a) \(\sqrt{\frac{\left(165-124\right)\left(165+124\right)}{164}}=\sqrt{\frac{41.289}{164}}=\sqrt{\frac{289}{4}}=\frac{17}{2}\)

b) tương tự ý a

c) \(\left(\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}\right)^2=7+4\sqrt{3}+7-4\sqrt{3}-2.\sqrt{7+4\sqrt{3}}.\sqrt{7-4\sqrt{3}}\)

\(=14-2\sqrt{\left(7+4\sqrt{3}\right)\left(7-4\sqrt{3}\right)}\)

\(=14-2\sqrt{49-48}\)

\(=14-2.1=12\)

\(\Rightarrow\sqrt{7+4\sqrt{3}}-\sqrt{7-4\sqrt{3}}=\sqrt{12}=2\sqrt{3}\)

Ta có:

a² = (1 - a)/(2√2) 

=> a⁴ + a + 1

= (1 - a)^2/8 + a + 1

= (1 - 2a + a^2 + 8a + 8)/8

= (a + 3)^2/8

=> VT = |a + 3|/(2√2) + a^2

Làm nốt

`a)\sqrt{16x+48}+\sqrt{x+3}=15`     `ĐK: x >= -3`

`<=>4\sqrt{x+3}+\sqrt{x+3}=15`

`<=>5\sqrt{x+3}=15`

`<=>\sqrt{x+3}=3`

`<=>x+3=9<=>x=6` (t/m).

`b)\sqrt{x^2-4}-3\sqrt{x-2}=0`     `ĐK: x >= 2`

`<=>\sqrt{x-2}(\sqrt{x+2}-3)=0`

`<=>[(\sqrt{x-2}=0),(\sqrt{x+2}=3):}`

`<=>[(x-2=0),(x+2=9):}<=>[(x=2(t//m)),(x=7(t//m)):}`

\(a)\)\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\)\(\sqrt{6-6\sqrt{6}+9}+\sqrt{24-12\sqrt{6}+9}\)

\(=\)\(\sqrt{\left(\sqrt{6}+3\right)}+\sqrt{\left(\sqrt{24}+3\right)}\)

\(=\)\(\left|\sqrt{6}+3\right|+\left|\sqrt{24}+3\right|\)

\(=\)\(\sqrt{6}+3+\sqrt{24}+3\)

\(=\)\(\sqrt{6}\left(1+\sqrt{4}\right)+9\)

\(=\)\(3\sqrt{6}+9\)

Chúc bạn học tốt ~ 

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

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

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

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

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

\(=\)\(1\)

Chúc bạn học tốt ~ 

PS : mới lớp 8 sai thì thông cảm >.< 

\(\sqrt{\left(\sqrt{9}-\sqrt{6}\right)^2}\) + \(\sqrt{\left(\sqrt{24}-\sqrt{9}\right)^2}\)

\(\sqrt{9}-\sqrt{6}\) + \(\sqrt{24}-\sqrt{9}\) vì  \(\sqrt{9}>\sqrt{6}\),\(\sqrt{24}>\sqrt{9}\)

\(3-\sqrt{6}\) + \(2\sqrt{6}-3\)

= \(\sqrt{6}\)

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

\(=\frac{12+6\sqrt{3}}{6}=2+\sqrt{3}\)

Xét \(A=\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}>0\)

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

\(\Rightarrow A=\sqrt{6+2\sqrt{2}}\)

\(\Rightarrow\sqrt{3+\sqrt{7}}+\sqrt{3-\sqrt{7}}-\sqrt{6+2\sqrt{2}}=\sqrt{6+2\sqrt{2}}-\sqrt{6+2\sqrt{2}}=0\)

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

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

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

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

\(=\frac{2+\sqrt{3}+1}{2-\sqrt{3}+1}\)(Vì \(\sqrt{3}>1>0\))

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

Nếu đề bài cho vô hạn dấu căn thì ta làm như sau :

Nhận xét : A > 0 

Ta có : \(A=\sqrt{2\sqrt{2\sqrt{2\sqrt{2\sqrt{...}}}}}\)

\(\Rightarrow A^2=2\sqrt{2\sqrt{2\sqrt{2\sqrt{.....}}}}=2A\)

\(\Rightarrow A^2-2A=0\Rightarrow A\left(A-2\right)=0\)

\(\Rightarrow\left[\begin{array}{nghiempt}A=0\left(\text{loại}\right)\\A=2\left(\text{nhận}\right)\end{array}\right.\)

Vậy A = 2