Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\frac{\sqrt{2}-1}{\left(\sqrt{1}+\sqrt{2}\right)\left(\sqrt{2}-1\right)}+\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{3}-\sqrt{2}\right)}+...+\frac{\sqrt{n}-\sqrt{n-1}}{\left(\sqrt{n-1}+\sqrt{n}\right)\left(\sqrt{n}-\sqrt{n-1}\right)}\)\(\Leftrightarrow A=\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{n}-\sqrt{n-1}}{n-\left(n-1\right)}\)
\(\Leftrightarrow A=\sqrt{n}-1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{1}{\sqrt{x}\left(x\sqrt{x}-1\right)}:\frac{\sqrt{x}+1}{\sqrt{x}\left(x+\sqrt{x}+1\right)}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x^3}-1\right)}.\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}.\frac{\sqrt{x}\left(x+\sqrt{x}+1\right)}{\sqrt{x}+1}\)
\(=\frac{1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{1}{x-1}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{\sqrt{640}\sqrt{34,3}}{\sqrt{567}}\)
\(= \frac{\sqrt{64.10}\sqrt{49.\frac{7}{10}}}{\sqrt{81.7}}\)
\(= \frac{\sqrt{64}\sqrt{10}\sqrt{49}\sqrt{\frac{7}{10}}}{\sqrt{81}\sqrt{7}}\)
\(= \frac{\sqrt{64}\sqrt{49}}{\sqrt{81}} . \frac{\sqrt{10}\sqrt{\frac{7}{10}}}{\sqrt{7}}\)
\(= \frac{8.7}{9} . \frac{\sqrt{10 . \frac{7}{10}}}{\sqrt{7}}\)
\(= \frac{56}{9} . \frac{\sqrt{7}}{\sqrt{7}}\)
\(= \frac{56}{9} . 1 = \frac{56}{9}\)
b) \(\sqrt{21,6}\sqrt{810}\sqrt{11^2−5^2}\)
\(= \sqrt{216.\frac{1}{10}}\sqrt{81.10}\sqrt{(11−5)(11+5)}\)
\(= \sqrt{36.6.\frac{1}{10}}\sqrt{81}\sqrt{10}\sqrt{6.16}\)
\(= \sqrt{36}\sqrt{6}\sqrt{\frac{1}{10}}\sqrt{81}\sqrt{10}\sqrt{6}\sqrt{16}\)
\(= (\sqrt{36}\sqrt{81}\sqrt{16}).(\sqrt{6}\sqrt{6}).(\sqrt{\frac{1}{10}}\sqrt{10})\)
\(= (6.9.4).\sqrt{6.6}.\sqrt{\frac{1}{10}.10}\)
\(= (54.4).\sqrt{36}.\sqrt{1}\)
\(= 216.6.1 = 1296\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Tớ làm nốt nè :3
\(1b.3\sqrt{2}+4\sqrt{8}-\sqrt{18}=3\sqrt{2}+8\sqrt{2}-3\sqrt{2}=8\sqrt{2}\)
\(c.\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{2-\sqrt{3}}=\dfrac{2-\sqrt{3}+2+\sqrt{3}}{\left(2-\sqrt{3}\right)\left(2+\sqrt{3}\right)}=4\)
\(2a.\sqrt{4x^2-4x+1}=3\)
\(\Leftrightarrow4x^2-4x+1=9\)
\(\Leftrightarrow4x^2+4x-8x-8=0\)
\(\Leftrightarrow4\left(x+1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
\(b.\sqrt{4x-4}-\sqrt{9x-9}+5\sqrt{x-1}=7\left(x\ge1\right)\)
\(\Leftrightarrow2\sqrt{x-1}-3\sqrt{x-1}+5\sqrt{x-1}=7\)
\(\Leftrightarrow4\sqrt{x-1}=7\)
\(\Leftrightarrow\sqrt{x-1}=\dfrac{7}{4}\)
\(\Leftrightarrow x=\dfrac{65}{16}\)
c. Sai đề.
\(5\sqrt{12a}+\sqrt{49a}-\sqrt{64a}\) = \(10\sqrt{3a}+7a-8a\) = \(10\sqrt{3a}-a\)