Bài toán này lỗi rồi, sao vận tốc về bé hơn vận tốc đi mà thời gian về lại ít hơn thười gian đi 30 phút được

a)ĐKXĐ:\(x\ne0,2\)

\(A=\left(\dfrac{-x}{2-x}+\dfrac{3x-2}{x\left(2-x\right)}\right):\left(\dfrac{\left(x+2\right)\left(x-1\right)+\left(4-x\right)x}{x\left(x-2\right)}\right)=\left(\dfrac{-x^2}{x\left(2-x\right)}+\dfrac{3x-2}{x\left(2-x\right)}\right):\dfrac{x^2+x-2+4x-x^2}{x\left(x-2\right)}=\dfrac{-x^2+3x-2}{x\left(2-x\right)}:\dfrac{5x-2}{x\left(x-2\right)}=\dfrac{-\left(x-1\right)\left(x-2\right)}{x\left(2-x\right)}.\dfrac{x\left(x-2\right)}{5x-2}=\dfrac{\left(x-1\right)\left(x-2\right)}{x\left(x-2\right)}.\dfrac{x\left(x-2\right)}{5x-2}=\dfrac{\left(x-1\right)\left(x-2\right)}{5x-2}=\dfrac{x^2-3x+2}{5x-2}\)b)\(x^2-5x+6=0\Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)

Với x = 2, ta có A = 0

Với x = 3, ta có A = 2/13

a)Thay x = 3, ta có

\(A=\dfrac{2.3}{3-1}=\dfrac{6}{2}=3\)

b)ĐKXĐ:\(x\ne2,-2\)\(B=\dfrac{x}{x+2}-\dfrac{x^2+8}{x^2-4}+\dfrac{3}{x-2}=\dfrac{x^2-2x+3x+6}{x^2-4}-\dfrac{x^2+8}{x^2-4}=\dfrac{x^2+x+6-x^2-8}{x^2-4}=\dfrac{x-2}{x^2-4}=\dfrac{1}{x+2}\)

c)\(A.B=1\Leftrightarrow3.\dfrac{1}{x+2}=1\Leftrightarrow3=x+2\Leftrightarrow x=1\left(tm\right)\)