\(\sqrt{x+1+\sqrt{x+\dfrac{3}{4}}}=x+1\)

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

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

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

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

\(\Leftrightarrow\left\{{}\begin{matrix}x+\dfrac{1}{2}\ge0\\\left(x+\dfrac{1}{2}\right)^2=x+\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x^2+x+\dfrac{1}{4}=x+\dfrac{3}{4}\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\x^2-\dfrac{1}{2}=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\\left(x-\dfrac{\sqrt{2}}{2}\right)\left(x+\dfrac{\sqrt{2}}{2}\right)=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge-\dfrac{1}{2}\\\left[{}\begin{matrix}x=\dfrac{\sqrt{2}}{2}\left(nhận\right)\\x=\dfrac{-\sqrt{2}}{2}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

\(\Leftrightarrow x=\dfrac{\sqrt{2}}{2}\)

- Vậy phương trình có nghiệm duy nhất là \(x=\dfrac{\sqrt{2}}{2}\)

\(ĐKXĐ:x>0;x\ne4;x\ne9\)

\(A=\left(\dfrac{\sqrt{x}+3}{\sqrt{x}-2}-\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}+2}{x-5\sqrt{x}+6}\right):\left(\dfrac{x-2}{x-\sqrt{x}-2}-1\right)\)

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

\(=\dfrac{x-9-\left(x-4\right)+\sqrt{x}+2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}:\dfrac{\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}\)

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

\(=\dfrac{1}{\sqrt{x-2}}.\dfrac{\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)}{\sqrt{x}}=\dfrac{\sqrt{x}+1}{\sqrt{x}}\)

b. \(P=2A-\dfrac{1}{x}=\dfrac{2\sqrt{x}+2}{\sqrt{x}}-\dfrac{1}{x}=2+\dfrac{2}{\sqrt{x}}-\dfrac{1}{x}\)

- Đặt \(t=\dfrac{1}{\sqrt{x}}\left(t>0\right)\). Khi đó:

\(P=2+2t-t^2=-\left(t^2-2t+1\right)+3=-\left(t-1\right)^2+3\le3\)

- Dấu "=" xảy ra khi \(\left(t-1\right)^2=0\Leftrightarrow t=1\Leftrightarrow\dfrac{1}{\sqrt{x}}=1\Leftrightarrow x=1\left(tmĐKXĐ\right)\)

- Vậy \(MaxP=3\), đạt tại \(x=1\)

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

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]`