Kết quả gọn nhất của tích $\dfrac{{10{x^3}}}{{11{y^2}}}.\dfrac{{121{y^5}}}{{25x}}$  là

$3x(x - 2)$    

$x - 2$

$3{x^2}(x - 2)$

$3x{(x - 2)^2}$

$\dfrac{x}{{x + 1}} - \dfrac{2}{{x + 1}}$.

$\dfrac{{2x}}{{x + 1}} - \dfrac{2}{{x + 1}}$.

$\dfrac{x}{{x - 1}} + \dfrac{1}{{x - 1}}$.

$\dfrac{x}{{x + 1}} + \dfrac{1}{{ - x - 1}}$.

$4{\left( {x + 3} \right)^2}$

$4\left( {x - 3} \right)\left( {x + 3} \right)$

$\left( {x - 3} \right)\left( {x + 3} \right)$

$4{\left( {x - 3} \right)^2}$

$\dfrac{{ - {x^2} + 15}}{{2(x + 3)}}$

$\dfrac{{{x^2} - 15}}{{2(x + 3)}}$

$\dfrac{{ - {x^2} - 15}}{{2(x + 3)}}$

Cả A, B, C đều sai

$\dfrac{{2x - 9}}{{{x^2} - 9}}$

$\dfrac{{2x - 3}}{{{x^2} - 9}}$

$\dfrac{{2x - 9}}{{x - 3}}$

$\dfrac{{x - 6}}{{{x^2} - 9}}$

${\left( {x + 3} \right)^3}$

$3{\left( {x + 3} \right)^2}$

$3{\left( {x + 3} \right)^3}$

${\left( {x + 3} \right)^4}$