Question

A = \(\left(\dfrac{2x}{x^2-x-6}+\dfrac{x}{x-3}\right):\dfrac{x}{x-3}\)
B= \(\dfrac{x-1}{x-3}+\dfrac{x}{1-x}-\dfrac{2}{x^2-4x+3}\)

Answer 1

a: \(A=\dfrac{2x+x\left(x+2\right)}{\left(x-3\right)\cdot\left(x+2\right)}\cdot\dfrac{x-3}{x}\)

\(=\dfrac{2x+x^2+2x}{x+2}\cdot\dfrac{1}{x}=\dfrac{x^2+4x}{x+2}\cdot\dfrac{1}{x}=\dfrac{x+4}{x+2}\)

b: \(B=\dfrac{x^2-2x+1-x^2+3x-2}{\left(x-1\right)\left(x-3\right)}\)

\(=\dfrac{x-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{1}{x-3}\)

Answer 2

\(A=\left(\dfrac{2x}{x^2-x-6}+\dfrac{x}{x-3}\right):\dfrac{x}{x-3}\left(x\ne3;x\ne-2\right)\)

\(=\left[\dfrac{2x}{\left(x+2\right)\left(x-3\right)}+\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-3\right)}\right].\dfrac{x-3}{x}\)

\(=\dfrac{x^2+4x}{\left(x+2\right)\left(x-3\right)}.\dfrac{x-3}{x}\)

\(=\dfrac{x+4}{x+2}\)

Answer 3

\(đk:x\ne3;x\ne0\\ A=\left(\dfrac{2x}{x^2-3x+2x-6}+\dfrac{x}{x-3}\right):\dfrac{x}{x-3}\\ =\left(\dfrac{2x}{\left(x-3\right)\left(x+2\right)}+\dfrac{x}{x-3}\right):\dfrac{x}{x-3}\\ =\left(\dfrac{2x+x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}\right):\dfrac{x}{x-3}\\ =\dfrac{2x+x^2+2x}{\left(x-3\right)\left(x+2\right)}:\dfrac{x}{x-3}\\ =\dfrac{x^2+4x}{\left(x-3\right)\left(x+2\right)}:\dfrac{x}{x-3}\\ =\dfrac{x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}.\dfrac{x-3}{x}=1\)

\(đkx\ne3;x\ne1\\ B=\dfrac{x-1}{x-3}+\dfrac{x}{1-x}-\dfrac{2}{x^2-3x-x+3}\\ =\dfrac{x-1}{x-3}+\dfrac{x}{1-x}-\dfrac{2}{\left(x-3\right)\left(x-1\right)}\\ =\dfrac{\left(x-1\right)^2}{\left(x-3\right)\left(x-1\right)}+\dfrac{-x\left(x-3\right)-2}{\left(x-1\right)\left(x-3\right)}\\ =\dfrac{x^2-2x+1-x^2+3x-2}{\left(x-1\right)\left(x-3\right)}\\ =\dfrac{x+1}{\left(x-1\right)\left(x-3\right)}\)

Answer 4

\(A=\left(\dfrac{2x}{x^2-x-6}+\dfrac{x}{x-3}\right):\dfrac{x}{x-3}\left(ĐK:x\ne\left\{3;-2;0\right\}\right)\\ =\left(\dfrac{2x}{\left(x-3\right)\left(x+2\right)}+\dfrac{x}{x-3}\right).\dfrac{x-3}{x}\\ =\dfrac{2x+x\left(x+2\right)}{\left(x-3\right)\left(x+2\right)}.\dfrac{x-3}{x}\\ =\dfrac{2x+x^2+2x}{x\left(x+2\right)}\\ =\dfrac{x\left(x+4\right)}{x\left(x+2\right)}=\dfrac{x+4}{x+2}\)

\(B=\dfrac{x-1}{x-3}+\dfrac{x}{1-x}-\dfrac{2}{x^2-4x+3}\left(ĐK:x\ne\left\{3;1\right\}\right)\\ =\dfrac{x-1}{x-3}-\dfrac{x}{x-1}-\dfrac{2}{\left(x-1\right)\left(x-3\right)}\\ =\dfrac{\left(x-1\right)^2-x\left(x-3\right)-2}{\left(x-1\right)\left(x-3\right)}\\ =\dfrac{x^2-2x+1-x^2+3x-2}{\left(x-1\right)\left(x-3\right)}\\ =\dfrac{x-1}{\left(x-1\right)\left(x-3\right)}=\dfrac{1}{x-3}\)