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\(\frac{5\sqrt{7}-7\sqrt{5}+2\sqrt{70}}{\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{\sqrt{35}.\sqrt{35}}\)
\(=\frac{\sqrt{35}.(5\sqrt{7}-7\sqrt{5}+2\sqrt{70})}{35}\)
\(\sqrt{\frac{4}{3}}+\sqrt{12}-\frac{4}{3}\sqrt{\frac{3}{4}}\)
\(=\frac{\sqrt{4}}{\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{\sqrt{4}}\)
\(=\frac{2\sqrt{3}}{\sqrt{3}.\sqrt{3}}+\sqrt{12}-\frac{4}{3}\cdot\frac{\sqrt{3}}{2}\)
\(=\frac{2\sqrt{3}}{3}+2\sqrt{3}-\frac{2\sqrt{3}}{3}\)
\(=2\sqrt{3}\left(\frac{1}{3}+1-\frac{1}{3}\right)\)
\(=2\sqrt{3}\)
Đặt \(a=\sqrt[4]{5}\Leftrightarrow5=a^4\)
Ta cần chứng minh: \(\left(\frac{a+1}{a-1}\right)^4=\frac{3+2a}{3-2a}\)
Khai triển: \(VT=\left(\frac{a+1}{a-1}\right)^4=\frac{\left(a+1\right)^4}{\left(a-1\right)^4}\)
\(=\frac{2\left(3+2a\right).\left(1+a^2\right)}{2\left(3-2a\right).\left(1+a^2\right)}\)
\(\frac{3+2a}{3-2a}=VP\)(đpcm)
a) \(\sqrt{17}-4\) b) \(\sqrt{3}\) c) \(\frac{\sqrt{2}}{2}\) d)\(\frac{\sqrt{x}+1}{\sqrt{x}-1}\) e) \(x-\sqrt{5}\)
f) \(4+2\sqrt{3}\) g) \(3+2\sqrt{2}\) h) \(x+\sqrt{x}+1\) i) \(\frac{3\sqrt{5}-\sqrt{15}}{10}\)
k) \(\sqrt{5}+\sqrt{6}\) i) 5 h) 0 l) \(\sqrt{5}+\sqrt{3}\) m) \(\frac{20\sqrt{3}}{3}\) d) 0