Đáp án:
B3:
1) \(\left\{ \begin{array}{l}
y = 12\\
x = - 1
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
a)Thay:x = 16\\
\to A = \dfrac{{\sqrt {16} - 1}}{{\sqrt {16} + 1}} = \dfrac{3}{5}\\
b)B = \dfrac{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 1} \right) + 5\left( {\sqrt x + 1} \right) + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 2\sqrt x - 3 + 5\sqrt x + 5 + 4}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 7\sqrt x + 6}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\left( {\sqrt x + 6} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x + 6}}{{\sqrt x - 1}}\\
c)P = A.B = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}.\dfrac{{\sqrt x + 6}}{{\sqrt x - 1}}\\
= \dfrac{{\sqrt x + 6}}{{\sqrt x + 1}} = \dfrac{{\sqrt x + 1 + 5}}{{\sqrt x + 1}}\\
= 1 + \dfrac{5}{{\sqrt x + 1}}\\
P \in Z \to \dfrac{5}{{\sqrt x + 1}} \in Z\\
\to \sqrt x + 1 \in U\left( 5 \right)\\
\to \left[ \begin{array}{l}
\sqrt x + 1 = 5\\
\sqrt x + 1 = 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
\sqrt x = 4\\
\sqrt x = 0
\end{array} \right. \to \left[ \begin{array}{l}
x = 16\\
x = 0
\end{array} \right.\\
B3:\\
1)DK:x \ne - 2;y \ge 3\\
\left\{ \begin{array}{l}
\dfrac{{ - 2}}{{x + 2}} - 4\sqrt {y - 3} = - 14\\
\dfrac{2}{{x + 2}} - 3\sqrt {y - 3} = - 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
- 7\sqrt {y - 3} = - 21\\
\dfrac{2}{{x + 2}} - 3\sqrt {y - 3} = - 7
\end{array} \right.\\
\to \left\{ \begin{array}{l}
\sqrt {y - 3} = 3\\
\dfrac{2}{{x + 2}} = 2
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 12\\
x = - 1
\end{array} \right.\\
2)\left\{ \begin{array}{l}
x = 1 - my\\
m\left( {1 - my} \right) + 4y = 2\left( 1 \right)
\end{array} \right.\\
\left( 1 \right) \to m - {m^2}y + 4y = 2\\
\to \left( {4 - {m^2}} \right)y = 2 - m\\
\to y = \dfrac{{2 - m}}{{\left( {2 - m} \right)\left( {m + 2} \right)}} = \dfrac{1}{{m + 2}}\\
\to x = 1 - m.\dfrac{1}{{m + 2}} = \dfrac{{m + 2 - m}}{{m + 2}}\\
= \dfrac{2}{{m + 2}}\\
DK:m \ne \pm 2\\
Do:x + y > - 5\\
\to \dfrac{2}{{m + 2}} + \dfrac{1}{{m + 2}} > - 5\\
\to \dfrac{3}{{m + 2}} + 5 > 0\\
\to \dfrac{{3 + 5m + 10}}{{m + 2}} > 0\\
\to \dfrac{{13 + 5m}}{{m + 2}} > 0\\
\to \left[ \begin{array}{l}
m > - 2;m \ne 2\\
m < - \dfrac{{13}}{5}
\end{array} \right.
\end{array}\)
