Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(B=\dfrac{x^2-1}{x^2}:\dfrac{x^2+x+1}{x^3}=\dfrac{x^2-1}{x^2}\cdot\dfrac{x^3}{x^2+x+1}=\dfrac{x\left(x^2-1\right)}{x^2+x+1}\)
\(9,\dfrac{2}{x^2-2x}=\dfrac{6}{3x\left(x-2\right)};\dfrac{x}{3x-6}=\dfrac{x^2}{3x\left(x-2\right)}\\ 10,\dfrac{x}{x-5}=\dfrac{x}{x-5};x+1=\dfrac{\left(x+1\right)\left(x-5\right)}{x-5}\\ 11,-3=\dfrac{-3\left(x^2+x+5\right)}{x^2+x+5}\\ 12,\dfrac{x}{2x-8}=\dfrac{x^2}{2x\left(x-4\right)};\dfrac{x+1}{4x-x^2}=\dfrac{-2\left(x+1\right)}{2x\left(x-4\right)}\)
a: \(\dfrac{x}{2x^2+7x-15}=\dfrac{x}{\left(x+5\right)\left(2x-3\right)}=\dfrac{x^2-2x}{\left(x+5\right)\left(x-2\right)\left(2x-3\right)}\)
\(\dfrac{x+2}{x^2+3x-10}=\dfrac{x+2}{\left(x+5\right)\left(x-2\right)}=\dfrac{\left(x+2\right)\left(2x-3\right)}{\left(2x-3\right)\left(x+5\right)\left(x-2\right)}\)
\(\dfrac{1}{x+5}=\dfrac{\left(2x-3\right)\left(x-2\right)}{\left(2x-3\right)\left(x-2\right)\left(x+5\right)}\)
b: \(\dfrac{1}{-x^2+3x-2}=\dfrac{-1}{\left(x-1\right)\left(x-2\right)}=\dfrac{-\left(x+6\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x+6\right)\left(x-3\right)}\)
\(\dfrac{1}{x^2+5x-6}=\dfrac{1}{\left(x+6\right)\left(x-1\right)}=\dfrac{\left(x-2\right)\left(x-3\right)}{\left(x+6\right)\left(x-1\right)\left(x-2\right)\left(x-3\right)}\)
\(\dfrac{1}{-x^2+4x-3}=\dfrac{-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{-\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-3\right)\left(x+6\right)\left(x-2\right)}\)
c: \(\dfrac{3}{x^3-1}=\dfrac{3}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{2x}{x^2+x+1}=\dfrac{2x\left(x-1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(\dfrac{x}{x-1}=\dfrac{x\left(x^2+x+1\right)}{\left(x-1\right)\left(x^2+x+1\right)}\)
\(A=\dfrac{\dfrac{x}{x-1}-\dfrac{x+1}{x}}{\dfrac{x}{x+1}-\dfrac{x-1}{x}}=\dfrac{\dfrac{x^2-\left(x^2-1\right)}{x\left(x-1\right)}}{\dfrac{x^2-\left(x^2-1\right)}{x\left(x+1\right)}}=\dfrac{\dfrac{1}{x\left(x-1\right)}}{\dfrac{1}{x\left(x+1\right)}}\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\left\{0;\pm1\right\}\\A=\dfrac{x+1}{x-1}\end{matrix}\right.\)