\(P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
\(=\left(\dfrac{-x-2}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
a) Đkxđ: \(x-2\ne0\Leftrightarrow x\ne2\)
\(x+2\ne0\Leftrightarrow x\ne-2\)
\(2-x\ne0\Leftrightarrow x\ne2\)
\(x-3\ne0\Leftrightarrow x\ne3\)
b) \(P=\left(\dfrac{-x-2}{x-2}+\dfrac{4x^2}{\left(x-2\right)\left(x+2\right)}+\dfrac{x-2}{x+2}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\left(\dfrac{-\left(x+2\right)\left(x+2\right)+4x^2+\left(x-2\right)\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\right).\dfrac{x^2\left(2-x\right)}{x\left(x-3\right)}\)
\(=\dfrac{-x\left(-x^2-4x-4+4x^2+x^2-4x+4\right)}{\left(x+2\right)\left(x-3\right)}\)
\(=\dfrac{-x\left(4x^2-8x\right)}{\left(x+2\right)\left(x-3\right)}\)
Chắc là mình làm sai rồi. Mình không làm nữa :v. Sorry bạn nha :<<
Bài 8:
a) ĐKXĐ: \(x\notin\left\{0;2;-2;3\right\}\)
b) Ta có: \(P=\left(\dfrac{2+x}{2-x}+\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)
\(=\left(\dfrac{\left(2+x\right)^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{4x^2}{\left(2-x\right)\left(2+x\right)}-\dfrac{\left(2-x\right)^2}{\left(2+x\right)\left(2-x\right)}\right):\dfrac{x\left(x-3\right)}{x^2\left(2-x\right)}\)
\(=\dfrac{x^2+4x+4-4x^2-\left(x^2-4x+4\right)}{\left(2-x\right)\left(2+x\right)}:\dfrac{x-3}{x\left(2-x\right)}\)
\(=\dfrac{-3x^2+4x+4-x^2+4x-4}{\left(2-x\right)\left(2+x\right)}\cdot\dfrac{x\left(2-x\right)}{x-3}\)
\(=\dfrac{-4x^2+8x}{2+x}\cdot\dfrac{x}{x-3}\)
\(=\dfrac{x\left(-4x^2+8x\right)}{\left(x+2\right)\left(x-3\right)}\)
c) Ta có: |x-5|=2
\(\Leftrightarrow\left[{}\begin{matrix}x-5=2\\x-5=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=7\left(nhận\right)\\x=3\left(loại\right)\end{matrix}\right.\)
Thay x=7 vào biểu thức \(P=\dfrac{x\left(-4x^2+8x\right)}{\left(x+2\right)\left(x-3\right)}\), ta được:
\(P=\dfrac{7\cdot\left(-4\cdot7^2+8\cdot7\right)}{\left(7+2\right)\left(7-3\right)}=\dfrac{7\cdot\left(-4\cdot28+56\right)}{9\cdot4}\)
\(\Leftrightarrow P=\dfrac{7\cdot\left(-112+56\right)}{36}=\dfrac{7\cdot\left(-56\right)}{36}=\dfrac{-98}{9}\)
Vậy: Với |x-5|=2 thì \(P=-\dfrac{98}{9}\)
