Lời giải:
a) ĐKXĐ:
\(\left\{\begin{matrix} 2-x\neq 0\\ x^2-4\neq 0\\ 2+x\neq 0\\ x^2-3x\neq 0\\ 2x^2-x^3\neq 0\end{matrix}\right.\) \(\Leftrightarrow \left\{\begin{matrix} x\neq 2\\ x\neq -2\\ x\neq 3\\ x\neq 0\end{matrix}\right.\)
Rút gọn:
\(P=\left ( \frac{2+x}{2-x}+\frac{4x^2}{x^2-4}-\frac{2-x}{2+x} \right ):\frac{x^2-3x}{2x^2-x^3}\)
\(P=\left ( \frac{(2+x)^2}{4-x^2}-\frac{4x^2}{4-x^2}-\frac{(2-x)^2}{4-x^2} \right ):\frac{x-3}{x(2-x)}\)
\(P=\frac{(x+2)^2-4x^2-(2-x)^2}{4-x^2}.\frac{x(2-x)}{x-3}\)
\(=\frac{4x(2-x)}{4-x^2}.\frac{x(2-x)}{x-3}=\frac{4x^2(2-x)}{(x+2)(x-3)}\)
b) Có: \(|x-5|=2\Leftrightarrow \) \(\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\\x=3\end{matrix}\right.\)
TH $x=3$ loại do không thỏa mãn ĐKXĐ
\(x=7\Rightarrow P=\frac{-245}{9}\)
d) Để P<0 thì \(\frac{4x^2(2-x)}{(x+2)(x-3)}< 0\)
\(\Leftrightarrow \frac{2-x}{(x+2)(x-3)}< 0\) (do \(4x^2>0\) )
Khi đó xảy ra các TH:
1. \(\left\{\begin{matrix} 2-x> 0\\ (x+2)(x-3)< 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x< 2\\ -2< x< 3\end{matrix}\right.\)
\(\Leftrightarrow -2< x< 2\) kêt hợp với \(x\neq 0\)
2. \(\left\{\begin{matrix} 2-x< 0\\ (x+2)(x-3)> 0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x>2\\ x< -2 \text{or}x>3\end{matrix}\right.\)
\(\Leftrightarrow x>3\)
câu b cái / .../ có nghĩa là j , giá trị tuyệt đối hay cái j?
Có @Akai Haruma trả lời r , nhg vì bn ấy nhắn tin nhờ mk lm chi tiết phân rút gọn nên mk lm lại nhé .
\(P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
\(=\left(\dfrac{-x-2}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{2-x}{x+2}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
\(=\left(\dfrac{-\left(x+2\right)^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{\left(2-x\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
\(=\dfrac{-x^2-4x-4+4x^2+x^2-4x+4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x^2-3x}{2x^2-x^3}\)
\(=\dfrac{4x^2-8x}{\left(x-2\right)\left(x+2\right)}:\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}:\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{4x\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}.\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{4x^2\left(2-x\right)}{\left(x+2\right)\left(x-3\right)}=\dfrac{8x^2-4x^3}{x^2-x-6}\)
