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a: Ta có: \(4\sqrt{3a}-3\sqrt{12a}+\dfrac{6\sqrt{a}}{3}-2\sqrt{20a}\)
\(=4\sqrt{3a}-6\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)
\(=-2\sqrt{3a}+2\sqrt{2a}-4\sqrt{5a}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(15\sqrt{\dfrac{4}{3}}-5\sqrt{48}+2\sqrt{12}-6\sqrt{\dfrac{1}{3}}\)
\(=\sqrt{15^2\cdot\dfrac{4}{3}}-5\cdot4\sqrt{3}+2\cdot2\sqrt{3}-\sqrt{6^2\cdot\dfrac{1}{3}}\)
\(=\sqrt{\dfrac{225\cdot4}{3}}-20\sqrt{3}+4\sqrt{3}-\sqrt{\dfrac{36}{3}}\)
\(=\sqrt{75\cdot4}-16\sqrt{3}-\sqrt{12}\)
\(=10\sqrt{3}-16\sqrt{3}-2\sqrt{3}\)
\(=-8\sqrt{3}\)
b) \(\dfrac{15}{\sqrt{6}+1}-\dfrac{3}{\sqrt{7}-\sqrt{2}}-15\sqrt{6}+3\sqrt{7}\)
\(=\dfrac{15\left(\sqrt{6}-1\right)}{\left(\sqrt{6}+1\right)\left(\sqrt{6}-1\right)}-\dfrac{3\left(\sqrt{7}+\sqrt{2}\right)}{\left(\sqrt{7}-\sqrt{2}\right)\left(\sqrt{7}+\sqrt{2}\right)}-15\sqrt{6}+3\sqrt{7}\)
\(=\dfrac{15\left(\sqrt{6}-1\right)}{6-1}-\dfrac{3\sqrt{7}+3\sqrt{2}}{7-2}-15\sqrt{6}+3\sqrt{7}\)
\(=3\left(\sqrt{6}-1\right)-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)
\(=3\sqrt{6}-3-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}-15\sqrt{6}+3\sqrt{7}\)
\(=-12\sqrt{6}-3+3\sqrt{7}-\dfrac{3\sqrt{7}+3\sqrt{2}}{5}\)
\(=\dfrac{-60\sqrt{6}-15+15\sqrt{7}-3\sqrt{7}-3\sqrt{2}}{5}\)
\(=\dfrac{-60\sqrt{6}-15+12\sqrt{7}-3\sqrt{2}}{5}\)
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\(A=\sqrt{x}+1\) (đã thu gọn)
\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)
\(A=x-\sqrt{x}+1=\sqrt{x}\cdot\sqrt{x}-\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)+1\)
\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)
\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)
\(A=1-\sqrt{x}\) (đã thu gọn)
\(A=x-2\sqrt{x}-1=\sqrt{x}\left(\sqrt{x}-2\right)-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{4}+\sqrt{4}+\sqrt{6}+\sqrt{8}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+2\right)}{2+\sqrt{2}+\sqrt{3}}\)
=1+căn 2
\(\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{16}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(\sqrt{2}+\sqrt{4}\right)+\left(\sqrt{6}+\sqrt{3}\right)+\left(\sqrt{4}+\sqrt{8}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\sqrt{2}\left(1+\sqrt{2}\right)+\sqrt{3}\left(1+\sqrt{2}\right)+\sqrt{4}\left(1+\sqrt{2}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=\dfrac{\left(1+\sqrt{2}\right)\left(\sqrt{2}+\sqrt{3}+\sqrt{4}\right)}{\sqrt{2}+\sqrt{3}+\sqrt{4}}\)
\(=1+\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{\dfrac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}}+\sqrt{\dfrac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}}\)
\(=\sqrt{\dfrac{\left(\sqrt{3}-\sqrt{2}\right)^2}{3-2}}+\sqrt{\dfrac{\left(\sqrt{3}+\sqrt{2}\right)^2}{3-2}}\)
\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}\)
\(=\sqrt{3}-\sqrt{2}+\sqrt{3}+\sqrt{2}=2\sqrt{3}\)
\(3333333\hept{\begin{cases}\\\end{cases}}\hept{\begin{cases}\\\end{cases}}3\)