Question

[Loan Tran]

Giúp mình bài 1;2 thôi ạ

Answer 1

[Nguyễn Lê Phước Thịnh]

2:

a: ĐKXĐ: \(x\notin\left\{0;-4\right\}\)

\(\dfrac{6}{x^2+4x}+\dfrac{3}{2x+8}\)

\(=\dfrac{6}{x\left(x+4\right)}+\dfrac{3}{2\left(x+4\right)}\)

\(=\dfrac{12+3x}{2x\left(x+4\right)}=\dfrac{3\left(x+4\right)}{2x\left(x+4\right)}=\dfrac{3}{2x}\)

b: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

\(\dfrac{x+1}{x-2}+\dfrac{x-2}{x+2}+\dfrac{x-14}{x^2-4}\)

\(=\dfrac{\left(x+1\right)\cdot\left(x+2\right)+\left(x-2\right)^2+x-14}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x^2+3x+2+x^2-4x+4+x-14}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{2x^2-8}{x^2-4}=2\)

c: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

\(\dfrac{2}{x+1}+\dfrac{-4}{1-x}+\dfrac{5x+1}{1-x^2}\)

\(=\dfrac{2}{x+1}+\dfrac{4}{x-1}-\dfrac{5x+1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{2x-2+4x+4-5x-1}{\left(x+1\right)\left(x-1\right)}=\dfrac{x+1}{\left(x+1\right)\left(x-1\right)}=\dfrac{1}{x-1}\)

d: ĐKXĐ: \(x\ne\pm y\)

\(\dfrac{x}{x^2+xy}+\dfrac{x-3y}{y^2-x^2}+\dfrac{x}{xy-x^2}\)

\(=\dfrac{x}{x\left(x+y\right)}-\dfrac{x-3y}{\left(x-y\right)\left(x+y\right)}-\dfrac{x}{x\left(x-y\right)}\)

\(=\dfrac{1}{x+y}-\dfrac{x-3y}{\left(x-y\right)\left(x+y\right)}-\dfrac{1}{x-y}\)

\(=\dfrac{x-y-x+3y-x-y}{\left(x-y\right)\left(x+y\right)}=\dfrac{-x+y}{\left(x-y\right)\left(x+y\right)}=\dfrac{-1}{x+y}\)

e: ĐKXĐ: \(\left\{{}\begin{matrix}x< >0\\y< >0;x\ne y\end{matrix}\right.\)

\(\dfrac{y}{x^2-xy}+\dfrac{x}{y^2-xy}\)

\(=\dfrac{y}{x\left(x-y\right)}-\dfrac{x}{y\left(x-y\right)}\)

\(=\dfrac{y^2-x^2}{xy\left(x-y\right)}=\dfrac{-\left(x-y\right)\left(x+y\right)}{xy\left(x-y\right)}=\dfrac{-x-y}{xy}\)

f: ĐKXĐ: x<>1

\(\dfrac{11x-4}{x-1}+\dfrac{10x+4}{2-2x}\)

\(=\dfrac{11x-4}{x-1}-\dfrac{5x+2}{x-1}\)

\(=\dfrac{11x-4-5x-2}{x-1}=\dfrac{6x-6}{x-1}=6\)