bài 1:

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

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

\(=\left(3-\sqrt{2}\right)\left(2+\sqrt{3}\right)\)\(do2>\sqrt{3}\)

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

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

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

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

\(=\sqrt{15}-\sqrt{6}+5-\sqrt{10}\)

c)\(\left(2+\sqrt{5}\right)\sqrt{9-4\sqrt{5}}\)

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

\(=\left(2+\sqrt{5}\right)\left(\sqrt{5}-2\right)do\sqrt{5}>2\)

\(=5-4\)

\(=1\left(hđt.3\right)\)

d)\(\left(\sqrt{6}+\sqrt{10}\right)\sqrt{4-\sqrt{15}}\)

\(=\sqrt{2}\left(\sqrt{3}+\sqrt{5}\right)\sqrt{4-\sqrt{15}}\)

\(=\left(\sqrt{3}+\sqrt{5}\right)\sqrt{8-2\sqrt{15}}\)

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

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

\(=5-3\)

\(=2\)

e)\(\sqrt{2}\left(\sqrt{8}-\sqrt{32}+3\sqrt{18}\right)\)

\(=\sqrt{2}\left(2\sqrt{2}-4\sqrt{2}+9\sqrt{2}\right)\)

\(=2\left(2-4+9\right)\)

\(=2.7=14\)

f)\(\sqrt{2}\left(\sqrt{2}-\sqrt{3-\sqrt{5}}\right)\)

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

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

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

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

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

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

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

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

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

\(=\left(2-\sqrt{6+2\sqrt{5}}\right)+2\sqrt{5}\)

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

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

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

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

bài 2)

a) \(\sqrt{4x^2-4x+1}=5\)

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

\(\Leftrightarrow2x-1=5\)hoặc \(\Leftrightarrow2x-1=-5\)

\(\Leftrightarrow x=3\)hoặc \(\Leftrightarrow x=-2\)

Vậy x = 3 hoặc x = -2

Bài 1: 

a: \(=\left|5-\sqrt{3}\right|-\left|\sqrt{3}-2\right|\)

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

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

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

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

\(=\dfrac{20-18}{\sqrt{2}}=\sqrt{2}\)

c: \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

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

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

d: \(A=\left(\sqrt{5}-1\right)\cdot\sqrt{6+2\sqrt{5}}\)

\(=\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)=5-1=4\)

Bài 1 )

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

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

Bài 2)

a)\(\sqrt{36x^2-12x+1}=5\)

\(\Leftrightarrow36x^2-12x+1=25\)

\(\Leftrightarrow36x^2-12x+1=25\)

\(\Leftrightarrow\left(6x\right)^2-2.6x+1=25\)

\(\Leftrightarrow\left(6x-1\right)^2=25\)

\(\Rightarrow6x-1=5\)

\(\Leftrightarrow6x=6\)

\(\Rightarrow x=1\)

b)\(\sqrt{x-5}-2\sqrt{4x-20}-\frac{1}{3}\sqrt{9x-45}=12\)

\(\Leftrightarrow\sqrt{x-5}-2\sqrt{4.\left(x-5\right)}-\frac{1}{3}\sqrt{9.\left(x-5\right)}=12\)

\(\Leftrightarrow\sqrt{x-5}-4\sqrt{\left(x-5\right)}-\sqrt{\left(x-5\right)}=12\)

\(\Leftrightarrow-4\sqrt{\left(x-5\right)}=12\)

\(\Rightarrow\)ko tồn tại giá trị nào của x trong biểu thức này

P/s tham khảo nha

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

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

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

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

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

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

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

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

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

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

=\(1+1\)=\(2\)

2) a) \(\sqrt{36x^2-12x+1}=5\)

<=>\(\sqrt{\left(6x\right)^2-2.6x.1+1^2}=5\)

<=>\(\sqrt{\left(6x-1\right)^2}=5\)

<=>\(|6x-1|=5\)

Nếu \(6x-1>=0\)=> \(6x>=1\)=>\(x>=\frac{1}{6}\)

Nên \(|6x-1|=6x-1\)

Ta có \(|6x-1|=5\)

<=> \(6x-1=5\)

<=> \(6x=6\)

<=> \(x=1\)(thỏa)

Nếu \(6x-1< 0\)=> \(6x< 1\)=>\(x< \frac{1}{6}\)

Nên \(|6x-1|=-\left(6x-1\right)=1-6x\)

Ta có \(|6x-1|=5\)

<=> \(1-6x=5\)

<=> \(-6x=4\)

<=> \(x=\frac{4}{-6}=\frac{-2}{3}\)(thỏa)

Vậy \(x=1\)và \(x=\frac{-2}{3}\)

b) \(\sqrt{x-5}-2\sqrt{4x-20}-\frac{1}{3}\sqrt{9x-45}=12\)

<=>\(\sqrt{x-5}-2\sqrt{4\left(x-5\right)}-\frac{1}{3}\sqrt{9\left(x-5\right)}=12\)

<=>\(\sqrt{x-5}-2.2\sqrt{x-5}-\frac{1}{3}.3\sqrt{x-5}=12\)

<=>\(\sqrt{x-5}-4\sqrt{x-5}-\sqrt{x-5}=12\)

<=>\(-4\sqrt{x-5}=12\)

<=> \(\sqrt{x-5}=-3\)

<=> \(\left(\sqrt{x-5}\right)^2=\left(-3\right)^2\)

<=>\(x-5=9\)

<=>\(x=14\)

Vậy x=14

Kết bạn với mình nhá

a)

DK: x\(\ge\)-2,x\(\ge\)\(\dfrac{1}{2}\)

=>\(\sqrt{4\left(x+2\right)}-\sqrt{2x-1}+\sqrt{9\left(x+2\right)}=0\)

\(\Leftrightarrow2\sqrt{x+2}-\sqrt{2x-1}+3\sqrt{x+2}=0\)

\(\Leftrightarrow5\sqrt{x+2}-\sqrt{2x-1}=0\)

\(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\)

<=>25x+50=2x-1

=>23x=-51

=>x=\(-\dfrac{51}{23}\)(ko thỏa mãn dk)

=> phương trình vô nghiệm..

b)

ĐKXĐ:\(x\ge1,x\ge-1\)

\(\Leftrightarrow\sqrt{\left(x+1\right)\left(x-1\right)}-3\sqrt{x-1}=0\)

\(\Leftrightarrow\sqrt{x-1}\left(\sqrt{x+1}-3\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)(nhận)

Vậy S={1;8}

c) ĐKXĐ:

\(x\ge0\)

\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)

\(\Leftrightarrow-11\sqrt{2x}=-11\)

\(\Leftrightarrow\sqrt{2x}=1\)

\(\Leftrightarrow2x=1\\ \Leftrightarrow x=\dfrac{1}{2}\)

Câu a :\(\sqrt{4x+8}-2\sqrt{2x-1}+\sqrt{9x+18}=0\) ( ĐK : \(x\ge\dfrac{1}{2}\) )

\(\Leftrightarrow\sqrt{4x+8}+\sqrt{9x+18}=\sqrt{2x-1}\)

\(\Leftrightarrow2\sqrt{x+2}+3\sqrt{x+2}=\sqrt{2x-1}\)

\(\Leftrightarrow5\sqrt{x+2}=\sqrt{2x-1}\)

\(\Leftrightarrow25\left(x+2\right)=2x-1\)

\(\Leftrightarrow25x+50=2x-1\)

\(\Leftrightarrow23x=-51\)

\(\Leftrightarrow x=-\dfrac{51}{23}< -\dfrac{1}{2}\)

Vậy phương trình vô nghiệm .

Câu b :

\(\sqrt{x^2-1}-\sqrt{9\left(x-1\right)}=0\) ( ĐK : \(x\ge1\) )

\(\Leftrightarrow\sqrt{\left(x-1\right)\left(x+1\right)}-3\sqrt{\left(x-1\right)}=0\)

\(\Leftrightarrow\sqrt{\left(x-1\right)}\left(\sqrt{x+1}-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x-1}=0\\\sqrt{x+1}-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=8\end{matrix}\right.\)

Vậy \(S=\left\{1;8\right\}\)

Câu c : \(\left(3-\sqrt{2x}\right)\left(2-3\sqrt{2x}\right)=6x-5\) ( ĐK : \(x\ge\dfrac{5}{6}\) )

\(\Leftrightarrow6-9\sqrt{2x}-2\sqrt{2x}+6x=6x-5\)

\(\Leftrightarrow-11\sqrt{2x}+11=0\)

\(\Leftrightarrow-11\left(\sqrt{2x}-1\right)=0\)

\(\Leftrightarrow\sqrt{2x}-1=0\)

\(\Leftrightarrow x=\dfrac{1}{2}\left(TMĐK\right)\)

Vậy \(S=\left\{\dfrac{1}{2}\right\}\)

Chúc bạn học tốt

Bài 2 :

ĐKXĐ : \(\left\{{}\begin{matrix}x\le3\\x\le5\end{matrix}\right.\)

=> \(x\le3\)

Ta có : \(\sqrt{3-x}+\sqrt{5-x}=2\)

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

=> \(3-x=4-4\sqrt{5-x}+5-x\)

=> \(-4\sqrt{5-x}=-6\)

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

=> \(x=2,75\) ( TM )

Ta có : \(A=\sqrt{3-2,75}-\sqrt{5-2,75}=-1\)

Vậy ...