\(M=\frac{3x+3\sqrt{x}-3}{x+\sqrt{x}-2}-\frac{\sqrt{x}+1}{\sqrt{x}+2}+\frac{\sqrt{x}-2}{\sqrt{x}}.\left(\frac{1}{1-\sqrt{x}}-1\right)\)

\(M=\frac{3x+3\sqrt{x}-3}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)  \(+\frac{\sqrt{x}-2}{\sqrt{x}}.\frac{\sqrt{x}}{\sqrt{x}-1}\)

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

\(M=\frac{3x+3\sqrt{x}-3-x+1+x-4}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

\(M=\frac{3x+3\sqrt{x}-6}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}\)

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

\(M=3\)

b) \(\sqrt{x}=M\)

\(\Leftrightarrow x=M^2\)

thay vào ta có: 

\(x=3^2\)

\(x=9\)

c) \(M=3\in N\)

\(\Rightarrow x=3\)

d) \(M>1\Leftrightarrow x>1\)

Mình bị nhầm

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

Để P\(\in Z\) thì \(\sqrt{x}-1\inƯ\left(2\right)\in\left\{\pm1;\pm2\right\}\)

Vì \(\sqrt{x}-1\ge-1\)

Vậy \(\sqrt{x}-1\in\left\{\pm1;2\right\}\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x}-1=-1\\\sqrt{x}-1=2\\\sqrt{x}-1=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=0\left(tm\right)\\x=4\left(tm\right)\\x=9\left(tm\right)\end{matrix}\right.\)

Vậy x=0, x=4,x=9 thì P\(\in Z\)

a)

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

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

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

b)

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

Vì 1 \(\in Z\) nên

Để P \(\in\) Z thì \(2⋮\sqrt{x}-1=>\sqrt{x}-1\in\) Ư₍₂₎ = { -2;-1;1;2 }

=> \(\sqrt{x}\) = { -1;0;2;3 }

=> x ={0;4;9} thỏa mãn đkxđ

Vậy, ...............

a) ĐKXĐ: x \(\ge\)0; x \(\ne\)4; x \(\ne\)9

Ta có: \(P=\frac{\sqrt{x}+2}{\sqrt{x}-3}-\frac{\sqrt{x}+1}{\sqrt{x}-2}-\frac{3\left(\sqrt{x}+1\right)}{x-5\sqrt{x}+6}\)

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

\(P=\frac{x-4-x+2\sqrt{x}+3-3\sqrt{x}-3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

P = \(\frac{-4+2\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

P = \(\frac{2\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)

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

b) Ta có: P < -1 <=> \(\frac{2}{\sqrt{x}-3}< -1\) <=> \(\frac{2}{\sqrt{x}-3}+1< 0\)

<=> \(\frac{2+\sqrt{x}-3}{\sqrt{x}-3}< 0\) <=> \(\frac{\sqrt{x}-1}{\sqrt{x}-3}< 0\)

TH1: \(\hept{\begin{cases}\sqrt{x}-1< 0\\\sqrt{x}-3>0\end{cases}}\) <=> \(\hept{\begin{cases}x< 1\\x>9\end{cases}}\)(loại)

TH2: \(\hept{\begin{cases}\sqrt{x}-1>0\\\sqrt{x}-3< 0\end{cases}}\) <=> \(\hept{\begin{cases}x>1\\x< 9\end{cases}}\)

Kết hợp vs đk => S = {x|1  < x < 9 và x \(\ne\)4}

c) Để P nguyên <=> 2 \(⋮\)\(\sqrt{x}-3\) <=> \(\sqrt{x}-3\inƯ\left(2\right)=\left\{1;-1;2;-2\right\}\)

Lập bảng: tự làm

@Edogawa Conan phân số thứ 2 bạn bị sai rồi \(\left(\sqrt{x}+1\right)\left(\sqrt{x}-3\right)=x+2\sqrt{x}-3\)

trước phân số là dấu "-" phải đổi dấu