\(\sqrt{242}.\sqrt{26}.\sqrt{130}.\sqrt{0,9}-\left(\sqrt{2}-1\right)\left(\sqrt{2}+1\right)\)

\(=\sqrt{121}.\sqrt{2}.\sqrt{2}.\sqrt{13}.\sqrt{13}.\sqrt{10}.\sqrt{0,9}-\left(2-1\right)\)

\(=11.2.13.\sqrt{9}-1=286.3-1=857\)

\(\frac{3-\sqrt{6}}{\sqrt{12}-\sqrt{8}}-\frac{\sqrt{15}-\sqrt{5}}{2\sqrt{12}-4}+\frac{\sqrt{17-4\sqrt{15}}}{4}\)

\(=\frac{\sqrt{3}\left(\sqrt{3}-\sqrt{2}\right)}{2\left(\sqrt{3}-\sqrt{2}\right)}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{4\left(\sqrt{3}-1\right)}+\frac{\sqrt{\left(2\sqrt{3}-\sqrt{5}\right)^2}}{4}\)

\(=\frac{\sqrt{3}}{2}-\frac{\sqrt{5}}{4}+\frac{2\sqrt{3}-\sqrt{5}}{4}\)

\(=\sqrt{3}-\frac{\sqrt{5}}{4}\)

giúp nốt mình câu e đi bạn !

`a)sqrt{4+sqrt7}-sqrt{4-sqrt7}`

`=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}`

`=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}`

`=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7-1)^2/2}`

`=(sqrt7+1)/sqrt2-(sqrt7-1)/sqrt2`

`=2/sqrt2=sqrt2`

`b)sqrt{4--sqrt15}-sqrt{4+sqrt15}`

`=sqrt{(8-2sqrt15)/2}-sqrt{(8+2sqrt15)/2}`

`=sqrt{(5-2sqrt{5.3}+3)/2}-sqrt{(5+2sqrt{5.3}+3)/2}`

`=sqrt{(sqrt5-sqrt3)^2/2}-sqrt{(sqrt5+sqrt3)^2/2}`

`=(sqrt5-sqrt3)/sqrt2-(sqrt5+sqrt3)/sqrt2`

`=(-2sqrt3)/sqrt2=-sqrt6`

`c)sqrt{2+sqrt3}+sqrt{2-sqrt3}`

`=sqrt{(4+2sqrt3)/2}+sqrt{(4-2sqrt3)/2}`

`=sqrt{(3+2sqrt3+1)/2}+sqrt{(3-2sqrt3+1)/2}`

`=sqrt{(sqrt3+1)^2/2}+sqrt{(sqrt3-1)^2/2}`

`=(sqrt3+1)/sqrt2+(sqrt3-1)/sqrt2`

`=(2sqrt3)/sqrt2=sqrt6`

`d)sqrt{9+sqrt17}-sqrt{9-sqrt17}`

`=sqrt{(18+2sqrt17)/2}-sqrt{(18-2sqrt17)/2}`

`=sqrt{(17+2sqrt17+1)/2}-sqrt{(17-2sqrt17+1)/2}`

`=sqrt{(sqrt17+1)^2/2}-sqrt{(sqrt17-1)^2/2}`

`=(sqrt17+1)/sqrt2-(sqrt17-1)/sqrt2`

`=2/sqrt2=sqrt2`

a: Ta có: \(\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)

\(=\dfrac{\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{7}+1-\sqrt{7}+1}{\sqrt{2}}=\sqrt{2}\)

b: Ta có: \(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\)

\(=\dfrac{\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)

\(-11\)

3: Ta có: \(\sqrt{4x+1}=x+1\)

\(\Leftrightarrow x^2+2x+1=4x+1\)

\(\Leftrightarrow x\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(nhận\right)\end{matrix}\right.\)

4: Ta có: \(2\sqrt{x-1}+\dfrac{1}{3}\sqrt{9x-9}=15\)

\(\Leftrightarrow3\sqrt{x-1}=15\)

\(\Leftrightarrow x-1=25\)

hay x=26

5: Ta có: \(\sqrt{4x^2-12x+9}=7\)

\(\Leftrightarrow\left|2x-3\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}2x-3=7\\2x-3=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=10\\2x=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\\x=-2\end{matrix}\right.\)

Câu 1,2 bạn đã đăng và có lời giải rồi

Câu 3:

\(=\frac{(\sqrt{3})^2+(2\sqrt{5})^2-2.\sqrt{3}.2\sqrt{5}}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{(\sqrt{3}-2\sqrt{5})^2}{\sqrt{2}(\sqrt{3}-2\sqrt{5})}=\frac{\sqrt{3}-2\sqrt{5}}{\sqrt{2}}\)