Question

[22 - Đỗ Nhật Minh - 6A17]

Answer 1

[Dang Tung]

\(\left(a\right):ĐKXĐ:\left\{{}\begin{matrix}x+3\ne0\\3-x\ne0\\x^2-9=\left(x-3\right)\left(x+3\right)\ne0\\x+2\ne0\end{matrix}\right.\\ < =>x\ne\left\{\pm3;-2\right\}\)

\(P=\left(\dfrac{2x-1}{x+3}-\dfrac{x}{3-x}-\dfrac{3-10x}{x^2-9}\right):\dfrac{x+2}{x-3}\\ =\left[\dfrac{2x-1}{x+3}+\dfrac{x}{x-3}-\dfrac{3-10x}{\left(x-3\right)\left(x+3\right)}\right].\dfrac{x-3}{x+2}\\ =\dfrac{\left(2x-1\right)\left(x-3\right)+x\left(x+3\right)-\left(3-10x\right)}{\left(x-3\right)\left(x+3\right)}.\dfrac{x-3}{x+2}\\ =\dfrac{2x^2-x-6x+3+x^2+3x-3+10x}{\left(x+3\right)\left(x+2\right)}\\ =\dfrac{3x^2+6x}{\left(x+3\right)\left(x+2\right)}=\dfrac{3x\left(x+2\right)}{\left(x+3\right)\left(x+2\right)}\\ =\dfrac{3x}{x+3}\)

Answer 2

[Dang Tung]

\(\left(b\right):x^2-4=0< =>x^2=4\\ < =>\left[{}\begin{matrix}x=2\left(TM\right)\\x=-2\left(KTM\right)\end{matrix}\right.\)

Với \(x=2=>P=\dfrac{3.2}{2+3}=\dfrac{6}{5}\)