1: Ta có: \(\sqrt{7-3\sqrt{5}}+\sqrt{7+3\sqrt{5}}\)

\(=\frac{\sqrt{14-6\sqrt{5}}+\sqrt{14+6\sqrt{5}}}{\sqrt{2}}\)

\(=\frac{\sqrt{9-2\cdot3\cdot\sqrt{5}+5}+\sqrt{9+2\cdot3\cdot\sqrt{5}+5}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(3-\sqrt{5}\right)^2}+\sqrt{\left(3+\sqrt{5}\right)^2}}{\sqrt{2}}\)

\(=\frac{\left|3-\sqrt{5}\right|+\left|3+\sqrt{5}\right|}{\sqrt{2}}\)

\(=\frac{3-\sqrt{5}+3+\sqrt{5}}{\sqrt{2}}\)(Vì \(3>\sqrt{5}>0\))

\(=\frac{6}{\sqrt{2}}=\sqrt{18}=3\sqrt{2}\)

2) Ta có: \(\sqrt{6-\sqrt{35}}+\sqrt{6+\sqrt{35}}\)

\(=\frac{\sqrt{12-2\sqrt{35}}+\sqrt{12+2\sqrt{35}}}{\sqrt{2}}\)

\(=\frac{\sqrt{7-2\cdot\sqrt{7}\cdot\sqrt{5}+5}+\sqrt{7+2\cdot\sqrt{7}\cdot\sqrt{5}+5}}{\sqrt{2}}\)

\(=\frac{\sqrt{\left(\sqrt{7}-\sqrt{5}\right)^2}+\sqrt{\left(\sqrt{7}+\sqrt{5}\right)^2}}{\sqrt{2}}\)

\(=\frac{\left|\sqrt{7}-\sqrt{5}\right|+\left|\sqrt{7}+\sqrt{5}\right|}{\sqrt{2}}\)

\(=\frac{\sqrt{7}-\sqrt{5}+\sqrt{7}+\sqrt{5}}{\sqrt{2}}\)(Vì \(\sqrt{7}>\sqrt{5}>0\))

\(=\frac{2\sqrt{7}}{\sqrt{2}}=\sqrt{14}\)

căn 5+căn 7-căn 5+căn 7- căn 7

= - căn 7

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\(B=50-3\sqrt{98}+2\sqrt{8}+3\sqrt{32}-5\sqrt{18}\)

\(=50-3.\sqrt{7^2.2}+2\sqrt{2^2.2}+3\sqrt{4^2.2}-5\sqrt{3^2.2}\)

\(=50-3.7\sqrt{2}+2.2\sqrt{2}+3.4\sqrt{2}-5.3\sqrt{2}\)

\(=50-21\sqrt{2}+4\sqrt{2}+12\sqrt{2}-15\sqrt{2}\)

\(=50+\sqrt{2}.\left(-21+4+12-15\right)\)

\(=50+\sqrt{2}.\left(-20\right)\)

\(=50-20\sqrt{2}\)

\(C=\left(\sqrt{3}+\sqrt{5}+\sqrt{7}\right)\left(\sqrt{3}+\sqrt{5}-\sqrt{7}\right)\)

\(=\left(\sqrt{3}+\sqrt{5}\right)^2-\sqrt{7}^2\)

\(=\sqrt{3}^2+2.\sqrt{3}.\sqrt{5}+\sqrt{5}^2-7\)

\(=2\sqrt{15}+3+5-7\)

\(=2\sqrt{15}+1\)

giúp mình với ạ :<<

\(\left(\sqrt{7}+\sqrt{5}\right)\cdot\left(\sqrt{7}+\sqrt{5}\right)\)

\(=\left(\sqrt{7}+\sqrt{5}\right)^2\)

\(=7+2\sqrt{35}+5\)

\(=12+2\sqrt{35}\)

\(\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}+\sqrt{5}\right)\)

\(=\left(\sqrt{7}+\sqrt{5}\right)^2\)

\(=12+2\sqrt{35}\)

bn ghi đề rõ hơn ik bn

a) Ta có: \(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{12}{3-\sqrt{3}}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{1}+\dfrac{12\left(3+\sqrt{3}\right)}{6}\)

\(=\sqrt{3}+1-6-3\sqrt{3}+6+2\sqrt{3}\)

\(=1\)

b) Ta có: \(\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\sqrt{7}+\sqrt{5}}-\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{2}+\sqrt{7}-\sqrt{3}\)

=0

\(\left(5\sqrt{3}-2\sqrt{7}\right)\left(5\sqrt{3}+2\sqrt{7}\right)\)

\(=\left(5\sqrt{3}\right)^2-\left(2\sqrt{7}\right)^2\)

\(=25\cdot3-4\cdot7\)

\(=75-28\)

\(=47\)

a) Ta có: \(\dfrac{2}{\sqrt{3}-1}+\dfrac{3}{\sqrt{3}-2}+\dfrac{12}{3-\sqrt{3}}\)

\(=\dfrac{2\left(\sqrt{3}+1\right)}{2}-\dfrac{3\left(2+\sqrt{3}\right)}{1}+\dfrac{12\left(3+\sqrt{3}\right)}{6}\)

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

\(=-2\sqrt{3}-5+6+2\sqrt{3}\)

=1

b) Ta có: \(\dfrac{1}{\sqrt{3}-\sqrt{2}}-\dfrac{2}{\sqrt{7}+\sqrt{5}}-\dfrac{3}{\sqrt{5}-\sqrt{2}}+\dfrac{4}{\sqrt{7}+\sqrt{3}}\)

\(=\sqrt{3}+\sqrt{2}-\sqrt{7}+\sqrt{5}-\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{3}\)

\(=\sqrt{2}-\sqrt{3}\)