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

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

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

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

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

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

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

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

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