Đáp án:
B3:
2) \(\left[ \begin{array}{l}
m = 0\\
m = - 2
\end{array} \right.\)
Giải thích các bước giải:
\(\begin{array}{l}
B1:\\
1)Thay:x = 25\\
\to A = \dfrac{{\sqrt {25} - 1}}{{\sqrt {25} }} = \dfrac{4}{5}\\
2)B = \dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right) + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x - 1 + 1 - \sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x \left( {\sqrt x - 1} \right)}}{{\sqrt x \left( {\sqrt x + 1} \right)}}\\
= \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}}\\
3)Xet:P - 1 = \dfrac{{\sqrt x - 1}}{{\sqrt x + 1}} - 1\\
= \dfrac{{\sqrt x - 1 - \sqrt x - 1}}{{\sqrt x + 1}} = \dfrac{{ - 2}}{{\sqrt x + 1}}\\
Do:\sqrt x + 1 > 0\forall x > 0;x \pm 1\\
\to \dfrac{{ - 2}}{{\sqrt x + 1}} < 0\\
\to P < 1\\
B3:\\
1)\left\{ \begin{array}{l}
9x - 3y = 2\\
12x - 3y = 3
\end{array} \right.\\
\to \left\{ \begin{array}{l}
3x = 1\\
4x - y = 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{1}{3}\\
y = \dfrac{1}{3}
\end{array} \right.\\
2)\left\{ \begin{array}{l}
y = 1 - x\\
mx - 1 + x = 2m
\end{array} \right.\\
\to \left\{ \begin{array}{l}
y = 1 - x\\
\left( {m + 1} \right)x = 2m + 1
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2m + 1}}{{m + 1}}\\
y = 1 - \dfrac{{2m + 1}}{{m + 1}} = \dfrac{{m + 1 - 2m - 1}}{{m + 1}}
\end{array} \right.\\
\to \left\{ \begin{array}{l}
x = \dfrac{{2m + 1}}{{m + 1}} = \dfrac{{2\left( {m + 1} \right) - 1}}{{m + 1}} = 2 - \dfrac{1}{{m + 1}}\\
y = \dfrac{{ - m}}{{m + 1}} = \dfrac{{ - \left( {m + 1} \right) + 1}}{{m + 1}} = - 1 + \dfrac{1}{{m + 1}}
\end{array} \right.\\
DK:m \pm - 1\\
Do:x \in Z;y \in Z\\
\to \dfrac{1}{{m + 1}} \in Z\\
\to m + 1 \in U\left( 1 \right)\\
\to \left[ \begin{array}{l}
m + 1 = 1\\
m + 1 = - 1
\end{array} \right.\\
\to \left[ \begin{array}{l}
m = 0\\
m = - 2
\end{array} \right.
\end{array}\)
