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a, \(=>3-\sqrt{2}+\sqrt{50}=3-\sqrt{2}+5\sqrt{2}=3+4\sqrt{2}\)
b, \(=>\dfrac{\sqrt[3]{125.5}}{\sqrt[3]{5}}-\sqrt[3]{\left(-4\right).2}=\sqrt[3]{125}-\sqrt[3]{\left(-2\right)^3}\)
\(=5-\left(-2\right)=7\)
c, \(=>\sqrt{6}.\sqrt{\dfrac{6}{2}}-\sqrt{2}-3\sqrt{4.2}=\sqrt{6}.\sqrt{3}-\sqrt{2}-6\sqrt{2}\)
\(=\sqrt{18}-7\sqrt{2}=3\sqrt{2}-7\sqrt{2}=-4\sqrt{2}\)
d, \(=>\dfrac{\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-\dfrac{2}{\sqrt{3}-1}=\sqrt{3}-\dfrac{2}{\sqrt{3}-1}\)
\(=\dfrac{3-\sqrt{3}-2}{\sqrt{3}-1}=\dfrac{1-\sqrt{3}}{\sqrt{3}-1}=-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\sqrt{72}-5\sqrt{2}+3\sqrt{12}\\ =6\sqrt{2}-5\sqrt{2}+6\sqrt{3}\\ =\sqrt{2}+6\sqrt{3}\\ b.6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\\ =3\sqrt{2}-\sqrt{2}-5\sqrt{2}\\ =-3\sqrt{2}\\ c.\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\\ =2+1+\sqrt{3}-\sqrt{3}\\ =3\\ d.\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\\ =4+3+4\\ =11\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(A=6\sqrt{3}+10\sqrt{3}-12\sqrt{3}=4\sqrt{3}\)
b: \(B=7\sqrt{3}+5\sqrt{3}-12\sqrt{3}=0\)
c: \(=12\sqrt{2}-6+3\left(9-4\sqrt{2}\right)=12\sqrt{2}-6+27-12\sqrt{2}=21\)
d: \(=2\sqrt{5}-5\sqrt{5}-4\sqrt{5}+11\sqrt{5}=4\sqrt{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}\)
\(=\sqrt{2}\left(3-12+8-5\right)=-6\sqrt{2}\)
\(b,=\left|\sqrt{2}-\sqrt{3}\right|+3\sqrt{2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}=\sqrt{3}+2\sqrt{2}\)
\(c,=\sqrt{5}+\sqrt{5}+\dfrac{5}{\sqrt{5}}-1=3\sqrt{5}-1\)
\(d,=\sqrt{3-2.2\sqrt{3}+4}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(1+\sqrt{3}\right)^2}\)
\(=2-\sqrt{3}+1+\sqrt{3}=2\)
a) \(3\sqrt{2}-4\sqrt{18}+2\sqrt{32}-\sqrt{50}=3\sqrt{2}-4\sqrt{9.2}+2\sqrt{16.2}-\sqrt{25.2}\)
\(=3\sqrt{2}-12\sqrt{2}+8\sqrt{2}-5\sqrt{2}=-6\sqrt{2}\)
b) \(\sqrt{\left(\sqrt{2}-\sqrt{3}\right)^2}+\sqrt{18}=\left|\sqrt{2}-\sqrt{3}\right|+\sqrt{9.2}=\sqrt{3}-\sqrt{2}+3\sqrt{2}\)
\(=2\sqrt{2}+\sqrt{3}\)
c) \(5\sqrt{\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{45}+\dfrac{5-\sqrt{5}}{\sqrt{5}}=\sqrt{25.\dfrac{1}{5}}+\dfrac{1}{3}\sqrt{9.5}+\dfrac{\sqrt{5}\left(\sqrt{5}-1\right)}{\sqrt{5}}\)
\(=\sqrt{5}+\sqrt{5}+\sqrt{5}-1=3\sqrt{5}-1\)
d) \(\sqrt{7-4\sqrt{3}}+\sqrt{\left(1+\sqrt{3}\right)^2}=\sqrt{2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2}+\left|\sqrt{3}+1\right|\)
\(=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{3}+1=\left|2-\sqrt{3}\right|+\sqrt{3}+1=2-\sqrt{3}+\sqrt{3}+1=3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a. \(\sqrt{48}-2\sqrt{32}-\sqrt{75}+3\sqrt{50}\) = \(4\sqrt{3}-2.4\sqrt{2}-5\sqrt{3}+3.5\sqrt{2}\)
= \(4\sqrt{3}-8\sqrt{2}-5\sqrt{3}+15\sqrt{2}\) = \(-\sqrt{3}+7\sqrt{2}\)
b. \(\sqrt{20}-15\sqrt{\dfrac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}\) = \(2\sqrt{5}-3.5.\sqrt{\dfrac{1}{5}}+\left|1-\sqrt{5}\right|\)
= \(2\sqrt{5}-3\sqrt{25.\dfrac{1}{5}}+\sqrt{5}-1\) = \(2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1\) = \(-1\)
c. \(\dfrac{3}{3+2\sqrt{3}}+\dfrac{3}{3-2\sqrt{3}}\) = \(\dfrac{3\left(3-2\sqrt{3}\right)+3\left(3+2\sqrt{3}\right)}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\)
= \(\dfrac{9-6\sqrt{3}+9+6\sqrt{3}}{\left(3+2\sqrt{3}\right)\left(3-2\sqrt{3}\right)}\) = \(\dfrac{18}{9-12}=\dfrac{18}{-3}=-6\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a, Ta có : \(A=\sqrt{2-\sqrt{3}}-\sqrt{2+\sqrt{3}}\)
\(\Rightarrow A^2=2-\sqrt{3}+2+\sqrt{3}-2\sqrt{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}\)
\(=4-2\sqrt{4-3}=4-2=2\)
\(\Rightarrow A=-\sqrt{2}\)
b, Ta có : \(B=\sqrt{3+\sqrt{5}}+\sqrt{7-3\sqrt{5}}-\sqrt{2}\)
\(\Rightarrow B\sqrt{2}=\sqrt{6+2\sqrt{5}}+\sqrt{14-6\sqrt{5}}-2\)
\(=\sqrt{5+2\sqrt{5}+1}+\sqrt{9-2.3\sqrt{5}+5}-2\)
\(=\sqrt{5}+1+3-\sqrt{5}-2=2\)
\(\Rightarrow B=\sqrt{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{6-2}+\dfrac{3.\left(\sqrt{6}-\sqrt{5}\right)}{6-5}\\ =\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}+3\left(\sqrt{6}-\sqrt{5}\right)\\ =\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}+3\sqrt{6}-3\sqrt{5}\\ =4\sqrt{6}-2\sqrt{5}\)
\(b,=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{5-2}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}-\dfrac{\sqrt{2}.\sqrt{2}}{\sqrt{2}\sqrt{4+\sqrt{15}}}\\ =\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}-\dfrac{1}{\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}}-\dfrac{2}{\sqrt{8+2.\sqrt{3}.\sqrt{5}}}\\ =\sqrt{5}+\sqrt{2}-\dfrac{1}{\left|\sqrt{3}-\sqrt{2}\right|}-\dfrac{2}{\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}}\\ =\sqrt{5}+\sqrt{2}-\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\left|\sqrt{5}+\sqrt{3}\right|}\)
\(=\sqrt{5}+\sqrt{2}-\dfrac{\sqrt{3}+\sqrt{2}}{3-2}-\dfrac{2.\left(\sqrt{5}-\sqrt{3}\right)}{5-3}\\ =\sqrt{5}+\sqrt{2}-\sqrt{3}-\sqrt{2}-\dfrac{2.\left(\sqrt{5}-\sqrt{3}\right)}{2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{3}-\sqrt{2}-\sqrt{5}+\sqrt{3}\\ =0\)
a: \(=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}+\dfrac{4\left(\sqrt{6}-\sqrt{2}\right)}{4}+\dfrac{3\left(\sqrt{6}-\sqrt{5}\right)}{1}\)
\(=\sqrt{5}+\sqrt{2}+\sqrt{6}-\sqrt{2}+3\sqrt{6}-3\sqrt{5}\)
\(=-2\sqrt{5}+4\sqrt{6}\)
b: \(=\dfrac{3\left(\sqrt{5}+\sqrt{2}\right)}{3}-\dfrac{1}{\sqrt{5-2\sqrt{6}}}+\dfrac{2}{\sqrt{8+2\sqrt{15}}}\)
\(=\sqrt{5}+\sqrt{2}-\dfrac{1}{\sqrt{3}-\sqrt{2}}+\dfrac{2}{\sqrt{5}+\sqrt{3}}\)
\(=\sqrt{5}+\sqrt{2}+\sqrt{5}-\sqrt{3}-\sqrt{3}-\sqrt{2}\)
=2căn 5-2căn 3
\(\sqrt{m+2\sqrt{m-1}}+\sqrt{m-2\sqrt{m-1}}\)
=\(\sqrt{\left(\sqrt{m-1}\right)^2+2\sqrt{m-1}+1}+\sqrt{\left(\sqrt{m-1}\right)^2-2\sqrt{m-1}+1}\)
=\(\sqrt{\left(\sqrt{m-1}+1\right)^2}+\sqrt{\left(\sqrt{m-1}-1\right)^2}\)
=\(\sqrt{m-1}+1+\sqrt{m-1}-1=2\sqrt{m-1}\)