đề là rút gọn các biểu thức sau

nhờ mọi người giải giúp mình. cảm ơn mn nhìu

a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

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

\(=\dfrac{2\sqrt{5}+2}{2\sqrt{5}+2}=1\)

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

=2căn 5-2-2căn 5

=-2

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

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

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

a: =2-căn 3-2-căn 3

=-2căn 3

b: \(=\dfrac{1}{\sqrt{2}}\left(\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}\right)\)

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

c: \(A=\sqrt{4-\sqrt{10-2\sqrt{5}}}-\sqrt{4+\sqrt{10-2\sqrt{5}}}\)

=>\(A^2=4-\sqrt{10-2\sqrt{5}}+4+\sqrt{10-2\sqrt{5}}+2\cdot\sqrt{16-10+2\sqrt{5}}\)

\(\Leftrightarrow A^2=8+2\left(\sqrt{5}+1\right)=10+2\sqrt{5}\)

=>\(A=\sqrt{10+2\sqrt{5}}\)

\(1.\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{5-2.\sqrt{2}.\sqrt{5}+2}-\sqrt{5+2.\sqrt{5}.\sqrt{2}+2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}=\text{|}\sqrt{5}-\sqrt{2}\text{|}-\text{|}\sqrt{5}+\sqrt{2}\text{|}=-2\sqrt{2}\)\(2.\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}=\sqrt{8+2.2\sqrt{2}.\sqrt{5}+5}+\sqrt{8-2.2\sqrt{2}.\sqrt{5}+5}=\sqrt{\left(2\sqrt{2}+\sqrt{5}\right)^2}+\sqrt{\left(2\sqrt{2}-\sqrt{5}\right)^2}=\text{|}2\sqrt{2}+\sqrt{5}\text{|}+\text{|}2\sqrt{2}-\sqrt{5}\text{|}=4\sqrt{2}\)\(3.\left(\sqrt{3}+\sqrt{5}\right)\sqrt{7-2\sqrt{10}}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\left(\sqrt{3}+\sqrt{5}\right)\text{|}\sqrt{5}-\sqrt{2}\text{|}=\left(\sqrt{3}+\sqrt{5}\right)\left(\sqrt{5}-\sqrt{2}\right)\)

\(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}=-2\sqrt{2}\)

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

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

\(\sqrt{6-4\sqrt{2}}+\sqrt{22-12\sqrt{2}}=\sqrt{\left(2-\sqrt{2}\right)^2}+\sqrt{\left(3\sqrt{2}-2\right)^2}\)

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

1: Ta có: \(\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}\)

\(=\sqrt{5}-\sqrt{2}-\sqrt{5}-\sqrt{2}\)

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

2: Ta có: \(\sqrt{4-2\sqrt{3}}+\sqrt{4+2\sqrt{3}}\)

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

\(=2\sqrt{3}\)

\(VT=\sqrt{7-2\sqrt{10}}-\sqrt{7+2\sqrt{10}}=\sqrt{2-2\sqrt{10}+5}-\sqrt{2+2\sqrt{10}+5}=\sqrt{\left(\sqrt{2}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}=\left|\sqrt{2}-\sqrt{5}\right|-\left|\sqrt{2}+\sqrt{5}\right|=\sqrt{5}-\sqrt{2}-\sqrt{2}-\sqrt{5}=-2\sqrt{2}=VP\)

Sửa đề: \(\sqrt{7-2\sqrt{10}}-\sqrt{14+4\sqrt{10}}-2\sqrt{22-4\sqrt{10}}+3\cdot\sqrt{13-4\sqrt{10}}\)

Ta có: \(\sqrt{7-2\sqrt{10}}-\sqrt{14+4\sqrt{10}}-2\sqrt{22-4\sqrt{10}}+3\cdot\sqrt{13-4\sqrt{10}}\)

\(=\sqrt{5-2\cdot\sqrt5\cdot\sqrt2+2}-\sqrt{10+2\cdot\sqrt{10}\cdot2+4}-2\cdot\sqrt{20-2\cdot2\sqrt5\cdot\sqrt2+2}+3\cdot\sqrt{8-2\cdot2\sqrt2\cdot\sqrt5+5}\)

\(=\sqrt{\left(\sqrt5-\sqrt2\right)^2}-\sqrt{\left(\sqrt{10}+2\right)^2}-2\cdot\sqrt{\left(2\sqrt5-\sqrt2\right)^2}+3\cdot\sqrt{\left(2\sqrt2-\sqrt5\right)^2}\)

\(=\sqrt5-\sqrt2-\sqrt{10}-2-2\left(2\sqrt5-\sqrt2\right)+3\left(2\sqrt2-\sqrt5\right)\)

\(=\sqrt5-\sqrt2-\sqrt{10}-2-4\sqrt5+2\sqrt2+6\sqrt2-3\sqrt5=-6\sqrt5+7\sqrt2-\sqrt{10}\) -2

a: Ta có: \(A=\sqrt{12-3\sqrt7}-\sqrt{12+3\sqrt7}\)

=>\(A^2=12-3\sqrt7+12+3\sqrt7-2\cdot\sqrt{\left(12-3\sqrt7\right)\left(12+3\sqrt7\right)}\)

=>\(A^2=24-2\cdot\sqrt{144-63}=24-2\cdot\sqrt{81}=24-2\cdot9=24-18=6\)

=>\(A=\sqrt6\)

b: \(B=\sqrt{4+\sqrt{10+2\sqrt5}}+\sqrt{4-\sqrt{10+2\sqrt5}}\)

=>\(B^2=4+\sqrt{10+2\sqrt5}+4-\sqrt{10+2\sqrt5}+2\cdot\sqrt{4^2-\left(10+2\sqrt5\right)}\)

=>\(B^2=8+2\cdot\sqrt{16-10-2\sqrt5}=8+2\cdot\sqrt{6-2\sqrt5}\)

=>\(B^2=8+2\cdot\sqrt{\left(\sqrt5-1\right)^2}=8+2\left(\sqrt5-1\right)=6+2\sqrt5=\left(\sqrt5+1\right)^2\)

=>\(B=\sqrt5+1\)

c: \(C=\sqrt{3-\sqrt5}+\sqrt{3+\sqrt5}\)

=>\(C^2=3-\sqrt5+3+\sqrt5+2\cdot\sqrt{\left(3-\sqrt5\right)\left(3+\sqrt5\right)}\)

=>\(C^2=6+2\cdot\sqrt{9-5}=6+2\cdot\sqrt4=6+2\cdot2=10\)

=>\(C=\sqrt{10}\)

`F=sqrt{(3-sqrt2)^2}+sqrt{(1-sqrt2)^2}``

`=3-sqrt2+sqrt2-1=2`

`G=sqrt{(5+sqrt7)^2}-sqrt{(2-sqrt7)^2}`

`=5+sqrt7-(sqrt7-2)`

`=5+sqrt7-sqrt7+2=2`

`H=sqrt{(3-sqrt{10})^2}+sqrt{(2-sqrt{10})^2}`

`=sqrt{10}-3+sqrt{10}-2`

`=2\sqrt{10}-5`

\(F=\left|3-\sqrt{2}\right|+\left|1-\sqrt{2}\right|=3-\sqrt{3}+\sqrt{2}-1=2\)

\(G=\left|5+\sqrt{7}\right|-\left|2-\sqrt{7}\right|=5+\sqrt{7}-\sqrt{7}+2=7\)

\(H=\left|3-\sqrt{10}\right|+\left|2-\sqrt{10}\right|=\sqrt{10}-3+\sqrt{10}-2=2\sqrt{10}-5\)

\(A=\sqrt{7-2\sqrt{10}}+\sqrt{7+2\sqrt{10}}\)

\(A^2=\left(7+2\sqrt{10}+7-2\sqrt{10}\right)+2\sqrt{\left(7-2\sqrt{10}\right)\left(7+2\sqrt{10}\right)}\)

\(=14+2\sqrt{49-40}=14+6=20\)

Khi đó:\(A=\sqrt{20}\)

Các câu còn lại bạn làm nốt nhé

a) đặt \(A=\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

nhân cả hai vế với \(\sqrt{2}\), ta được:

\(\sqrt{2}A=\sqrt{2}\sqrt{4-\sqrt{7}}-\sqrt{2}\sqrt{4+\sqrt{7}}\)

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

\(=\sqrt{\left(1-\sqrt{7}\right)^2}-\sqrt{\left(1+ \sqrt{7}\right)^2}\)

\(=\left|1-\sqrt{7}\right|-\left|1+\sqrt{7}\right|\)

\(=\sqrt{7}-1-\sqrt{7}-1\)

\(=-2\)

\(\Rightarrow A=-\frac{2}{\sqrt{2}}=-\sqrt{2}\)

a) \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}\)

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

\(=\frac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}}{\sqrt{2}}\)

\(=\frac{\sqrt{7}-1-\sqrt{7}-1}{\sqrt{2}}\)

\(=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)