\(Q=\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}+\dfrac{2\sqrt{x}}{x-4}\left(dk:x\ge0,x\ne4\right)\\ =\dfrac{2}{2+\sqrt{x}}+\dfrac{1}{2-\sqrt{x}}-\dfrac{2\sqrt{x}}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{2\left(2-\sqrt{x}\right)+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{4-2\sqrt{x}+2+\sqrt{x}-2\sqrt{x}}{4-x}\\ =\dfrac{-3\sqrt{x}+6}{\left(2-\sqrt{x}\right)\left(2+\sqrt{x}\right)}\\ =\dfrac{-3\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\\ =\dfrac{3}{\sqrt{x}+2}\)

\(b,Q=\dfrac{6}{5}\Leftrightarrow\dfrac{3}{\sqrt{x}+2}=\dfrac{6}{5}\Rightarrow15-6\left(\sqrt{x}+2\right)=0\Rightarrow15-6\sqrt{x}-12=0\)

\(\Rightarrow-6\sqrt{x}=-3\Rightarrow\sqrt{x}=\dfrac{1}{2}\Rightarrow x=\dfrac{1}{4}\left(tm\right)\)

Vậy \(x=\dfrac{1}{4}\)thỏa mãn đề bài.

1. P = \(\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\)                       ĐKXĐ: \(x\ne-3\),  \(x\ne2\)

       = \(\frac{x+2}{x+3}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{1}{x-2}\)

       = \(\frac{x^2-4}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}-\frac{x+3}{x-2}\)

       = \(\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

       = \(\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

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

       = \(\frac{x-4}{x-2}\)

2. P=\(\frac{-3}{4}\)

<=> \(\frac{x-4}{x-2}=\frac{-3}{4}\)

<=> 4 ( x - 4 ) = -3  ( x - 2 )

<=> 4x - 16 = -3x + 6

<=> 7x = 2 

<=> x = \(\frac{22}{7}\)

3. \(x^2-9=0\)

<=> ( x -3 ) ( x + 3 ) = 0

<=> \(\orbr{\begin{cases}x=3\left(tm\right)\\x=-3\left(ktm\right)\end{cases}}\)

-> P = \(\frac{3-4}{3-2}\) = -1

M = 1/(x+1).(x+2) + 1/(x+2).(x+3) + 1/(x+3).(x+4) + 1/(x+4).(x+5) + 1/x+5

    = 1/x+1 - 1/x+2 + 1/x+2 - 1/x+3 + 1/x+3 - 1/x+4 + 1/x+4 - 1/x+5 + 1/x+5 = 1/x+1

k mk nha

a) đk : \(x\ne2;-3\)

\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{x^2+x-6}-\frac{x+3}{\left(x-2\right)\left(x+3\right)}\)

\(=\frac{x^2-4-5-x-3}{x^2+x-6}\)

\(=\frac{x^2-x-12}{x^2+x-6}\)

\(=\frac{x^2-4x+3x-12}{x^2+3x-2x-6}\)

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

b)

A>0.

\(\frac{x-4}{x-2}>0\)

th1 : 

x-4>0 và x-2>0

<=> x>4

th2 : x-4 <0 và x-2 < 0

<=> x<2

Vậy để A>0 thì x>4 hoặc x<2

a) \(A=\frac{x+2}{x+3}-\frac{5}{x^2+x-6}+\frac{1}{2-x}\) \(\left(ĐKXĐ:x\ne2;-3\right)\)

\(A=\frac{\left(x+2\right)\left(x-2\right)}{\left(x+3\right)\left(x-2\right)}-\frac{5}{\left(x+3\right)\left(x-2\right)}+\frac{-1\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}\)

\(A=\frac{x^2-4-5-x-3}{\left(x+3\right)\left(x-2\right)}\)

\(A=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}\)

\(A=\frac{\left(x^2-4x\right)+\left(3x-12\right)}{\left(x+3\right)\left(x-2\right)}\)

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

\(A=\frac{x-4}{x-2}\)

b) Để  \(A>0\)thì  \(\frac{x-4}{x-2}>0\)

\(\Rightarrow\)(x - 4) ; (x - 2) cùng dấu

* hoặc  \(\hept{\begin{cases}x-4>0\\x-2>0\end{cases}}\Leftrightarrow\hept{\begin{cases}x>4\\x>2\end{cases}}\Leftrightarrow x>4\)

* hoặc  \(\hept{\begin{cases}x-4< 0\\x-2< 0\end{cases}}\Leftrightarrow\hept{\begin{cases}x< 4\\x< 2\end{cases}}\Leftrightarrow x< 2\)

Vậy  \(\orbr{\begin{cases}x>4\\x< 2\end{cases}}\)

a)\(A=\left(\frac{x+y}{x-2y}+\frac{3y}{2y-x}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)

\(=\left(\frac{x+y-3y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)

\(=\left(\frac{x-2y}{x-2y}-3xy\right).\frac{x+1}{3xy-1}+\frac{x^2}{x+1}\)

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

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

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

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

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

b) Thay \(x=-3,y=2014\)vào (1) ta được:

\(A=\frac{-2.\left(-3\right)-1}{-3+1}=\frac{-5}{2}\)

Vậy \(A=\frac{-5}{2}\)với x=-3 và y=2014