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a) Ta có: \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=\left(-\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}\)
\(=-2+2\sqrt{5}-\sqrt{5}\)
\(=-2+\sqrt{5}\)
b) \(\left(\frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right)\div\frac{1}{8}\)
\(=\left(\frac{\sqrt{2}}{4}-\frac{3\sqrt{2}}{2}+8\sqrt{2}\right)\cdot8\)
\(=\frac{27\sqrt{2}}{4}\cdot8\)
\(=54\sqrt{2}\)
a, \(\frac{1}{\left(\sqrt{3}+\sqrt{2}\right)^2}\) +\(\frac{1}{\left(\sqrt{3}-\sqrt{2}\right)^2}\) =\(\frac{\left(\sqrt{3}+\sqrt{2}\right)^2+\left(\sqrt{3}-\sqrt{2}\right)^2}{\left(\sqrt{3}+\sqrt{2}\right)^2\left(\sqrt{3}-\sqrt{2}\right)^2}\)
\(=\frac{10}{1}=10\)
mấy câu còn lại bạn tự làm nốt nhé mk ban rồi
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
Mk năm nay mới lên lớp 9 thôi nhưng cũng biết chút!Mk giải ho bạn câu 1 còn lại bạn tự giải nhé!
1,\(\frac{1}{1+\sqrt{5}}\)+\(\frac{1}{\sqrt{5}-1}\)
=\(\frac{1}{\sqrt{5}+1}\)+\(\frac{1}{\sqrt{5}-1}\)
=\(\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)+\(\frac{\sqrt{5}+1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}\)
=\(\frac{\sqrt{5}-1}{5-1}\)+\(\frac{\sqrt{5}+1}{5-1}\)
=\(\frac{\sqrt{5}-1}{4}\)+\(\frac{\sqrt{5}+1}{4}\)
=\(\frac{\sqrt{5}-1+\sqrt{5}+1}{4}\)
=\(\frac{2\sqrt{5}}{4}\)
=\(\frac{\sqrt{5}}{2}\)
\( \begin{align} & 1)\dfrac{1}{\sqrt{5}-1}-\dfrac{1}{\sqrt{5}+1}=\dfrac{\sqrt{5}+1}{{{\left( \sqrt{5} \right)}^{2}}-1}-\dfrac{\sqrt{5}-1}{{{\left( \sqrt{5} \right)}^{2}}-1}=\dfrac{\sqrt{5}+1-\sqrt{5}+1}{4}=\dfrac{1}{2} \\ & 2)\dfrac{1}{\sqrt{5}-2}+\dfrac{1}{\sqrt{5}+2}=\sqrt{5}+2+\sqrt{5}-2=2\sqrt{5} \\ & 3)\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left( \sqrt{2}+1 \right)}{1+\sqrt{2}}=\sqrt{2} \\ & 4)\dfrac{2}{4-3\sqrt{2}}-\dfrac{2}{4+3\sqrt{2}}=-4-3\sqrt{2}+4-3\sqrt{2}=-6\sqrt{2} \\ \end{align} \)
Ta có:
\(\sqrt{\frac{1}{1^2}+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{\left(1+n+n^2\right)^2}{n^2\left(n+1\right)^2}}\)
\(=\frac{1+n+n^2}{n\left(n+1\right)}=1+\frac{1}{n}-\frac{1}{n+1}\)
Áp dụng bài toán được
\(A=\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{2021^2}+\frac{1}{2022^2}}\)
\(=1+\frac{1}{2}-\frac{1}{3}+1+\frac{1}{3}+\frac{1}{4}+...+1+\frac{1}{2021}-\frac{1}{2022}\)
\(=2020+\frac{1}{2}-\frac{1}{2022}=\)