\(\sqrt{21-12\sqrt{3}}\)

\(=\sqrt{12-2.2\sqrt{3}.3+9}\)

\(=\sqrt{\left(\sqrt{12}-3\right)^2}\)

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

\(\sqrt{21-12\sqrt{3}}=\sqrt{\left(2\sqrt{3}-3\right)^2}=\left|2\sqrt{3}-3\right|=2\sqrt{3}-3\)

Đề bài ko rõ nên làm 2 trường hợp :

+) \(A=\dfrac{\sqrt{x}-1}{\sqrt{x}}\) ( Điều kiện : x > 0 )

Trừ A đi 1 ta có :

\(\dfrac{\sqrt{x}-1}{\sqrt{x}}-1=\dfrac{\sqrt{x}-1-\sqrt{x}}{\sqrt{x}}=\dfrac{-1}{\sqrt{x}}< 0\left(\sqrt{x}>0\right)\)

\(\Leftrightarrow A-1< 0\Leftrightarrow A< 1\)

+) \(A=\dfrac{\sqrt{x-1}}{\sqrt{x}}\) ( Điều kiện : x > 1 )

Trừ A đi 1 , ta được : 

\(\dfrac{\sqrt{x-1}}{\sqrt{x}}-1=\dfrac{\sqrt{x-1}-\sqrt{x}}{\sqrt{x}}\) 

Nhận xét : Dễ thấy rằng , với x > 1 thì \(\sqrt{x-1}< \sqrt{x}\)

\(\Rightarrow\sqrt{x-1}-\sqrt{x}< 0\)

Mặt khác \(\sqrt{x}>0\left(x>1\right)\)

\(\Rightarrow A< 1\)

\(\dfrac{4}{a-1}>2\)

\(\dfrac{4}{a-1}-2>0\)

\(\dfrac{4-2\left(a-1\right)}{a-1}>0\)

\(\dfrac{6-2a}{a-1}>0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}6-2a>0\\a-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}6-2a< 0\\a-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}1< a< 3\\a>3;a< 1\left(L\right)\end{matrix}\right.\)

ĐKXĐ: \(a\ne1\)

Ta có: \(\dfrac{4}{a-1}>2\Leftrightarrow\dfrac{4}{a-1}-2>0\)

\(\Leftrightarrow\dfrac{2\left(a-3\right)}{1-a}>0\Leftrightarrow\)\(\left[{}\begin{matrix}\left\{{}\begin{matrix}a-3>0\\1-a>0\end{matrix}\right.\\\left[{}\begin{matrix}a-3< 0\\1-a< 0\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}a>3\\a< 1\end{matrix}\right.\\\left\{{}\begin{matrix}a< 3\\a>1\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow1< a< 3\)

Vậy...

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

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

\(=\sqrt{\left(\sqrt{3}+2\right)^2}=\sqrt{3}+2\left(vì\left(\sqrt{3}+2>0\right)\right)\)

a/

\(\sqrt{7+4\sqrt{3}}=\sqrt{2^2+2.2\sqrt{3}+\left(\sqrt{3}\right)^2}=\)

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

b/

\(\sqrt{21-12\sqrt{3}}=\sqrt{3^2-2.3.2\sqrt{3}+\left(2\sqrt{3}\right)^2}=\)

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

d/ \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\)

\(=\sqrt{3^2+2.3.\sqrt{6}+\left(\sqrt{6}\right)^2}=\sqrt{\left(3+\sqrt{6}\right)^2}=3+\sqrt{6}\)

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

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

\(=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}=\sqrt{5}-\sqrt{2}\left(vì\sqrt{5}-\sqrt{2}>0\right)\)

Lời giải:

$\sqrt{11-2\sqrt{10}}=\sqrt{10-2\sqrt{10}+1}=\sqrt{(\sqrt{10}-1)^2}$

$=|\sqrt{10}-1|=\sqrt{10}-1$

P/s: Lần sau bạn lưu ý đăng đầy đủ yêu cầu đề bài.

\(\sqrt{11-2\sqrt{10}}=\sqrt{10-2\sqrt{10}+1}\\ =\sqrt{\sqrt{10}^2-2.\sqrt{10}.1+1^2}\\ =\sqrt{\left(\sqrt{10}-1\right)^2}\\ =\left|\sqrt{10}-1\right|=\sqrt{10}-1\)

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

\(ĐK:x\ge1\)

\(\Leftrightarrow3x+1+x-1-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

\(\Leftrightarrow4x-2\sqrt{\left(3x+1\right)\left(x-1\right)}=4\)

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

\(\Leftrightarrow\left(3x+1\right)\left(x-1\right)=4x^2-8x+4\)

\(\Leftrightarrow3x^2-3x+x-1=4x^2-8x+4\)

\(\Leftrightarrow x^2-6x+5=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=5\\x=1\end{matrix}\right.\) `(tm)`

Vậy \(S=\left\{5;1\right\}\)

`b)`\(\dfrac{\sqrt{x+27}+\sqrt{27-x}}{\sqrt{27+x}-\sqrt{27-x}}=\dfrac{27}{x}\)

\(ĐK:-27\le x\le27;x\ne0\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}+\sqrt{27-x}\right)\left(\sqrt{x+27}-\sqrt{27-x}\right)}{\left(\sqrt{27+x}-\sqrt{27-x}\right)\left(\sqrt{27+x}-\sqrt{27-x}\right)}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{\left(\sqrt{x+27}\right)^2-\left(\sqrt{27-x}\right)^2}{\left(\sqrt{27+x}-\sqrt{27-x}\right)^2}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{2x}{54-2\sqrt{\left(27+x\right)\left(27-x\right)}}=\dfrac{27}{x}\)

\(\Leftrightarrow\dfrac{x}{27-\sqrt{27^2-x^2}}=\dfrac{27}{x}\)
\(\Leftrightarrow x^2=27^2-\sqrt{27^2-x^2}\)

\(\Leftrightarrow27^2-x^2-\sqrt{27^2-x^2}=0\)

\(\Leftrightarrow\sqrt{27^2-x^2}\left(\sqrt{27^2-x^2}-1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}27^2-x^2=0\\\sqrt{27^2-x^2}=1\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\pm27\\x=\pm\sqrt{27^2-1}\end{matrix}\right.\)

Thế vào pt ta được \(x=\pm27\) là thỏa mãn

Vậy \(S=\left\{\pm27\right\}\)

a) ĐKXĐ: \(x\ge1\)

Ta có: \(\sqrt{3x+1}-\sqrt{x-1}=2\)

\(\Leftrightarrow\sqrt{3x+1}-4+2-\sqrt{x-1}=0\)

\(\Leftrightarrow\dfrac{3x+1-16}{\sqrt{3x+1}+4}+\dfrac{4-x+1}{2+\sqrt{x-1}}=0\)

\(\Leftrightarrow\dfrac{3\left(x-5\right)}{\sqrt{3x+1}+4}+\dfrac{5-x}{2+\sqrt{x-1}}=0\)

\(\Leftrightarrow\left(x-5\right)\left(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}\right)=0\)

+ Nếu \(x-5=0\Leftrightarrow x=5\left(tmđkxđ\right)\)

+ Nếu \(\dfrac{3}{\sqrt{3x+1}+4}-\dfrac{1}{2+\sqrt{x-1}}=0\Leftrightarrow\dfrac{3}{\sqrt{3x+1}+4}=\dfrac{1}{2+\sqrt{x-1}}\)

\(\Leftrightarrow3\sqrt{x-1}+2=\sqrt{3x+1}\Rightarrow6x-6+12\sqrt{x-1}=0\)

\(\sqrt{x-1}\left(6\sqrt{x-1}+12\right)=0\Leftrightarrow\sqrt{x-1}=0\)

\(\Leftrightarrow x=1\left(tmđkxđ\right)\)

Vậy ....