Đáp án:
a) \(\dfrac{{4y}}{{\sqrt y - 3}}\)
b) y=1
Giải thích các bước giải:
\(\begin{array}{l}
a)DK:y > 0;y \ne 4\\
A = \left( {\dfrac{{4\sqrt y }}{{2 + \sqrt y }} + \dfrac{{8y}}{{4 - y}}} \right):\left( {\dfrac{{\sqrt y - 1}}{{y - 2\sqrt y }} - \dfrac{2}{{\sqrt y }}} \right)\\
= \dfrac{{4\sqrt y \left( {2 - \sqrt y } \right) + 8y}}{{\left( {2 - \sqrt y } \right)\left( {2 + \sqrt y } \right)}}:\dfrac{{\sqrt y - 1 - 2\left( {\sqrt y - 2} \right)}}{{\sqrt y \left( {\sqrt y - 2} \right)}}\\
= \dfrac{{8\sqrt y - 4y + 8y}}{{\left( {2 - \sqrt y } \right)\left( {2 + \sqrt y } \right)}}.\dfrac{{\sqrt y \left( {\sqrt y - 2} \right)}}{{\sqrt y - 1 - 2\sqrt y + 4}}\\
= \dfrac{{4y + 8\sqrt y }}{{\left( {2 - \sqrt y } \right)\left( {2 + \sqrt y } \right)}}.\dfrac{{\sqrt y \left( {\sqrt y - 2} \right)}}{{3 - \sqrt y }}\\
= \dfrac{{4\sqrt y \left( {\sqrt y + 2} \right)}}{{\left( {2 - \sqrt y } \right)\left( {2 + \sqrt y } \right)}}.\dfrac{{\sqrt y \left( {\sqrt y - 2} \right)}}{{3 - \sqrt y }}\\
= \dfrac{{4\sqrt y \left( {\sqrt y + 2} \right)}}{{\left( {\sqrt y - 2} \right)\left( {2 + \sqrt y } \right)}}.\dfrac{{\sqrt y \left( {\sqrt y - 2} \right)}}{{\sqrt y - 3}}\\
= \dfrac{{4y}}{{\sqrt y - 3}}\\
b)A = - 2\\
\to \dfrac{{4y}}{{\sqrt y - 3}} = - 2\\
\to 4y = - 2\sqrt y + 6\\
\to 4y + 2\sqrt y - 6 = 0\\
\to 2\left( {\sqrt y - 1} \right)\left( {2\sqrt y + 3} \right) = 0\\
\to \sqrt y - 1 = 0\left( {do:2\sqrt y + 3 > 0\forall y > 0} \right)\\
\to y = 1
\end{array}\)
( t sửa biểu thức thứ 2 mẫu \({y\sqrt y }\) thành \({\sqrt y }\) bài mới làm đc nha )
