10) Thời gian để xe tải đi hết 300km:
\(\dfrac{300}{x}\left(h\right)\left(x\ne0\right)\)
Thời gian để xe khách đi hết 300km:
\(\dfrac{300}{y}\left(h\right)\left(y\ne0\right)\)
Xe khách đi ít hơn thời gian xe tải đi là:
\(\dfrac{300}{y}-\dfrac{300}{x}\)
\(=\dfrac{300x}{xy}-\dfrac{300y}{xy}\)
\(=\dfrac{300x-300y}{xy}\)
\(=\dfrac{300\left(x-y\right)}{xy}\)
\(1,\dfrac{x}{x+y}-\dfrac{y}{y-x}\left(xy\ne0\right)\\ =\dfrac{x}{x+y}+\dfrac{y}{x-y}\\ =\dfrac{x\left(x-y\right)}{x^2-y^2}+\dfrac{y\left(x+y\right)}{x^2-y^2}\\ =\dfrac{x^2-xy+xy+y^2}{x^2-y^2}\\ =\dfrac{x^2+y^2}{x^2-y^2}\\ 2,\dfrac{1}{x-1}-\dfrac{2}{x^2-1}\left(x\ge0;x\ne1\right)\\ =\dfrac{1}{x-1}-\dfrac{2}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x+1-2}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{x-1}{\left(x-1\right)\left(x+1\right)}\\ =\dfrac{1}{x+2}\)
\(3,\dfrac{x+2}{x^2+xy}-\dfrac{y-2}{xy+y^2}\left(x,y\ne0\right)\\ =\dfrac{x+2}{x\left(x+y\right)}-\dfrac{y-2}{y\left(x+y\right)}\\ =\dfrac{y\left(x+2\right)}{xy\left(x+y\right)}-\dfrac{x\left(y-2\right)}{xy\left(x+y\right)}\\ =\dfrac{xy+2y-xy+2x}{xy\left(x+y\right)}\\ =\dfrac{2y+2x}{xy\left(x+y\right)}\\=\dfrac{2\left(y+x\right)}{xy\left(x+y\right)}\\ =\dfrac{2}{xy}\)
\(4,\dfrac{1}{2x^2-3x}-\dfrac{1}{4x^2-9}\left(x\ne\pm\dfrac{3}{2}\right)\\ =\dfrac{1}{x\left(2x-3\right)}-\dfrac{1}{\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{2x+3}{x\left(2x-3\right)\left(2x+3\right)}-\dfrac{x}{x\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{2x+3-x}{x\left(2x-3\right)\left(2x+3\right)}\\ =\dfrac{x+3}{x\left(2x-3\right)\left(2x+3\right)}\)
