot\dfrac{\sqrt{b}}{b}\)

\(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}+\dfrac{\sqrt{a}}{\sqrt{b}}\)

\(=\dfrac{\sqrt{ab}}{b}+\sqrt{ab}+\dfrac{\sqrt{ab}}{b}\)

\(=\sqrt{ab}\left(\dfrac{2}{b}+1\right)\)

10: \(A=5\sqrt{a}+6\sqrt{\dfrac{a}{4}}-a\sqrt{\dfrac{4}{a}}+\sqrt{5}\)

\(=5\sqrt{a}+6\cdot\dfrac{\sqrt{a}}{2}-a\cdot\dfrac{2}{\sqrt{a}}+\sqrt{5}\)

\(=5\sqrt{a}+3\sqrt{a}-2\sqrt{a}+\sqrt{5}\)

\(=6\sqrt{a}+\sqrt{5}\)

11: \(A=-\sqrt{36a}-\dfrac{1}{3}\cdot\sqrt{54a}+\dfrac{1}{5}\cdot\sqrt{150a}\)

\(=-6\sqrt{a}-\dfrac{1}{3}\cdot3\sqrt{6a}+\dfrac{1}{5}\cdot5\sqrt{6a}\)

\(=-6\sqrt{a}-\sqrt{6a}+\sqrt{6a}\)

\(=-6\sqrt{a}\)

12: \(A=5\sqrt{2a}-\sqrt{50a}-2\sqrt{a^3}+4\sqrt{32a}\)

\(=5\sqrt{2a}-5\sqrt{2a}-2a\sqrt{a}+4\cdot4\sqrt{2a}\)

\(=-2a\sqrt{a}+16\sqrt{2a}\)

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