Đáp án:
c. \(\dfrac{1}{{x + \sqrt 7 }}\)
Giải thích các bước giải:
\(\begin{array}{*{20}{l}}
{a.A = \dfrac{2}{{{x^2} - {y^2}}}.\sqrt {\dfrac{{9\left( {{x^2} + 2xy + {y^2}} \right)}}{4}} }\\
{ = \dfrac{2}{{\left( {x - y} \right)\left( {x + y} \right)}}.\sqrt {\dfrac{{9{{\left( {x + y} \right)}^2}}}{4}} = \dfrac{2}{{\left( {x - y} \right)\left( {x + y} \right)}}.\dfrac{{3\left| {x + y} \right|}}{2}}\\
\begin{array}{l}
= \dfrac{{3\left| {x + y} \right|}}{{\left( {x - y} \right)\left( {x + y} \right)}}\\
\to \left[ \begin{array}{l}
A = \dfrac{{3\left( {x + y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}}\left( {DK:x > - y} \right)\\
A = \dfrac{{ - 3\left( {x + y} \right)}}{{\left( {x - y} \right)\left( {x + y} \right)}}\left( {DK:x < - y} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
A = \dfrac{3}{{x + y}}\\
A = - \dfrac{3}{{x + y}}
\end{array} \right.
\end{array}\\
{b.\dfrac{1}{{2x - 1}}.\sqrt {25{x^4} - 100{x^5} + 100{x^2}} }\\
{ = \dfrac{{\sqrt {25\left( {{x^4} - 4{x^5} + 4{x^2}} \right)} }}{{2x - 1}}}\\
{ = \dfrac{{5\sqrt {{x^4} - 4{x^5} + 4{x^2}} }}{{2x - 1}}}\\
{c.DK:x \ne {\rm{\;}} - \sqrt 7 }\\
{\dfrac{{x + \sqrt 7 }}{{{x^2} + 2x\sqrt 7 {\rm{\;}} + 7}}}\\
{ = \dfrac{{x + \sqrt 7 }}{{{{\left( {x + \sqrt 7 } \right)}^2}}}}\\
{ = \dfrac{1}{{x + \sqrt 7 }}}
\end{array}\)
