Bài 2:
a) ĐKXĐ: \(\left\{{}\begin{matrix}2x-4\ne0\\2x+4\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow x\ne\pm2\)
b) \(A=\dfrac{x}{2x-4}-\dfrac{x-2}{2x+4}+\dfrac{8}{x^2-4}\)
\(=\dfrac{x}{2\left(x-2\right)}-\dfrac{x-2}{2\left(x+2\right)}+\dfrac{8}{\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)}-\dfrac{\left(x-2\right)^2}{2\left(x-2\right)\left(x+2\right)}+\dfrac{16}{2\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{x^2+2x-x^2+4x-4+16}{2\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{6x+12}{2\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{6\left(x+2\right)}{2\left(x+2\right)\left(x-2\right)}\)
\(=\dfrac{3}{x-2}\)
Thay `x=-4` vào A ta có:
\(A=\dfrac{3}{-4-2}=\dfrac{3}{-6}=-\dfrac{1}{2}\)
c) \(A=\dfrac{3}{x-2}\)
Để A nguyên thì 3 ⋮ x - 2
`=>x-2∈Ư(3)={1;-1;3;-3}`
`=>x∈{3;1;5;-1}`
bài 1:
a: ĐKXĐ: \(x\notin\left\{0;2\right\}\)
\(A=\dfrac{2x^3-4x^2}{3x^2-6x}=\dfrac{2x^2\left(x-2\right)}{3x\left(x-2\right)}=\dfrac{2x^2}{3x}=\dfrac{2x}{3}\)
b:
ĐKXĐ: \(\left(a-1\right)^2+b^2>0\)
\(B=\dfrac{2a+2b-2}{\left(a-1\right)^2-b^2}\)
\(=\dfrac{2\left(a+b-1\right)}{\left(a-1-b\right)\left(a-1+b\right)}\)
\(=\dfrac{2}{a-b-1}\)
c:
ĐKXĐ: \(\left\{{}\begin{matrix}x\ne-1\\y\ne-x\end{matrix}\right.\)
\(C=\dfrac{x^2-xy+x-y}{x^2+xy+x+y}\)
\(=\dfrac{x\left(x-y\right)+\left(x-y\right)}{x\left(x+y\right)+\left(x+y\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+1\right)}{\left(x+y\right)\left(x+1\right)}\)
\(=\dfrac{x-y}{x+y}\)
d:
ĐKXĐ: \(x\notin\left\{-2;1\right\}\)
\(D=\dfrac{x^2-5x+4}{x^2+x-2}\)
\(=\dfrac{\left(x-1\right)\left(x-4\right)}{\left(x+2\right)\left(x-1\right)}\)
\(=\dfrac{x-4}{x+2}\)
