Câu 10:
a: ĐKXĐ: \(x\notin\left\{2;-2;-3\right\}\)
\(A=\left(\dfrac{1}{x+2}+\dfrac{5}{x-2}+\dfrac{4}{x^2-4}\right):\dfrac{6}{x+3}\)
\(=\left(\dfrac{1}{x+2}+\dfrac{5}{x-2}+\dfrac{4}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{6}{x+3}\)
\(=\dfrac{x-2+5\left(x+2\right)+4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+3}{6}\)
\(=\dfrac{x+2+5x+10}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{x+3}{6}\)
\(=\dfrac{6x+12}{6\left(x+2\right)}\cdot\dfrac{x+3}{x-2}=\dfrac{x+3}{x-2}\)
b: Để A nguyên thì \(x+3⋮x-2\)
=>\(x-2+5⋮x-2\)
=>\(5⋮x-2\)
=>\(x-2\in\left\{1;-1;5;-5\right\}\)
=>\(x\in\left\{3;1;7;-3\right\}\)
Kết hợp ĐKXĐ, ta được: \(x\in\left\{3;1;7\right\}\)
Bài 11:
a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
\(A=\left(\dfrac{x+2}{x-2}-\dfrac{1}{x+2}-\dfrac{x-4}{4-x^2}\right):\dfrac{1}{x^2-4}\)
\(=\left(\dfrac{x+2}{x-2}-\dfrac{1}{x+2}+\dfrac{x-4}{\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{1}\)
\(=\dfrac{\left(x+2\right)^2-x+2+x-4}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{\left(x-2\right)\left(x+2\right)}{1}\)
\(=\left(x+2\right)^2-2\)
\(=x^2+4x+2\)
b: Để A=14 thì \(x^2+4x+2=14\)
=>\(x^2+4x-12=0\)
=>(x+6)(x-2)=0
=>\(\left[{}\begin{matrix}x=-6\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)
