\(\left(5\sqrt{7}-\sqrt{11}\right)\left(5\sqrt{7}+\sqrt{11}\right)=\left(5\sqrt{7}\right)^2-\left(\sqrt{11}\right)^2=245-11=234\)

164

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

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

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

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

15-6=11

6+33=39

12-6=2

dễ

\(\sqrt{4-\sqrt{15}}=\dfrac{\sqrt{2}.\sqrt{4-\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{8-2\sqrt{15}}}{\sqrt{2}}=\dfrac{\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}}{\sqrt{2}}=\dfrac{\left|\sqrt{5}-\sqrt{3}\right|}{\sqrt{2}}=\dfrac{\sqrt{5}-\sqrt{3}}{\sqrt{2}}=\dfrac{\sqrt{10}-\sqrt{6}}{2}\)

ngu thế

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

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

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

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

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

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

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