\(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-14\sqrt{2}+14\sqrt{2}\)

=21

`a)(\sqrt{14}-3\sqrt{2})^2+6\sqrt{28}`

`=14-12\sqrt{7}+18+12\sqrt{7}=32`

`b)2\sqrt{20}-3\sqrt{20}+\sqrt{125}`

`=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}`

`=3\sqrt{5}`.

a) \(\left(\sqrt{14}-3\sqrt{2}\right)^2-6\sqrt{28}\)

\(=\left(\sqrt{14}\right)^2-2\cdot\sqrt{14}\cdot3\sqrt{2}+\left(3\sqrt{2}\right)^2+6\sqrt{28}\)

\(=14-6\sqrt{28}+18+6\sqrt{28}\)

\(=14+18\)

\(=32\)

b) \(2\sqrt{20}-3\sqrt{20}+\sqrt{125}\)

\(=2\cdot2\sqrt{5}-3\cdot2\sqrt{5}+5\sqrt{5}\)

\(=4\sqrt{5}-6\sqrt{5}+5\sqrt{5}\)

\(=3\sqrt{5}\)

a) \(=2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1=-1\)

b) \(=\left[\sqrt{14}+\dfrac{\sqrt{6}\left(\sqrt{2}+\sqrt{5}\right)}{\sqrt{2}+\sqrt{5}}\right].\sqrt{\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)^2}\)

\(=\left(\sqrt{14}+\sqrt{6}\right)\left(\sqrt{\dfrac{7}{2}}-\sqrt{\dfrac{3}{2}}\right)\)

\(=\sqrt{49}-\sqrt{21}+\sqrt{21}-\sqrt{9}\)

\(=7-3=4\)

a) \(\sqrt{20}-15\sqrt{\dfrac{1}{5}}+\sqrt{\left(1-\sqrt{5}\right)^2}=2\sqrt{5}-3\sqrt{5}+\sqrt{5}-1=-1\)

b) \(\left(\dfrac{14}{\sqrt{14}}+\dfrac{\sqrt{12}+\sqrt{30}}{\sqrt{2}+\sqrt{5}}\right).\sqrt{5-\sqrt{21}}=\left(\sqrt{14}+\sqrt{6}\right)\sqrt{5-\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\sqrt{10-2\sqrt{21}}=\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}-\sqrt{3}\right)=4\)

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

Câu 3:

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

a) Ta có: \(M=\dfrac{2}{\sqrt{7}-\sqrt{6}}-\sqrt{28}+\sqrt{54}\)

\(=\dfrac{2\left(\sqrt{7}+\sqrt{6}\right)}{\left(\sqrt{7}-\sqrt{6}\right)\left(\sqrt{7}+\sqrt{6}\right)}-2\sqrt{7}+3\sqrt{6}\)

\(=2\sqrt{7}+2\sqrt{6}-2\sqrt{7}+3\sqrt{6}\)

\(=5\sqrt{6}\)

b) Ta có: \(N=\left(2-\sqrt{3}\right)\left(\sqrt{26+15\sqrt{3}}\right)-\left(2+\sqrt{3}\right)\sqrt{26-15\sqrt{3}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{52+30\sqrt{3}}-\left(2+\sqrt{3}\right)\sqrt{52-30\sqrt{3}}}{\sqrt{2}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{27+2\cdot3\sqrt{3}\cdot5+25}-\left(2+\sqrt{3}\right)\sqrt{27-2\cdot3\sqrt{3}\cdot5+25}}{\sqrt{2}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\sqrt{\left(3\sqrt{3}+5\right)^2}-\left(2+\sqrt{3}\right)\sqrt{\left(3\sqrt{3}-5\right)^2}}{\sqrt{2}}\)

\(=\dfrac{\left(2-\sqrt{3}\right)\left(3\sqrt{3}+5\right)-\left(2+\sqrt{3}\right)\left(3\sqrt{3}-5\right)}{\sqrt{2}}\)

\(=\dfrac{6\sqrt{3}+10-9-5\sqrt{3}-\left(6\sqrt{3}-10+9-5\sqrt{3}\right)}{\sqrt{2}}\)

\(=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}\)

\(=\dfrac{2}{\sqrt{2}}=\sqrt{2}\)

1:

a: \(\sqrt{25}+\sqrt{49}=5+7=12\)

b: \(\sqrt{121}-\sqrt{81}=11-9=2\)

2: x>-2

=>2x>-4

=>2x+1>-3

=>Với x>-2 thì \(\sqrt{2x+1}\) chưa chắc có nghĩa

3:

a: \(\sqrt{\left(\sqrt{3}-1\right)^2}-\sqrt{3}\)

\(=\left|\sqrt{3}-1\right|-\sqrt{3}\)

\(=\sqrt{3}-1-\sqrt{3}=-1\)

b: \(\left(\sqrt{28}-2\sqrt{14}+\sqrt{7}\right)\cdot\sqrt{7}+7\sqrt{8}\)

\(=\left(3\sqrt{7}-2\sqrt{14}\right)\cdot\sqrt{7}+14\sqrt{2}\)

\(=21-14\sqrt{2}+14\sqrt{2}=21\)

c:

\(\dfrac{\sqrt{27}-\sqrt{108}+\sqrt{12}}{\sqrt{3}}\)

\(=\dfrac{3\sqrt{3}-6\sqrt{3}+2\sqrt{3}}{\sqrt{3}}=3+2-6=-1\)

b: Ta có: \(\left(\sqrt{7-3\sqrt{5}}\right)\cdot\left(7+3\sqrt{5}\right)\cdot\left(3\sqrt{2}+\sqrt{10}\right)\)

\(=\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)\left(7+3\sqrt{5}\right)\)

\(=4\left(7+3\sqrt{5}\right)\)

\(=28+12\sqrt{5}\)

Lời giải:

a. 

$A=\sqrt{8+\sqrt{55}}-\sqrt{8-\sqrt{55}}-\sqrt{125}$
$\sqrt{2}A=\sqrt{16+2\sqrt{55}}-\sqrt{16-2\sqrt{55}}-\sqrt{250}$

$=\sqrt{(\sqrt{11}+\sqrt{5})^2}-\sqrt{(\sqrt{11}-\sqrt{5})^2}-5\sqrt{10}$

$=|\sqrt{11}+\sqrt{5}|-|\sqrt{11}-\sqrt{5}|-5\sqrt{10}$

$=2\sqrt{5}-5\sqrt{10}$

$\Rightarrow A=\sqrt{10}-5\sqrt{5}$

b.

$B=\sqrt{7-3\sqrt{5}}.(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$B\sqrt{2}=\sqrt{14-6\sqrt{5}}(7+3\sqrt{5})(3\sqrt{2}+\sqrt{10})$

$=\sqrt{(3-\sqrt{5})^2}(7+3\sqrt{5}).\sqrt{2}(3+\sqrt{5})$

$=(3-\sqrt{5})(7\sqrt{2}+3\sqrt{10})(3+\sqrt{5})$

$=(3^2-5)(7\sqrt{2}+3\sqrt{10})$

$=4(7\sqrt{2}+3\sqrt{10})=28\sqrt{2}+12\sqrt{10}$

$\Rightarrow B=28+12\sqrt{5}$

c.

$C=\sqrt{2}(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{6+\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{12+2\sqrt{35}}$

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})\sqrt{(\sqrt{7}+\sqrt{5})^2}

$=(\sqrt{7}-\sqrt{5})(6-\sqrt{35})(\sqrt{7}+\sqrt{5})$

$=(7-5)(6-\sqrt{35})$

$=2(6-\sqrt{35})=12-2\sqrt{35}$

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

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

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

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

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

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

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

giúp nốt mình câu e đi bạn !