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

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

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

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

a.ĐKXĐ \(\hept{\begin{cases}x\ne-3\\x\ne2\end{cases}}\)

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

=\(\frac{\left(x+2\right)\left(x-2\right)-5-\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}=\frac{x^2-x-12}{\left(x+3\right)\left(x-2\right)}=\frac{\left(x-4\right)\left(x+3\right)}{\left(x+3\right)\left(x-2\right)}\)

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

b. Để A >0  thì \(\frac{x-4}{x-2}\) >0 \(\Rightarrow\orbr{\begin{cases}x< 2\\x>4\end{cases}}\)

Kết hợp ĐK thì \(\orbr{\begin{cases}x< 2,x\ne-3\\x>4\end{cases}}\)

c. \(A=\frac{x-4}{x-2}=1+\frac{-2}{x-2}\)

Để A nguyên thì \(x-2\inƯ\left(-2\right)=\left\{-2;-1;1;2\right\}\)

\(\Rightarrow x\in\left\{0,1,3,4\right\}\)

Khi thay vào A, để A dương thì \(x\in\left\{0;1\right\}\)

Vậy để A nguyên dương thì \(x\in\left\{0;1\right\}\)

Câu c, có thể nói kết hợp với điều kiện giải được trong câu b, ta tìm được \(x\in\left\{0;1\right\}\)

\(a,x\ne2;x\ne-2;x\ne0\)

\(b,A=\left(\frac{x}{x^2-4}+\frac{2}{2-x}+\frac{1}{x+2}\right):\frac{6}{x+2}\)

\(=\frac{x-2\left(x+2\right)+x-2}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}:\frac{6}{x+2}\)

\(=\frac{-6}{\left(x-2\right)\left(x+2\right)}.\frac{x+2}{6}\)

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

\(c,\)Để A > 0 thi \(\frac{1}{2-x}>0\Leftrightarrow2-x>0\Leftrightarrow x< 2\)

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

a)

DK:tồn tại P \(\hept{\begin{cases}x\ne0\\x\ne-+6\\x\ne3\end{cases}}\)

\(P=\left(\frac{x}{\left(x-6\right)\left(x+6\right)}-\frac{x-6}{x\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}\\ \)

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

\(P=\left(\frac{x^2-\left(x^2-12x+36\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}\)

\(P=\left(\frac{12\left(x-3\right)}{x\left(x-6\right)\left(x+6\right)}\right).\frac{x\left(x+6\right)}{2\left(x-3\right)}=\frac{6}{x-6}\)

b)6/(x-6)=1=> x-6=6=> x=12

c)x-6<0=> x<6

dieu kien xac  dinh cua bieu thuc tren la x khac -+6,x khac 3