11) \(\frac{3}{\sqrt{6}-\sqrt{3}}+\frac{4}{\sqrt{7}+\sqrt{3}}\)
12) \(\frac{6}{3\sqrt{2}+2\sqrt{3}}\)
13) \(\left(\sqrt{75}-3\sqrt{2}-\sqrt{12}\right)\left(\sqrt{3}+\sqrt{2}\right)\)
14)\(\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}\)
15)\(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}+\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}-\frac{\sqrt{5}+1}{\sqrt{5}-1}\)
16)\(\frac{\sqrt{2}}{2\sqrt{3}+4\sqrt{2}}\)
17) \(\frac{1}{4-3\sqrt{2}}-\frac{1}{4+3\sqrt{2}}\)
18)\(\frac{6}{\sqrt{2}-\sqrt{3}+3}\)
19)\(\frac{\sqrt{3+2\sqrt{2}}+\sqrt{3-2\sqrt{2}}}{\sqrt{3+2\sqrt{2}}-\sqrt{3-2\sqrt{2}}}\)
20)\(\sqrt{24}+6\sqrt{\frac{2}{3}}+\frac{10}{\sqrt{6}-1}\)
21)\(2\sqrt{40\sqrt{12}}-2\sqrt{\sqrt{75}}-3\sqrt{5\sqrt{58}}\)
22)\(4\sqrt{20}-3\sqrt{125}+5\sqrt{45}-15\sqrt{\frac{1}{5}}\)
23)\(\left(3\sqrt{8}-2\sqrt{12}+\sqrt{20}\right):\left(3\sqrt{18}-2\sqrt{27}+\sqrt{45}\right)\)
24)\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{6}-2}-\frac{12}{3-\sqrt{6}}\right)\left(\sqrt{6}+11\right)\)
25)\(\left(\sqrt{7}-\sqrt{5}\right)^2+2\sqrt{35}\)
26)\(\frac{\sqrt{6}+\sqrt{14}}{2\sqrt{3}+\sqrt{28}}+\frac{3\sqrt{45}+\sqrt{243}}{\sqrt{5}+\sqrt{3}}\)
27)\(\frac{1}{\sqrt{7-\sqrt{24}}+1}-\frac{1}{\sqrt{7+\sqrt{24}}-1}\)
28)\(\frac{1}{2+\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{2}{3+\sqrt{3}}\)
29)\(\frac{3+\sqrt{5}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
30)\(\left(15\sqrt{50}+5\sqrt{200}-3\sqrt{450}\right):\sqrt{10}\)
31)\(\left(\frac{2}{\sqrt{3}-1}+\frac{3}{\sqrt{3}-2}+\frac{15}{3-\sqrt{3}}\right).\frac{1}{\sqrt{3}+5}\)
32)\(\frac{5+\sqrt{5}}{5-\sqrt{5}}+\frac{5-\sqrt{5}}{5+\sqrt{5}}-\sqrt{10}\)
2. a) \(ĐKXĐ:x\ge\frac{1}{3}\)
\(\sqrt{3x-1}=4\)\(\Rightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)\(\Leftrightarrow3x=17\)\(\Leftrightarrow x=\frac{17}{3}\)( thỏa mãn ĐKXĐ )
Vậy \(x=\frac{17}{3}\)
b) \(ĐKXĐ:x\ge1\)
\(\sqrt{x-1}=x-1\)\(\Rightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}}\)( thỏa mãn ĐKXĐ )
Vậy \(x=1\)hoặc \(x=2\)
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
Vì \(6>1\)\(\Leftrightarrow\sqrt{6}>\sqrt{1}=1\)\(\Rightarrow\sqrt{6}-1>0\)
\(6>4\)\(\Rightarrow\sqrt{6}>\sqrt{4}=2\)\(\Rightarrow\sqrt{6}-2>0\)
\(\Rightarrow\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|=\left(\sqrt{6}-1\right)-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2=1\)
hay \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}=1\)
2a) \(\sqrt{3x-1}=4\)( ĐKXĐ : \(x\ge\frac{1}{3}\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{3x-1}\right)^2=4^2\)
\(\Leftrightarrow3x-1=16\)
\(\Leftrightarrow3x=17\)
\(\Leftrightarrow x=\frac{17}{3}\)( tmđk )
Vậy phương trình có nghiệm duy nhất là x = 17/3
b) \(\sqrt{x-1}=x-1\)( ĐKXĐ : \(x\ge1\))
Bình phương hai vế
\(\Leftrightarrow\left(\sqrt{x-1}\right)^2=\left(x-1\right)^2\)
\(\Leftrightarrow x-1=x^2-2x+1\)
\(\Leftrightarrow x^2-2x+1-x+1=0\)
\(\Leftrightarrow x^2-3x+2=0\)
\(\Leftrightarrow x^2-x-2x+2=0\)
\(\Leftrightarrow x\left(x-1\right)-2\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=1\\x=2\end{cases}\left(tmđk\right)}\)
Vậy phương trình có hai nghiệm là x = 1 hoặc x = 2
3. \(\sqrt{7-2\sqrt{6}}-\sqrt{10-4\sqrt{6}}\)
\(=\sqrt{6-2\sqrt{6}+1}-\sqrt{6-4\sqrt{6}+4}\)
\(=\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot1+1^2}-\sqrt{\left(\sqrt{6}\right)^2-2\cdot\sqrt{6}\cdot2+2^2}\)
\(=\sqrt{\left(\sqrt{6}-1\right)^2}-\sqrt{\left(\sqrt{6}-2\right)^2}\)
\(=\left|\sqrt{6}-1\right|-\left|\sqrt{6}-2\right|\)
\(=\sqrt{6}-1-\left(\sqrt{6}-2\right)\)
\(=\sqrt{6}-1-\sqrt{6}+2\)
\(=1\)