Question

cho A=\(\left(\dfrac{2-x}{x+3}-\dfrac{3-x}{x+2}+\dfrac{2-x}{x^2+5x+6}\right):\left(1-\dfrac{x}{x-1}\right)\)

rút gọn 

tìm x để A =0 

              A>0

Answer 1

ĐKXĐ: \(x\ne-3,x\ne-2,x\ne1\)

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

\(=\dfrac{-\left(x+3\right)}{\left(x+3\right)\left(x+2\right)}.\left(1-x\right)=\dfrac{x-1}{x+2}\)

\(A=0\Leftrightarrow\dfrac{x-1}{x+2}=0\Leftrightarrow x=1\left(ktm\right)\Leftrightarrow S=\varnothing\)