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

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

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

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

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

\(a,ĐK:x\ne1;x\ne-1\\ b,C=\dfrac{x^2+x+x^2+1}{2\left(x-1\right)\left(x+1\right)}=\dfrac{2x^2+2x+1}{2x^2-2}\\ c,C=-\dfrac{1}{2}\Leftrightarrow2-2x^2=2x^2+2x+1\\ \Leftrightarrow4x^2+2x-1=0\\ \Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{5}-1}{4}\\x=\dfrac{-\sqrt{5}-1}{4}\end{matrix}\right.\\ d,C>0\Leftrightarrow2x^2-2>0\left(2x^2+2x+1>0\right)\\ \Leftrightarrow\left(x-1\right)\left(x+1\right)>0\\ \Leftrightarrow\left[{}\begin{matrix}x>1\\x< -1\end{matrix}\right.\)

ĐKXĐ : \(\left\{{}\begin{matrix}x-3\ne0\\x+3\ne0\\9-x^2\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne3\\x\ne-3\end{matrix}\right.\)

a, \(A=\dfrac{x-5}{x-3}-\dfrac{2x}{x+3}-\dfrac{2x^2-x+15}{9-x^2}\)

\(=\dfrac{\left(x-5\right)\left(x+3\right)}{\left(x-3\right)\left(x+3\right)}-\dfrac{2x\left(x-3\right)}{\left(x-3\right)\left(x+3\right)}+\dfrac{2x^2-x+15}{\left(x-3\right)\left(x+3\right)}\)

\(=\dfrac{x^2-2x-15-2x^2+6x+2x^2-x+15}{\left(x-3\right)\left(x+3\right)}\)

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

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

b, \(\left|x-1\right|=2\)

\(\Rightarrow\left[{}\begin{matrix}x-1=2\\x-1=-2\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=3\left(kot/m\right)\\x=-1\left(t/m\right)\end{matrix}\right.\)

Thay x =- 1 vào biểu thức A ,có :

\(\dfrac{-1}{-1+3}=\dfrac{-1}{2}\)

Vậy tại x = -1 gtri của bt A là -1/2

Vậy tại x = 3 biểu thức A ko có giá trị

c,\(\dfrac{x}{x+3}=\dfrac{x+3-3}{x+3}=1-\dfrac{3}{x+3}\)

Để A có giá trị nguyên

\(\Leftrightarrow\dfrac{3}{x+3}\) là số nguyên

\(\Leftrightarrow3⋮x+3\)

\(\Leftrightarrow x+3\inƯ\left(3\right)=\left\{1;-1;3;-3\right\}\)

Vậy \(x\in\left\{0;-2;-4;-6\right\}\) thì A có giá trị nguyên

Chócứsủa Đoànngườicứđi Sốngchođángđichúcmàymaymắn

a/ ĐKXĐ: x khác -1

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

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

b/ |x + 1| = 2

\(\Leftrightarrow\left[{}\begin{matrix}x+1=2\\x+1=-2\end{matrix}\right.\)\(\Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-3\left(ktm\right)\end{matrix}\right.\)

Với x = 1 P = \(\dfrac{1+1}{1+3}=\dfrac{2}{4}=\dfrac{1}{2}\)

c/ \(\dfrac{x+1}{x+3}=\dfrac{x+3-2}{x+3}=\dfrac{x+3}{x+3}-\dfrac{2}{x+3}=1-\dfrac{2}{x+3}\)

ĐỂ P nguyên thì \(\dfrac{2}{x+3}\in Z\Leftrightarrow x+3\inƯ\left(2\right)\)

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

=> \(x=\left\{-5;-4;-2;-1\right\}\) (tm)

Vậy............

a: ĐKXĐ:\(x\notin\left\{2;0\right\}\)

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

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

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

c: Thay x=2017 vào C, ta được:

\(C=\dfrac{2017+1}{2\cdot2017}=\dfrac{1009}{2017}\)

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

b: \(C=\dfrac{x}{2x-2}+\dfrac{x^2+1}{2-2x^2}\)

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

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

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

c: \(C=-\dfrac{1}{2}\)

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

=>2x+2=-2

=>2x=-4

=>x=-2(nhận)

d: Để C là số nguyên thì \(2x+2\inƯ\left(1\right)\)

=>\(2x+2\in\left\{1;-1\right\}\)

=>\(2x\in\left\{-1;-3\right\}\)

=>\(x\in\left\{-\dfrac{1}{2};-\dfrac{3}{2}\right\}\)

a, Để C có nghĩa thì \(\hept{\begin{cases}2x-2\ne0\\2-2x\ne0\end{cases}\Rightarrow}x\ne1\)

b, Với x khác 1 thì 

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

c, \(C=-0,5\Rightarrow\frac{x^2-x+1}{2-2x}=\frac{-1}{2}\)

\(\Rightarrow2\left(x^2-x+1\right)=\left(2-2x\right).\left(-1\right)\)

\(\Rightarrow2x^2-2x+2=-2+2x\)

\(\Rightarrow2x^2-2x+2+2-2x=0\)

\(\Rightarrow2x^2-4x+4=0\Rightarrow2\left(x^2-2x+2\right)=0\)

\(x^2-2x+2=\left(x-1\right)^2+1>0\forall x\)

Do đó: \(2\left(x^2-2x+2\right)>0\forall x\)

Vậy \(x\in\varnothing\)