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

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]`

Gọi CTHH của muối sắt clorua là FeCl_n \(\left(n\in\left\{2;3\right\}\right)\)

\(m_{FeCl_2}=20.32,5\%=6,5\left(g\right)\Rightarrow n_{FeCl_n}=\dfrac{6,5}{56+35,5n}\left(mol\right)\)

PTHH: \(FeCl_n+nAgNO_3\rightarrow nAgCl\downarrow+Fe\left(NO_3\right)_n\)

            \(\dfrac{6,5}{56+35,5n}--\rightarrow\dfrac{6,5n}{56+35,5n}\rightarrow\dfrac{6,5}{56+35,5n}\)

          \(Fe\left(NO_3\right)_n+\left(3-n\right)AgNO_3\rightarrow Fe\left(NO_3\right)_3+\left(3-n\right)Ag\downarrow\)

           \(\dfrac{6,5}{56+35,5n}------------\rightarrow\dfrac{6,5\left(3-n\right)}{56+35,5n}\)

=> \(108.\dfrac{6,5\left(3-n\right)}{56+35,5n}+143,5.\dfrac{6,5n}{56+35,5n}=17,22\)

=> n = 3

Vậy CTHH của muối sắt là FeCl₃

ĐKXĐ: x>=0; x<>9

TA có: \(B=\frac{2\sqrt{a}}{\sqrt{a}+3}-\frac{\sqrt{a}}{3-\sqrt{a}}-\frac{3a+3}{a-9}\)

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

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

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

P=A:B

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

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

\(=-\frac13\sqrt{a}-1=-\left(\frac13\sqrt{a}+1\right)\le-1\)

=>P<=-1 ∀x thỏa mãn ĐKXĐ