Question

10/(x-2) - 8/(x+1) помогите решить

Answer 1

$\frac{10}{x-2}-\frac{8}{x+1}=\frac{10(x+1)-8(x-2)}{(x-2)\cdot(x+1)}=\frac{10x+10-8x+16}{x^2+x-2x-2}=\frac{2x+26}{x^2-x-2}$

Answer 2

$\displaystyle\frac{ ^{(x+1)/} 10}{x-2} -\frac{^{(x-2)/} 8}{x+1} =\frac{10*(x+1)-8*(x-2)}{(x-2)(x+1)} =\frac{10x+10-8x+16}{x^2+x-2x-2} =\boxed{\frac{2x+26}{x^2-x-2} } .$