Câu 6:
ĐKXĐ: \(x\ne-\dfrac{1}{3}\)
Để \(\dfrac{9x+4}{3x+1}\in Z\) thì \(9x+4⋮3x+1\)
=>\(9x+3+1⋮3x+1\)
=>\(1⋮3x+1\)
=>\(3x+1\in\left\{1;-1\right\}\)
=>\(3x\in\left\{0;-2\right\}\)
=>\(x\in\left\{0;-\dfrac{2}{3}\right\}\)
mà x nguyên
nên x=0
Câu 2:
a: ĐKXĐ: \(x\notin\left\{2;-2;0\right\}\)
b: \(A=\left(\dfrac{1}{x+2}-\dfrac{2x}{4-x^2}+\dfrac{1}{x-2}\right)\cdot\dfrac{x^2-4x+4}{4x}\)
\(=\left(\dfrac{1}{x+2}+\dfrac{2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{1}{x-2}\right)\cdot\dfrac{\left(x-2\right)^2}{4x}\)
\(=\dfrac{x-2+2x+x+2}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{\left(x-2\right)^2}{4x}\)
\(=\dfrac{4x\left(x-2\right)}{4x\left(x+2\right)}=\dfrac{x-2}{x+2}\)
