Bài 2:
a: ĐKXĐ: \(x\ne-2\)
b: \(D=\dfrac{2x^2-4x+8}{x^3+8}\)
\(=\dfrac{2\left(x^2-2x+4\right)}{\left(x+2\right)\left(x^2-2x+4\right)}\)
\(=\dfrac{2}{x+2}\)
c: Khi x=2 thì \(D=\dfrac{2}{2+2}=\dfrac{2}{4}=\dfrac{1}{2}\)
Bài 1:
a: ĐKXĐ: \(x\notin\left\{-3;2\right\}\)
b: \(A=\dfrac{x+2}{x+3}-\dfrac{5}{x^2+x-6}+\dfrac{1}{2-x}\)
\(=\dfrac{x+2}{x+3}-\dfrac{5}{\left(x+3\right)\left(x-2\right)}-\dfrac{1}{x-2}\)
\(=\dfrac{\left(x+2\right)\left(x-2\right)-5-x-3}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-4-x-8}{\left(x-2\right)\left(x+3\right)}\)
\(=\dfrac{x^2-x-12}{\left(x-2\right)\left(x+3\right)}=\dfrac{\left(x-4\right)\left(x+3\right)}{\left(x-2\right)\left(x+3\right)}=\dfrac{x-4}{x-2}\)
c: A=-3/4
=>\(\dfrac{x-4}{x-2}=\dfrac{-3}{4}\)
=>\(4\left(x-4\right)=-3\left(x-2\right)\)
=>4x-16=-3x+6
=>4x+3x=16+6
=>7x=22
=>\(x=\dfrac{22}{7}\left(nhận\right)\)