\(a,P=\dfrac{\sqrt{x}+2+\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\dfrac{2-\sqrt{x}}{\sqrt{x}}=\dfrac{-2\sqrt{x}}{\sqrt{x}\left(\sqrt{x}+2\right)}=\dfrac{-2}{\sqrt{x}+2}\\ P=-\dfrac{3}{5}\Leftrightarrow\dfrac{2}{\sqrt{x}+2}=\dfrac{3}{5}\\ \Leftrightarrow3\sqrt{x}+6=10\Leftrightarrow\sqrt{x}=\dfrac{4}{3}\Leftrightarrow x=\dfrac{16}{9}\left(tm\right)\)

Vì \(A=\frac{x^2-2x+2014}{\left(x+1\right)^2}\)

\(\Rightarrow x^2-2x+2014=A\left(x+1\right)^2\)

\(\Leftrightarrow x^2-2x+2014=Ax^2+2Ax+A\)

\(\Leftrightarrow\left(1-A\right)x^2-2\left(A+1\right)x+\left(2014-A\right)=0\)

\(\Delta=4\left(A+1\right)^2-4\left(1-A\right)\left(2014-A\right)\)

\(=8068A-8052\)

Vì A có GTNN nên phương trình có nghiệm

\(\Leftrightarrow8068A-8052\ge0\Leftrightarrow A\ge\frac{2013}{2017}\)

Dấu "=" khi \(x=\frac{2015}{2}\)

a) đkxđ : \(x\ne0,x\ne2,x\ne-2\)

\(A=\dfrac{2x^2+4x}{x^3-4x}+\dfrac{x^2-4}{x^2+2x}+\dfrac{2}{2-x}\)

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

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

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

\(=\dfrac{x-2}{x}\)

b)đkxđ : \(x\ne0,x\ne2,x\ne-2\)

với x = 4 ta có :

\(A=\dfrac{4-2}{4}=\dfrac{1}{2}\)

c)đkxđ : \(x\ne0,x\ne2,x\ne-2\)

\(A=\dfrac{x-2}{x}=1-\dfrac{2}{x}\)

\(A\in Z\Leftrightarrow\dfrac{2}{x}\in Z\Leftrightarrow x\inƯ\left(2\right)\)

\(\left[{}\begin{matrix}x=2\left(ktmdk\right)\\x=-2\left(ktmdk\right)\\x=1\left(tmdk\right)\\x=-1\left(tmdk\right)\end{matrix}\right.\)

vậy A nguyên khi x=1 hoặc x=-1

a, Rút gọn Biểu thức:

A=\(\left(\dfrac{x+2}{2x-4}-\dfrac{x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)

= \(\left(\dfrac{x+2}{2x-4}+\dfrac{-x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)

= \(\left(\dfrac{x+2+-x-2}{2x-4+2x+4}\right):\dfrac{2x}{x2+2x}\)

= 0 \(:\dfrac{2x}{x2+2x}\)

b, \(\left(\dfrac{x+2}{2x-4}-\dfrac{x-2}{2x+4}\right):\dfrac{2x}{x2+2x}\)

Thay tất cả x= -4

=> \(\left(\dfrac{-4+2}{2-4-4}-\dfrac{-4-2}{2-4+4}\right):\dfrac{2.-4}{-4.2+2.-4}\)

= -16 : \(\dfrac{1}{3}\)

= -18

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

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

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

b. Áp dụng BĐT AM-GM

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

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

Vậy $M_{\min}=4$ khi $\sqrt{x}-1=\frac{1}{\sqrt{x}-1}$

$\Rightarrow \sqrt{x}-1=0$

$\Leftrightarrow x=1$

a: Khi x=1 thì\(P=\dfrac{1-2}{1+2}=\dfrac{-1}{2}\)

b: \(=\dfrac{3x+6+5x-6+2x^2-4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x^2+4x}{\left(x-2\right)\left(x+2\right)}=\dfrac{2x}{x-2}\)

c: \(P=A\cdot B=\dfrac{2x}{x-2}\cdot\dfrac{x-2}{x+1}=\dfrac{2x}{x+1}\)

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

P<=2

=>x+1>0

=>x>-1