![](https://rs.olm.vn/images/avt/0.png?1311)
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)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(=3\sqrt{3}-2\sqrt{3}+4\sqrt{3}-5\sqrt{3}=2\sqrt{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a) \sqrt{4x^2− 9} = 2\sqrt{x + 3}\)
\(ĐK:x\ge\dfrac{3}{2}\)
\(pt\Leftrightarrow4x^2-9=4\left(x+3\right)\)
\(\Leftrightarrow4x^2-9=4x+12\)
\(\Leftrightarrow4x^2-4x-21=0\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1-\sqrt{22}}{2}\left(l\right)\\x=\dfrac{1+\sqrt{22}}{2}\left(tm\right)\end{matrix}\right.\)
\(b)\sqrt{4x-20}+3.\sqrt{\dfrac{x-5}{9}}-\dfrac{1}{3}\sqrt{9x-45}=4\)
\(ĐK:x\ge5\)
\(pt\Leftrightarrow2\sqrt{x-5}+\sqrt{x-5}-\sqrt{x-5}=4\)
\(\Leftrightarrow2\sqrt{x-5}=4\Leftrightarrow\sqrt{x-5}=2\)
\(\Leftrightarrow x-5=4\Leftrightarrow x=9\left(tm\right)\)
\(c)\dfrac{2}{3}\sqrt{9x-9}-\dfrac{1}{4}\sqrt{16x-16}+27.\sqrt{\dfrac{x-1}{81}}=4\)
ĐK:x>=1
\(pt\Leftrightarrow2\sqrt{x-1}-\sqrt{x-1}+3\sqrt{x-1}=4\)
\(\Leftrightarrow4\sqrt{x-1}=4\Leftrightarrow\sqrt{x-1}=1\)
\(\Leftrightarrow x-1=1\Leftrightarrow x=2\left(tm\right)\)
\(d)5\sqrt{\dfrac{9x-27}{25}}-7\sqrt{\dfrac{4x-12}{9}}-7\sqrt{x^2-9}+18\sqrt{\dfrac{9x^2-81}{81}}=0\)
\(ĐK:x\ge3\)
\(pt\Leftrightarrow3\sqrt{x-3}-\dfrac{14}{3}\sqrt{x-3}-7\sqrt{x^2-9}+6\sqrt{x^2-9}=0\)
\(\Leftrightarrow-\dfrac{5}{3}\sqrt{x-3}-\sqrt{x^2-9}=0\Leftrightarrow\dfrac{5}{3}\sqrt{x-3}+\sqrt{x^2-9}=0\)
\(\Leftrightarrow(\dfrac{5}{3}+\sqrt{x+3})\sqrt{x-3}=0\)
\(\Leftrightarrow\sqrt{x-3}=0\) (vì \(\dfrac{5}{3}+\sqrt{x+3}>0\))
\(\Leftrightarrow x-3=0\Leftrightarrow x=3\left(nhận\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,\sqrt{2}.\sqrt{7+3\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{14+6\sqrt{5}}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\sqrt{5^2}+2.3\sqrt{5}+3^2}-\dfrac{4}{\sqrt{5}-1}\\ =\sqrt{\left(\sqrt{5}+3\right)^2}-\dfrac{4}{\sqrt{5}-1}\\ =\left|\sqrt{5}+3\right|-\dfrac{4}{\sqrt{5}-1}\\ =\dfrac{\left(\sqrt{5}+3\right)\left(\sqrt{5}-1\right)-4}{\sqrt{5}-1}\\ =\dfrac{2+2\sqrt{5}-4}{\sqrt{5}-1}\\ =\dfrac{-2+2\sqrt{5}}{\sqrt{5}-1}\\ =\dfrac{2\left(-1+\sqrt{5}\right)}{\sqrt{5}-1}\\ =2\)
\(c,\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\\ =3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{\sqrt{3}-3}{\sqrt{3}}\)
\(=\dfrac{3\sqrt{3}.\sqrt{3}-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{9-6+\sqrt{3}-3}{\sqrt{3}}\\ =\dfrac{\sqrt{3}}{\sqrt{3}}\\ =1\)
\(d,\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}\\ =\dfrac{\left(9-2\sqrt{3}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}{\left(3\sqrt{6}-2\sqrt{2}\right)\left(3\sqrt{6}+2\sqrt{2}\right)}\\ =\dfrac{27\sqrt{6}+18\sqrt{2}-18\sqrt{2}-4\sqrt{6}}{\left(3\sqrt{6}\right)^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{23\sqrt{6}}{54-8}\\ =\dfrac{23\sqrt{6}}{46}\\ =\dfrac{\sqrt{6}}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a.\sqrt{72}-5\sqrt{2}+3\sqrt{12}\\ =6\sqrt{2}-5\sqrt{2}+6\sqrt{3}\\ =\sqrt{2}+6\sqrt{3}\\ b.6\sqrt{\dfrac{1}{2}}-\dfrac{2}{\sqrt{2}}-5\sqrt{2}\\ =3\sqrt{2}-\sqrt{2}-5\sqrt{2}\\ =-3\sqrt{2}\\ c.\dfrac{\sqrt{8}-2}{\sqrt{2}-1}+\dfrac{2}{\sqrt{3}-1}-\dfrac{3}{\sqrt{3}}\\ =2+1+\sqrt{3}-\sqrt{3}\\ =3\\ d.\sqrt[3]{64}+\sqrt[3]{27}-2\sqrt[3]{-8}\\ =4+3+4\\ =11\)
2: \(\dfrac{\sqrt{12}-\sqrt{5}}{\sqrt{2}-1}-\dfrac{1}{\sqrt{5}-2}\)
\(=\left(2\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}+1\right)-\sqrt{5}-2\)
\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}-\sqrt{5}-2\)
\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-2\sqrt{5}-2\)
3: \(=2\cdot3\sqrt{3}-6\cdot\dfrac{1}{\sqrt{3}}+2-\sqrt{3}-3\sqrt{3}\)
\(=6\sqrt{3}-2\sqrt{3}+2-4\sqrt{3}=2\)
1) \(\dfrac{3+\sqrt{3}}{\sqrt{5}}-\dfrac{2}{\sqrt{3}-1}\)
\(=\dfrac{\sqrt{5}\cdot\left(3+\sqrt{3}\right)}{\sqrt{5}\cdot\sqrt{5}}-\dfrac{2\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)
\(=\dfrac{3\sqrt{5}+\sqrt{15}}{5}-\dfrac{2\left(\sqrt{3}-1\right)}{3-1}\)
\(=\dfrac{3\sqrt{5}+\sqrt{15}}{5}-\left(\sqrt{3}-1\right)\)
\(=\dfrac{3\sqrt{5}+\sqrt{15}-5\sqrt{3}+5}{5}\)
2) \(\dfrac{\sqrt{12}-\sqrt{5}}{\sqrt{2}-1}-\dfrac{1}{\sqrt{5}-2}\)
\(=\dfrac{\left(2\sqrt{3}-\sqrt{5}\right)\left(\sqrt{2}+1\right)}{\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)}-\dfrac{\sqrt{5}+2}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)
\(=\dfrac{2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}}{2-1}-\dfrac{\sqrt{5}+2}{5-4}\)
\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-\sqrt{5}-\left(\sqrt{5}+2\right)\)
\(=2\sqrt{6}+2\sqrt{3}-\sqrt{10}-2\sqrt{5}-2\)
3) \(2\sqrt{27}-6\sqrt{\dfrac{1}{3}}+\dfrac{1}{2}+\dfrac{\sqrt{3}-9}{\sqrt{3}}\)
\(=2\cdot3\sqrt{3}-\dfrac{6}{\sqrt{3}}+\dfrac{1}{2}+\dfrac{\sqrt{3}\left(1-3\sqrt{3}\right)}{\sqrt{3}}\)
\(=6\sqrt{3}-\dfrac{\sqrt{3}\cdot2\sqrt{3}}{\sqrt{3}}+\dfrac{1}{2}+1-3\sqrt{3}\)
\(=6\sqrt{3}-2\sqrt{3}+\dfrac{1}{2}+1-3\sqrt{3}\)
\(=\dfrac{1}{2}+1+\sqrt{3}\)
\(=\dfrac{3}{2}+\sqrt{3}\)