Bài 22:

1: \(\sqrt{3-\sqrt5}=\frac{\sqrt{6-2\sqrt5}}{\sqrt2}\)

\(=\frac{\sqrt{\left(\sqrt5-1\right)^2}}{\sqrt2}=\frac{\sqrt5-1}{\sqrt2}=\frac{\sqrt{10}-\sqrt2}{2}\)

2: \(\sqrt{7+3\sqrt5}\)

\(=\frac{\sqrt{14+6\sqrt5}}{\sqrt2}\)

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

3: \(\sqrt{9+\sqrt{17}}-\sqrt{9-\sqrt{17}}-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{18+2\sqrt{17}}-\sqrt{18-2\sqrt{17}}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{\left(\sqrt{17}+1\right)^2}-\sqrt{\left(\sqrt{17}-1\right)^2}\right)-2\)

\(=\frac{1}{\sqrt2}\left(\sqrt{17}+1-\sqrt{17}+1\right)-2=\frac{2}{\sqrt2}-2=\sqrt2-2\)

Bài 26:

1: \(\left|3-2x\right|=2\sqrt5\)

=>\(\left[\begin{array}{l}2x-3=2\sqrt5\\ 2x-3=-2\sqrt5\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3+2\sqrt5\\ 2x=3-2\sqrt5\end{array}\right.\Rightarrow x=\frac{3\pm2\sqrt5}{2}\)

2: \(\sqrt{x^2}=12\)

=>|x|=12

=>x=12 hoặc x=-12

3: \(\sqrt{x^2-2x+1}=7\)

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

=>|x-1|=7

=>\(\left[\begin{array}{l}x-1=7\\ x-1=-7\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8\\ x=-6\end{array}\right.\)

22,

1, Đặt √(3-√5) = A

=> √2A=√(6-2√5)

=> √2A=√(5-2√5+1)

=> √2A=|√5 -1|

=> A=\(\dfrac{\sqrt{5}-1}{\text{√2}}\)

=> A= \(\dfrac{\sqrt{10}-\sqrt{2}}{2}\)

2, Đặt √(7+3√5) = B

=> √2B=√(14+6√5)

 => √2B=√(9+2√45+5)

=> √2B=|3+√5|

=> B= \(\dfrac{3+\sqrt{5}}{\sqrt{2}}\)

=> B= \(\dfrac{3\sqrt{2}+\sqrt{10}}{2}\)

3, 

Đặt √(9+√17) - √(9-√17) -\(\sqrt{2}\)=C

=> √2C=√(18+2√17) - √(18-2√17) -\(2\)

=> √2C=√(17+2√17+1) - √(17-2√17+1) -\(2\)

=> √2C=√17+1- √17+1 -\(2\)

=> √2C=0

=> C=0

26,

|3-2x|=2\(\sqrt{5}\)

TH1: 3-2x ≥ 0 ⇔ x≤\(\dfrac{-3}{2}\)

3-2x=2\(\sqrt{5}\)

-2x=2\(\sqrt{5}\) -3

x=\(\dfrac{3-2\sqrt{5}}{2}\) (KTMĐK)

TH2: 3-2x < 0 ⇔ x>\(\dfrac{-3}{2}\)

3-2x=-2\(\sqrt{5}\)

-2x=-2√5 -3

x=\(\dfrac{3+2\sqrt{5}}{2}\) (TMĐK)

Vậy x=\(\dfrac{3+2\sqrt{5}}{2}\)

2, \(\sqrt{x^2}\)=12 ⇔ |x|=12 ⇔ x=12, -12

3, \(\sqrt{x^2-2x+1}\)=7

⇔ |x-1|=7 

TH1: x-1≥0 ⇔ x≥1

x-1=7 ⇔ x=8 (TMĐK)

TH2: x-1<0 ⇔ x<1

x-1=-7 ⇔ x=-6 (TMĐK)

Vậy x=8, -6

4, \(\sqrt{\left(x-1\right)^2}\)=x+3

⇔ |x-1|=x+3

TH1: x-1≥0 ⇔ x≥1

x-1=x+3 ⇔ 0x=4 (KTM)

TH2: x-1<0 ⇔ x<1

x-1=-x-3 ⇔ 2x=-2 ⇔x=-1 (TMĐK)

Vậy x=-1

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

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

a)= \(\frac{\sqrt{2}-1}{2-1}+\frac{\sqrt{3}-\sqrt{2}}{3-2}+...+\frac{\sqrt{100}-\sqrt{99}}{100-99}\)

=\(\sqrt{2}-1+\sqrt{3}-\sqrt{2}+...+\sqrt{100}-\sqrt{99}\)

= \(-1+\sqrt{100}\)

= -1 +10

=9

b)Ta có\(\left(\sqrt{n+1}-\sqrt{n}\right)\cdot\left(\sqrt{n+1}+\sqrt{n}\right)\)=n+1-n=1  (1)

Lại có:\(\frac{1}{\sqrt{n+1}+1}\cdot\left(\sqrt{n+1}+1\right)=1\)(2)

Từ (1) và (2)=>\(\left(\sqrt{n+1}-1\right)=\frac{1}{\sqrt{n+1}+1}\)

\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}\)

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

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

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

\(\sqrt{\left(2-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-1\right)^2}=|2-\sqrt{5}|-|\sqrt{5}-1|.\)

\(=\sqrt{5}-2-\sqrt{5}+1\)(Vì \(2=\sqrt{4}< \sqrt{5};1=\sqrt{1}< \sqrt{5}\))

\(=-1\)

\(\sqrt{5+2\sqrt{6}}+\sqrt{10-4\sqrt{6}}=\sqrt{2+2.\sqrt{2}\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{6}+6}\)

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

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

\(=\sqrt{2}+\sqrt{3}+\sqrt{6}-2\)( Vì \(\sqrt{6}>\sqrt{4}=2\))

\(\Leftrightarrow\sqrt{\left(\sqrt{x}+1\right)^2}=2\Leftrightarrow\sqrt{x}+1=2\Leftrightarrow\sqrt{x}=1\Leftrightarrow x=1\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}=3\Leftrightarrow\sqrt{x}-2=3\Leftrightarrow\sqrt{x}=5\Leftrightarrow x=25\) 

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

\(\Leftrightarrow x=10\)

 ĐKXĐ tự tìm\(b,\sqrt{x-4\sqrt{x}+4}=3\)

\(\Leftrightarrow\sqrt{\left(\sqrt{x}-2\right)^2}=3\)

\(\Leftrightarrow\sqrt{x}-2=3\)

\(\Leftrightarrow\sqrt{x}=5\)

\(\Rightarrow x=5^2=25\)

\(\frac{9+4\sqrt{2}}{21}\)

cho  P = \(\frac{\sqrt{x}+2}{\sqrt{x}+1}\)  , Tìm GTLN của P  

a, \(\sqrt{17-12\sqrt{2}}-\sqrt{17+12\sqrt{2}}\)

\(=\sqrt{17-2.3.2\sqrt{2}}-\sqrt{17+2.3.2\sqrt{2}}\)

\(=\sqrt{9-2.3.2\sqrt{2}+8}-\sqrt{9+2.3.2\sqrt{2}+8}\)

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

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

b, \(\sqrt{31-12\sqrt{3}}-\sqrt{31+12\sqrt{3}}\)

\(=\sqrt{31-2.2.3\sqrt{3}}-\sqrt{31+2.2.3\sqrt{3}}\)

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

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