1) Ta có: \(P=\left(\dfrac{1}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}+1}\right):\dfrac{\sqrt{x}}{x+\sqrt{x}}\)

\(=\dfrac{x+\sqrt{x}+1}{\sqrt{x}\left(\sqrt{x}+1\right)}\cdot\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\sqrt{x}}\)

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

Để \(P=\dfrac{7}{2}\) thì \(2x+2\sqrt{x}+2-7\sqrt{x}=0\)

\(\Leftrightarrow2x-4\sqrt{x}-\sqrt{x}+2=0\)

\(\Leftrightarrow2\sqrt{x}\left(\sqrt{x}-2\right)-\left(\sqrt{x}-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{1}{4}\end{matrix}\right.\)

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

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

b: Để P nguyên thì \(\sqrt{x}+1⋮\sqrt{x}-1\)

\(\Leftrightarrow\sqrt{x}-1\in\left\{-1;1;2\right\}\)

hay \(x\in\left\{0;4;9\right\}\)

\(a,P=\dfrac{x\sqrt{x}+26\sqrt{x}-19-2x-6\sqrt{x}+x-4\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\left(x\ge0;x\ne1\right)\\ P=\dfrac{x\sqrt{x}-x+16\sqrt{x}-16}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{\left(x+16\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}\\ P=\dfrac{x+16}{\sqrt{x}+3}\\ b,P=4\Leftrightarrow\dfrac{x+16}{\sqrt{x}+3}=4\\ \Leftrightarrow x+16=4\sqrt{x}+12\\ \Leftrightarrow x-4\sqrt{x}+4=0\Leftrightarrow\left(\sqrt{x}-2\right)^2=0\\ \Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)

\(c,P=\dfrac{x+16}{\sqrt{x}+3}=\dfrac{x-9+25}{\sqrt{x}+3}=\sqrt{x}-3+\dfrac{25}{\sqrt{x}+3}\\ P=\sqrt{x}+3+\dfrac{25}{\sqrt{x}+3}-6\ge2\sqrt{\left(\sqrt{x}+3\right)\cdot\dfrac{25}{\sqrt{x}+3}}-6=2\cdot5-6=4\\ P_{min}=4\Leftrightarrow\left(\sqrt{x}+3\right)^2=25\Leftrightarrow\sqrt{x}+3=5\left(\sqrt{x}+3>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)

\(d,x=3-2\sqrt{2}\Leftrightarrow\sqrt{x}=\sqrt{2}-1\\ \Leftrightarrow P=\dfrac{3-2\sqrt{2}+16}{\sqrt{2}-1+3}=\dfrac{19-2\sqrt{2}}{\sqrt{2}+2}\\ P=\dfrac{\left(19-2\sqrt{2}\right)\left(2-\sqrt{2}\right)}{2}=\dfrac{42-23\sqrt{2}}{2}\)

\(M=\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(x\ge0,x\ne1\right)\)

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

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

2) Thay x=9 vào M đã rút gọn ta được:

\(M=\dfrac{\sqrt{9}-1}{9+\sqrt{9}+1}=\dfrac{2}{13}\)

3) Có \(M=\dfrac{\sqrt{x}-1}{x+\sqrt{x}+1}\)

\(\Leftrightarrow x.M+\sqrt{x}\left(M-1\right)+1+M=0\) (*)

Tại x=0 pt (*) <=> M=-1  (1)

Tại x khác 0, coi pt (*) là pt bậc 2 ẩn \(\sqrt{x}\)

Pt (*) có nghiệm không âm <=> \(\left\{{}\begin{matrix}\Delta\ge0\\S\ge0\\P\ge0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}-3M^2-6M+1\ge0\\\dfrac{1-M}{M}\ge0\\\dfrac{1+M}{M}\ge0\end{matrix}\right.\)

\(\Rightarrow0< M\le\dfrac{-3+2\sqrt{3}}{3}\) (2)

Từ (1) (2)=>  \(M_{min}=-1\) <=> x=0

Lời giải:

ĐKXĐ: $x\geq 0; x\neq 1$

\(P=\frac{x+\sqrt{x}-(x+2)}{\sqrt{x}+1}:\left[\frac{\sqrt{x}(\sqrt{x}-1)}{(\sqrt{x}+1)(\sqrt{x}-1)}+\frac{\sqrt{x}-4}{(\sqrt{x}-1)(\sqrt{x}+1)}\right]\)

\(=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-\sqrt{x}+\sqrt{x}-4}{(\sqrt{x}-1)(\sqrt{x}+1)}\)

\(=\frac{\sqrt{x}-2}{\sqrt{x}+1}:\frac{x-4}{(\sqrt{x}-1)(\sqrt{x}+1)}=\frac{\sqrt{x}-2}{\sqrt{x}+1}.\frac{(\sqrt{x}-1)(\sqrt{x}+1)}{(\sqrt{x}-2)(\sqrt{x}+2)}\)

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

Với mọi $x\geq 0; x\neq 1$ thì $\sqrt{x}+2\geq 2$

$\Rightarrow \frac{3}{\sqrt{x}+2}\leq \frac{3}{2}$

$\Rightarrow P=1-\frac{3}{\sqrt{x}+2}\geq 1-\frac{3}{2}=\frac{-1}{2}$
Vậy $P_{\min}=\frac{-1}{2}$ khi $x=0$

Điều kiện xác định: \(x\ge0;x\ne9\)

1/ \(P=\dfrac{3\sqrt{x}+2}{\sqrt{x}+1}-\dfrac{2\sqrt{x}-3}{3-\sqrt{x}}-\dfrac{3\left(3\sqrt{x}-5\right)}{x-2\sqrt{x}-3}\)

\(=\dfrac{3\sqrt{x}+2}{\sqrt{x}+1}+\dfrac{2\sqrt{x}-3}{\sqrt{x}-3}-\dfrac{9\sqrt{x}-15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{\left(3\sqrt{x}+2\right)\left(\sqrt{x}-3\right)+\left(2\sqrt{x}-3\right)\left(\sqrt{x}+1\right)-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{3x-7\sqrt{x}-6+2x-\sqrt{x}-3-9\sqrt{x}+15}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}\\ =\dfrac{5x-17\sqrt{x}+6}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{\left(5\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)}=\dfrac{5\sqrt{x}-2}{\sqrt{x}+1}\)

b) Khi \(x=4+2\sqrt{3}\Rightarrow\sqrt{x}=\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Ta có \(P=\dfrac{5\left(\sqrt{3}+1\right)-2}{\sqrt{3}+1+1}=\dfrac{5\sqrt{3}+3}{\sqrt{3}+2}\)

c) \(P=\dfrac{5\sqrt{x}-2}{\sqrt{x}+1}=\dfrac{5\left(\sqrt{x}+1\right)-7}{\sqrt{x}+1}=5-\dfrac{7}{\sqrt{x}+1}\)

Ta có \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+1\ge1\Rightarrow P\ge5-\dfrac{7}{1}=-2\)

Dấu = xảy ra \(\Leftrightarrow\sqrt{x}=0\Leftrightarrow x=0\)

Vậy \(P_{min}=-2\) đạt được khi \(x=0\)