a) \(P=\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)+8x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\left(\sqrt{x}-1\right)-2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{3-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-2\right)}\\ =\dfrac{8\sqrt{x}+4x}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{3-\sqrt{x}}=\dfrac{4x}{\sqrt{x}-3}\)

\(\left(x\ge0;x\ne4;9\right)\)

b)\(P=-1\Leftrightarrow4x+\sqrt{x}-3=0\Leftrightarrow\sqrt{x}=\dfrac{3}{4}\Leftrightarrow x=\dfrac{9}{16}\)

c) bpt đưa về dạng \(4mx>x+1\Leftrightarrow\left(4x-1\right)x>1\)

Nếu \(4m-1\le0\) thì tập nghiệm không thể chứa mọi giá trị \(x>9\); Nếu \(4m-1>0\) thì tập nghiệm bpt là \(x>\dfrac{1}{4m-1}\). Do đó bpt tm mọi \(x>9\Leftrightarrow9\ge\dfrac{1}{4m-1}\) và \(4m-1>0\). ta có \(m\ge\dfrac{5}{18}\)

a: \(=\dfrac{4x-8\sqrt{x}+8x}{x-4}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

b: \(m\left(\sqrt{x}-3\right)\cdot B>x+1\)

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

=>\(-12m\cdot x\sqrt{x}+8xm>x\sqrt{x}+2x+\sqrt{x}+2\)

=>\(x\sqrt{x}\left(-12m-1\right)+x\left(8m-2\right)-\sqrt{x}-2>0\)

Để BPT luôn đúng thì m<-0,3

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

b: \(P=M\cdot N\)

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

\(=\dfrac{3x+3\sqrt{x}+6}{\sqrt{x}\left(\sqrt{x}+1\right)\left(x+\sqrt{x}+1\right)}\)

Cái này mình chỉ rút gọn được P thôi, còn P nguyên thì mình xin lỗi bạn rất nhiều nha

uk

a: Ta có: \(N=\dfrac{x^2-\sqrt{x}}{x+\sqrt{x}+1}-\dfrac{2x+\sqrt{x}}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

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

Bài 8:

\(M=1+\frac{4}{\sqrt{x}+1}\)

Để $M$ nguyên thì $\frac{4}{\sqrt{x}+1}$ nguyên 

Đặt $\frac{4}{\sqrt{x}+1}=t$ với $t$ là số nguyên dương 

$\Rightarrow \sqrt{x}+1=\frac{4}{t}$

$\sqrt{x}=\frac{4}{t}-1=\frac{4-t}{t}\geq 0$

$\Rightarrow 4-t\geq 0\Rightarrow t\leq 4$

Mà $t$ nguyên dương suy ra $t=1;2;3;4$

Kéo theo $x=9; 1; \frac{1}{9}; 0$

Kết hợp đkxđ nên $x=0; \frac{1}{9};9$

Bài 9:

$P=1+\frac{5}{\sqrt{x}+2}$

Để $P$ nguyên thì $\frac{5}{\sqrt{x}+2}$ nguyên 

Đặt $\frac{5}{\sqrt{x}+2}=t$ với $t\in\mathbb{Z}^+$

$\Leftrightarrow \sqrt{x}+2=\frac{5}{t}$

$\Leftrightarrow \sqrt{x}=\frac{5-2t}{t}\geq 0$

Với $t>0\Rightarrow 5-2t\geq 0$

$\Leftrightarrow t\leq \frac{5}{2}$

Vì $t$ nguyên dương suy ra $t=1;2$

$\Rightarrow x=9; \frac{1}{4}$ (thỏa đkxđ)

Bài 8: 

Để M nguyên thì \(\sqrt{x}+5⋮\sqrt{x}+1\)

\(\Leftrightarrow\sqrt{x}+1\inƯ\left(4\right)\)

\(\Leftrightarrow\sqrt{x}+1\in\left\{1;2;4\right\}\)

\(\Leftrightarrow\sqrt{x}\in\left\{0;1;3\right\}\)

hay \(x\in\left\{0;1;9\right\}\)

a: ĐKXĐ: x=0; x<>1

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

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

b: Sửa đề: P=1/M

P=1/4-x=-1/x-4

Để P nguyên thì x-4 thuộc {1;-1}

=>x thuộc {5;3}

a) \(P=A:B=\left(\dfrac{\sqrt{x}}{\sqrt{x}-1}-\dfrac{1}{x-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{x-1}\left(đk:x>0,x\ne1\right)\)

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

b) \(P\sqrt{x}=m+\sqrt{x}\)

\(\Leftrightarrow\dfrac{x-1}{\sqrt{x}}.\sqrt{x}=m+\sqrt{x}\)

\(\Leftrightarrow x-1=m+\sqrt{x}\)

\(\Leftrightarrow m=x-\sqrt{x}-1\)

ĐK: \(x\ge0;x\ne9\)

\(M=P:Q=\dfrac{\sqrt{x}}{\sqrt{x}-3}:\dfrac{\sqrt{x}+5}{3-\sqrt{x}}=-\dfrac{\sqrt{x}}{\sqrt{x}+5}\)

\(\Rightarrow\left|M\right|=\dfrac{\sqrt{x}}{\sqrt{x}+5}>\dfrac{1}{2}\)

\(\Leftrightarrow2\sqrt{x}>\sqrt{x}+5\)

\(\Leftrightarrow\sqrt{x}>5\)

\(\Leftrightarrow x>25\)

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

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

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

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

b: M=20

=>\(-4\left(x-4\right)\left(\sqrt{x}-2\right)=-60x+60\sqrt{x}-40\)

=>\(x\sqrt{x}-2x-4\sqrt{x}+8=-15x+15\sqrt{x}-10\)

=>\(x\sqrt{x}+13x-19\sqrt{x}+18=0\)

=>\(x\in\varnothing\)

a: \(M=\dfrac{\sqrt{x}}{\sqrt{x}-2}-\dfrac{4\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}-2\right)}\)

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

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

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

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

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

c: M>0

=>\(\dfrac{\sqrt{x}-2}{\sqrt{x}}>0\)

mà \(\sqrt{x}>0\)

nên \(\sqrt{x}-2>0\)

=>\(\sqrt{x}>2\)

=>x>4