Bài 3:
ĐKXĐ: \(x\notin\left\{2;-2\right\}\)
a) Ta có: \(A=\left(\dfrac{x+1}{x-2}+\dfrac{x}{x+2}+\dfrac{2x^2+3}{x^2-4}\right):\left(1-\dfrac{x-3}{x+2}\right)\)
\(=\left(\dfrac{\left(x+1\right)\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{x\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{2x^2+3}{\left(x-2\right)\left(x+2\right)}\right):\left(\dfrac{x+2}{x+2}-\dfrac{x-3}{x+2}\right)\)
\(=\dfrac{x^2+3x+2+x^2-2x+2x^2+3}{\left(x+2\right)\left(x-2\right)}:\dfrac{x+2-x+3}{x+2}\)
\(=\dfrac{4x^2+x+5}{\left(x+2\right)\left(x-2\right)}\cdot\dfrac{x+2}{5}\)
\(=\dfrac{4x^2+x+5}{5\left(x-2\right)}=\dfrac{4x^2+x+5}{5x-10}\)
b) Vì x=-1 thỏa mãn ĐKXĐ nên Thay x=-1 vào biểu thức \(A=\dfrac{4x^2+x+5}{5x-10}\), ta được:
\(A=\dfrac{4\cdot\left(-1\right)^2-1+5}{5\cdot\left(-1\right)-10}=\dfrac{4-1+5}{-5-10}=\dfrac{-8}{15}\)
Vậy: Khi x=-1 thì \(A=-\dfrac{8}{15}\)
c) Để A=-3 thì \(\dfrac{4x^2+x+5}{5x-10}=-3\)
\(\Leftrightarrow4x^2+x+5=-3\left(5x-10\right)\)
\(\Leftrightarrow4x^2+x+5=-15x+30\)
\(\Leftrightarrow4x^2+16x-25=0\)
\(\Leftrightarrow\left(2x\right)^2+2\cdot2x\cdot4+16-41=0\)
\(\Leftrightarrow\left(2x+4\right)^2=41\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+4=\sqrt{41}\\2x+4=-\sqrt{41}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2x=\sqrt{41}-4\\2x=-\sqrt{41}-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqrt{41}-4}{2}\left(nhận\right)\\x=\dfrac{-\sqrt{41}-4}{2}\left(nhận\right)\end{matrix}\right.\)
Vậy: Khi A=-3 thì \(x\in\left\{\dfrac{\sqrt{41}-4}{2};\dfrac{-\sqrt{41}-4}{2}\right\}\)
