Đề sai ạ ! Sửa nhé :
\(S=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^3-4x}{2x^2-x^3}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)}{x-2}+\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{x+2}\right):\frac{x\left(x^2-4\right)}{x^2\left(2-x\right)}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)^2+4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}\right):\frac{\left(x-2\right)\left(x+2\right)}{-x\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x+2\right)\left(x-2\right)}.\frac{-x}{\left(x+2\right)}\)
\(\Leftrightarrow S=\frac{-x\left(4x^2-8x\right)}{\left(x+2\right)^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-4x^2\left(x-2\right)}{\left(x+2\right)^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-4x^2}{\left(x+2\right)^2}\)
P/s : nếu làm theo đề của bạn, sẽ ra kq dài... Nên mik tiện sửa, còn nếu đề bạn đúng rồi thì mik sẽ làm lại ạ !
mình nhầm tí nhé bạn
\(\frac{x^2-3x}{2x^2-x^3}\)
P/s : Làm theo đề đã sửa ạ !
\(S=\left(\frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x}\right):\frac{x^2-3x}{2x^2-x^3}\)
\(\Leftrightarrow S=\left(\frac{-\left(x+2\right)}{x-2}+\frac{4x^2}{\left(x-2\right)\left(x+2\right)}+\frac{x-2}{2+x}\right):\frac{x\left(x-3\right)}{-x^2\left(x-2\right)}\)
\(\Leftrightarrow S=\frac{-\left(x+2\right)^2+4x^2+\left(x-2\right)^2}{\left(x-2\right)\left(x+2\right)}.\frac{-x\left(x-2\right)}{\left(x-3\right)}\)
\(\Leftrightarrow S=\frac{-x\left(-x^2-4x-4+4x^2+x^2-4x+4\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)\left(x-3\right)}.\)
\(\Leftrightarrow S=\frac{-x\left(4x^2-8x\right)}{\left(x+2\right)\left(x-3\right)}\)
\(\Leftrightarrow S=\frac{4x^2\left(2-x\right)}{\left(x+2\right)\left(x-3\right)}\)
