a: ĐKXĐ: \(\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

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

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

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

c: Để P<0 thì x-1<0

hay x<1

Kết hợp ĐKXĐ, ta được: 0<x<1

a) ĐKXĐ: \(x>0,x\ne1\)

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

c) \(P=\dfrac{x-1}{\sqrt{x}}< 0\Leftrightarrow x-1< 0\Leftrightarrow x< 1\)( do \(\sqrt{x}>0\))

Xem lại biểu thức P.

Mình phải đi ăn nên chiều mình làm nốt câu d nhé

a) Điều kiện để P được xác định là: \(x\ne1;x\ne-1\)

b) \(P=\left(\dfrac{x+1}{x-1}-\dfrac{x-1}{x+1}\right):\dfrac{2x}{5x-5}x-\dfrac{x^2-1}{x^2+2x+1}\)

\(P=\left(\dfrac{\left(x+1\right)\left(x-1\right)-\left(x+1\right)\left(x-1\right)}{\left(x+1\right)\left(x-1\right)}\right):\dfrac{2x}{5x-5}x-\dfrac{\left(x+1\right)\left(x-1\right)}{\left(x+1\right)^2}\)

\(P=0:\dfrac{2x}{5x-5}x-\dfrac{x-1}{x+1}\)

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

c) Theo đề ta có:

\(P=2\)

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

\(\Leftrightarrow-\left(x-1\right)=2x+2\)

\(\Leftrightarrow-x-2x=2-1\)

\(\Leftrightarrow-3x=1\)

\(\Leftrightarrow x=-\dfrac{1}{3}\)

d) \(P=-\dfrac{x-1}{x+1}\) nguyên khi:

\(\Leftrightarrow x-1⋮-\left(x+1\right)\)

\(\Leftrightarrow\left(x+1\right)-2⋮-\left(x+1\right)\)

\(\Leftrightarrow-2⋮-\left(x+1\right)\)

\(\Leftrightarrow2⋮x+1\)

\(\Rightarrow x+1\inƯ\left(2\right)\)

Vậy \(P\) nguyên khi \(x\in\left\{-2;0;-3;1\right\}\)

Cậu xem lại đề hộ tớ xem là nhân \(\dfrac{2x}{5x-5}\)hay là chia cho \(\dfrac{2x}{5x-5}\)nhé :>

Lời giải:
ĐK: $x\neq \pm 1$

a) 

\(A=\frac{(x+1)^2-(x-1)^2}{(x-1)(x+1)}.\frac{2x}{5(x-1)}=\frac{4x}{(x-1)(x+1)}.\frac{2x}{5(x-1)}=\frac{8x^2}{5(x-1)^2(x+1)}\)

b) Tại $x=4$ thì: $A=\frac{8.4^2}{5(4-1)^2(4+1)}=\frac{128}{225}$

c) 

$A=5$

$\Leftrightarrow \frac{8x^2}{5(x-1)^2(x+1)}=5$

$\Leftrightarrow 8x^2=25(x-1)^2(x+1)$

PT này không phải không giải được nhưng giải phức tạp và nghiệm cực xấu. Nên mình nghĩ bạn đã viết sai đề?

a: ĐKXĐ: x<>-1

b: \(P=\left(1-\dfrac{x+1}{x^2-x+1}\right)\cdot\dfrac{x^2-x+1}{x+1}\)

\(=\dfrac{x^2-x+1-x-1}{x^2-x+1}\cdot\dfrac{x^2-x+1}{x+1}=\dfrac{x^2-2x}{x+1}\)

c: P=2

=>x^2-2x=2x+2

=>x^2-4x-2=0

=>\(x=2\pm\sqrt{6}\)

Lời giải:

a) ĐKXĐ: \(\left\{\begin{matrix} x+1\neq 0\\ x-1\neq 0\\ 2-2x^2\neq 0\end{matrix}\right.\Leftrightarrow x\neq \pm 1\)

b) 

\(A=\left[\frac{x(x-1)}{(x-1)(x+1)}+\frac{x+1}{(x+1)(x-1)}+\frac{2x}{(x-1)(x+1)}\right].\frac{1}{x+1}=\frac{x^2+2x+1}{(x-1)(x+1)}.\frac{1}{x+1}\)

\(=\frac{(x+1)^2}{(x-1)(x+1)}.\frac{1}{x+1}=\frac{1}{x-1}\)

Để $A$ nguyên thì $1\vdots x-1$

$\Rightarrow x-1\in\left\{\pm 1\right\}$

$\Rightarrow x\in\left\{0;2\right\}$ (đều thỏa mãn đkxđ)

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

Ta có: \(A=\left(\dfrac{x}{x+1}+\dfrac{1}{x-1}-\dfrac{4x}{2-2x^2}\right):\left(x+1\right)\)

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

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

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

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

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

\(=\dfrac{1}{x-1}\)

b) Để A nguyên thì \(1⋮x-1\)

\(\Leftrightarrow x-1\inƯ\left(1\right)\)

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

hay \(x\in\left\{2;0\right\}\)

Kết hợp ĐKXĐ, ta được: \(x\in\left\{2;0\right\}\)

Vậy: Để A nguyên thì \(x\in\left\{2;0\right\}\)

a) ĐKXĐ: \(x\ne-10;x\ne0;x\ne-5\)

b) \(P=\dfrac{x^2+2x}{2x+20}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x^2+2x}{2\left(x+10\right)}+\dfrac{x-5}{x}+\dfrac{50-5x}{2x\left(x+5\right)}\)

\(=\dfrac{x\left(x^2+2x\right)\left(x+5\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{2\left(x-5\right)\left(x+10\right)}{2x\left(x+10\right)\left(x+5\right)}+\dfrac{\left(50-5x\right)\left(x+10\right)}{2x\left(x+5\right)\left(x+10\right)}\)

\(=\dfrac{x^4+7x^3+10x^2+2x^2+10x-100+500-5x^2}{2x\left(x+10\right)\left(x+5\right)}\)

\(=\dfrac{x^4+7x^3+7x^2+10x+400}{2x\left(x+10\right)\left(x+5\right)}\)

c) \(P=0\Rightarrow x^4+7x^3+7x^2+10x+400=0\Leftrightarrow...\)

Số xấu thì câu c, d làm cũng như không. Bạn xem lại đề.