\(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\\ =\sqrt{25+2.5.\sqrt{2}+2}-\sqrt{16+2.4.\sqrt{2}+2}\\ =\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}\\ =5+\sqrt{2}-4-\sqrt{2}\\ =1\)

Ta có: \(\sqrt{27+10\sqrt{2}}-\sqrt{18+8\sqrt{2}}\)

\(=\sqrt{25+2.5\sqrt{2}+2}-\sqrt{16+2.4\sqrt{2}+2}\)

\(=\sqrt{5^2+2.5\sqrt{2}+\left(\sqrt{2}\right)^2}-\sqrt{4^2+2.4\sqrt{2}+\left(\sqrt{2}\right)^2}\)

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

\(=\left|5+\sqrt{2}\right|-\left|4+\sqrt{2}\right|=5+\sqrt{2}-4-\sqrt{2}=1\)

Ta có : \(2x^2-9x+2=\left(6x-3\right)\sqrt{x^2+1}\)

\(\Leftrightarrow2\left(x^2+1\right)-9x=6x\sqrt{x^2+1}-3\sqrt{x^2+1}\)

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

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

\(\Leftrightarrow\sqrt{x^2+1}=3x\left(\text{vì }2\sqrt{x^2+1}+3>0\forall x\right)\)

\(\Leftrightarrow\left\{{}\begin{matrix}x>0\\8x^2=1\end{matrix}\right.\Leftrightarrow x=\dfrac{1}{2\sqrt{2}}\)

Với `x \ne 0,x \ne 1` có:

`A=([x\sqrt{x}]/[\sqrt{x}-1]-[x^2]/[x\sqrt{x}])(1/\sqrt{x}-1)^2`

`A=([x\sqrt{x}]/[\sqrt{x}-1]-x/\sqrt{x})([1-\sqrt{x}]/\sqrt{x})^2`

`A=[x^2-x(\sqrt{x}-1)]/[\sqrt{x}(\sqrt{x}-1)].[(\sqrt{x}-1)^2]/x`

`A=[x(x-\sqrt{x}-1)]/\sqrt{x}.[\sqrt{x}-1]/x`

`A=[x-\sqrt{x}-1]/[\sqrt{x}-1]`

a) \(\sqrt{11-4\sqrt{7}}=\sqrt{7-2.2.\sqrt{7}+4}=\sqrt{\left(\sqrt{7}-2\right)^2}=\left|\sqrt{7}-2\right|=\sqrt{7}-2\)

b) \(\sqrt{11-6\sqrt{2}}=\sqrt{9-2.3.\sqrt{2}+2}=\sqrt{\left(3-\sqrt{2}\right)^2}=\left|3-\sqrt{2}\right|=3-\sqrt{2}\)

c) \(\sqrt{16-2\sqrt{15}}=\sqrt{15-2.\sqrt{15}.1+1}=\sqrt{\left(\sqrt{15}-1\right)^2}=\left|\sqrt{15}-1\right|=\sqrt{15}-1\)

Các ý còn lại bạn làm tương tự. 

a, 13,2 và 17,4 

b, 0,483 và 0,125