Tải dùm em để em kt lại bài ạ

Đáp án:

b) x=49

Giải thích các bước giải:

\(\begin{array}{*{20}{l}}
{a)Q = \dfrac{{2\sqrt x {\rm{\;}} - 9 - \left( {\sqrt x {\rm{\;}} + 3} \right)\left( {\sqrt x {\rm{\;}} - 3} \right) + \left( {2\sqrt x {\rm{\;}} + 1} \right)\left( {\sqrt x {\rm{\;}} - 2} \right)}}{{\left( {\sqrt x {\rm{\;}} - 3} \right)\left( {\sqrt x {\rm{\;}} - 2} \right)}}}\\
{ = \dfrac{{2\sqrt x {\rm{\;}} - 9 - x + 9 + 2x - 3\sqrt x  - 2}}{{\left( {\sqrt x {\rm{\;}} - 3} \right)\left( {\sqrt x {\rm{\;}} - 2} \right)}}}\\
{ = \dfrac{{x - \sqrt x  - 2}}{{\left( {\sqrt x {\rm{\;}} - 3} \right)\left( {\sqrt x {\rm{\;}} - 2} \right)}}}\\
{ = \dfrac{{\left( {\sqrt x  - 2} \right)\left( {\sqrt x  + 1} \right)}}{{\left( {\sqrt x  - 3} \right)\left( {\sqrt x {\rm{\;}} - 2} \right)}}}\\
{ = \dfrac{{\sqrt x  + 1}}{{\sqrt x {\rm{\;}} - 3}}}\\
{b)Q = 2}\\
{ \to \dfrac{{\sqrt x  + 1}}{{\sqrt x  - 3}} = 2}\\
{ \to \sqrt x  + 1 = 2\sqrt x {\rm{\;}} - 6}\\
{ \to \sqrt x {\rm{\;}} = 7}\\
{ \to x = 49}\\
{}\\
{c)Q = \dfrac{{\sqrt x  + 1}}{{\sqrt x {\rm{\;}} - 3}} = \dfrac{{\sqrt x {\rm{\;}} - 3 + 4}}{{\sqrt x {\rm{\;}} - 3}} = 1 + \dfrac{4}{{\sqrt x {\rm{\;}} - 3}}}\\
{Q \in Z \Leftrightarrow \dfrac{4}{{\sqrt x {\rm{\;}} - 3}} \in Z}\\
{ \Leftrightarrow \sqrt x {\rm{\;}} - 3 \in U\left( 4 \right)}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{\sqrt x {\rm{\;}} - 3 = 4}\\
{\sqrt x {\rm{\;}} - 3 = {\rm{\;}} - 4\left( l \right)}\\
{\sqrt x {\rm{\;}} - 3 = 2}\\
{\sqrt x {\rm{\;}} - 3 = {\rm{\;}} - 2}\\
{\sqrt x {\rm{\;}} - 3 = 1}\\
{\sqrt x {\rm{\;}} - 3 = {\rm{\;}} - 1}
\end{array}} \right. \to \left[ {\begin{array}{*{20}{l}}
{\sqrt x {\rm{\;}} = 7}\\
{\sqrt x {\rm{\;}} = 5}\\
{\sqrt x {\rm{\;}} = 1}\\
{\sqrt x {\rm{\;}} = 4}\\
{\sqrt x {\rm{\;}} = 2}
\end{array}} \right.}\\
{ \to \left[ {\begin{array}{*{20}{l}}
{x = 49}\\
{x = 25}\\
{x = 1}\\
{x = 16}\\
{x = 4\left( l \right)}
\end{array}} \right.}
\end{array}\)