Đặt \(A=\sqrt{2-\sqrt{3}}\)

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

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

\(\Rightarrow A=\frac{\sqrt{3}+1}{\sqrt{2}}=\frac{\sqrt{6}+\sqrt{2}}{2}\)

sửa dòng 2 nhé :vvv 

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

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

\(\Rightarrow A=\frac{\sqrt{3}-1}{\sqrt{2}}=\frac{\sqrt{6}-\sqrt{2}}{2}\)

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{a+b+c}\)

\(\Leftrightarrow\frac{ab+bc+ca}{abc}=\frac{1}{a+b+c}\)

\(\Rightarrow\left(ab+bc+ca\right)\left(a+b+c\right)=abc\)

\(\Leftrightarrow ab^2+a^2b+ac^2+a^2c+bc^2+b^2c+2abc=0\)

\(\Leftrightarrow ab^2+a^2b+ac^2+bc^2+a^2c+abc+b^2c+abc=0\)

\(\Leftrightarrow\left(a+b\right)ab+c^2\left(a+b\right)+bc\left(a+b\right)+ac\left(a+b\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(c^2+ab+bc+ac\right)=0\)

\(\Leftrightarrow\left(a+b\right)\left(b+c\right)\left(c+a\right)=0\)

Vậy ta có các trường hợp: \(a=-b,c=0\)hoặc \(b=-c,a=0\)hoăc \(a=-c,b=0\).

Với từng trường hợp ta đều có đpcm. 

kq là 29

kq là 46

b, bạn kiểm tra lại đề nhé 

c, \(\frac{x\sqrt{x}-8+2x-4\sqrt{x}}{x-4}=\frac{\sqrt{x}\left(x-4\right)+2\left(x-4\right)}{x-4}\)

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

Phương trình có tích hệ số \(ac=-m^2< 0\) nên phương trình luôn có hai nghiệm phân biệt \(x_1,x_2\).

Theo định lí Viete; 

\(\hept{\begin{cases}x_1+x_2=2\\x_1x_2=-m^2\end{cases}}\)

\(x_1^2+2x_1x_2+2x_2=\left(x_1+x_2\right)^2-x_2^2+2x_2=4-x_2^2+2x_2=3\Leftrightarrow\orbr{\begin{cases}x_2=1+\sqrt{2}\\x_2=1-\sqrt{2}\end{cases}}\)

Suy ra \(x_1x_2=\left(1-\sqrt{2}\right)\left(1+\sqrt{2}\right)=-1\)suy ra \(m=\pm1\).

\(\sqrt{33+20\sqrt{2}}=\sqrt{33+2.5.2\sqrt{2}}\)

\(=\sqrt{25+2.5.2\sqrt{2}+\left(2\sqrt{2}\right)^2}=\sqrt{\left(5+2\sqrt{2}\right)^2}\)

\(=\left|5+2\sqrt{2}\right|=5+2\sqrt{2}\)vì \(5+2\sqrt{2}>0\)

\(\sqrt{27+12\sqrt{2}}=\sqrt{3+2.\sqrt{3}.2\sqrt{3}.\sqrt{2}+24}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}.2\sqrt{6}+\left(2\sqrt{6}\right)^2}\)

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

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

<=> \(\sqrt{\left(x-2\right)^2}=4\)

<=> \(\left|x-2\right|=4\)

<=> \(\orbr{\begin{cases}x-2=4\\x-2=-4\end{cases}}\)

<=> \(\orbr{\begin{cases}x=6\\x=-2\end{cases}}\)

Vậy S= {-2; 6}

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

\(\Leftrightarrow\left|x-2\right|=2\)

\(TH1:x\ge2\Rightarrow x=4\left(tm\right)\)

\(TH2:x< 2\Rightarrow x=0\left(tm\right)\)

KL: pt có 2 nghiệm ........

\(\Delta=\left(m-2\right)^2+36>0\) nên phương trình luôn có hai nghiệm phân biệt \(x_1,x_2\).

Theo định lí Viete: 

\(\hept{\begin{cases}x_1+x_2=2-m\\x_1x_2=-8\end{cases}}\)

\(A=\left(x_1^2-1\right)\left(x_2^2-4\right)=x_1^2x_2^2-4x_1^2-x_2^2+4=x_1^2x_2^2+4x_1x_2+4-\left(4x_1^2+x_2^2+4x_1x_2\right)\)

\(=\left(x_1x_2+2\right)^2-\left(2x_1+x_2\right)^2=\left(-8+2\right)^2-\left(2x_1+x_2\right)^2=36-\left(2x_1+x_2\right)^2\le36\)

Dấu \(=\)khi \(2x_1=-x_2\)suy ra \(m=4\).