Google
Câu hỏi: Có bao nhiêu cặp số nguyên $\left( x; y \right)$ thỏa mãn
${{\log }_{3}}\left( 5-\left| x \right|+2\left| y \right| \right)+2{{\log }_{2}}\left( 5-\left| x \right| \right)+3\ge {{\log }_{3}}\left| y \right|+{{\log }_{2}}{{\left( 5-\left| x \right|+3\left| y \right| \right)}^{2}}$
A. $50$.
B. $61$.
C. $60$.
D. $51$.
${{\log }_{3}}\left( 5-\left| x \right|+2\left| y \right| \right)+2{{\log }_{2}}\left( 5-\left| x \right| \right)+3\ge {{\log }_{3}}\left| y \right|+{{\log }_{2}}{{\left( 5-\left| x \right|+3\left| y \right| \right)}^{2}}$
A. $50$.
B. $61$.
C. $60$.
D. $51$.
Điều kiện: $\left\{ \begin{aligned}
& x; y\in \mathbb{Z} \\
& 5-\left| x \right|>0 \\
& y\ne 0 \\
\end{aligned} \right.$.
Ta có: ${{\log }_{3}}\left( 5-\left| x \right|+2\left| y \right| \right)+2{{\log }_{2}}\left( 5-\left| x \right| \right)+3\ge {{\log }_{3}}\left| y \right|+{{\log }_{2}}{{\left( 5-\left| x \right|+3\left| y \right| \right)}^{2}}$
$\Leftrightarrow {{\log }_{3}}\left( \dfrac{5-\left| x \right|}{\left| y \right|}+2 \right)+2{{\log }_{2}}\left( \dfrac{5-\left| x \right|}{5-\left| x \right|+3\left| y \right|} \right)+3\ge 0$
$\Leftrightarrow {{\log }_{3}}\left( \dfrac{5-\left| x \right|}{\left| y \right|}+2 \right)-2{{\log }_{2}}\left( 1+3\dfrac{\left| y \right|}{5-\left| x \right|} \right)+3\ge 0$ (*).
Đặt $t=\dfrac{5-\left| x \right|}{\left| y \right|}>0$,
$\left( * \right)\Leftrightarrow {{\log }_{3}}\left( t+2 \right)-2{{\log }_{2}}\left( 1+\dfrac{3}{t} \right)+3\ge 0$ (**).
Xét hàm số $h\left( t \right)={{\log }_{3}}\left( t+2 \right)-2{{\log }_{2}}\left( 1+\dfrac{3}{t} \right)+3$.
${h}'\left( t \right)=\dfrac{1}{\ln 3\left( t+2 \right)}+\dfrac{6}{{{t}^{2}}.\left( 1+\dfrac{3}{t} \right).\ln 2}>0, \forall t>0$, mặt khác $h\left( 1 \right)=0$.
Do đó $\left( ** \right)\Leftrightarrow t\ge 1\Leftrightarrow \dfrac{5-\left| x \right|}{\left| y \right|}\ge 1\Leftrightarrow \left| y \right|\le 5-\left| x \right|$.
Với $x=0\Rightarrow \left| y \right|\le 5\Rightarrow y\in \left( -5,-4,-3,-2,-1,1,2,3,4,5 \right)$, có $10$ cặp $\left( x; y \right)$ thỏa mãn.
Với $x=\pm 1\Rightarrow \left| y \right|\le 4\Rightarrow y\in \left( -4,-3,-2,-1,1,2,3,4 \right)$, có $16$ cặp $\left( x; y \right)$ thỏa mãn.
Với $x=\pm 2\Rightarrow \left| y \right|\le 3\Rightarrow y\in \left( -3,-2,-1,1,2,3 \right)$, có $12$ cặp $\left( x; y \right)$ thỏa mãn.
Với $x=\pm 3\Rightarrow \left| y \right|\le 2\Rightarrow y\in \left( -2,-1,1,2 \right)$, có $8$ cặp $\left( x; y \right)$ thỏa mãn.
Với $x=\pm 4\Rightarrow \left| y \right|\le 1\Rightarrow y\in \left( -1,1 \right)$, có $4$ cặp $\left( x; y \right)$ thỏa mãn.
Vậy có $50$ cặp $\left( x; y \right)$ thỏa mãn.
& x; y\in \mathbb{Z} \\
& 5-\left| x \right|>0 \\
& y\ne 0 \\
\end{aligned} \right.$.
Ta có: ${{\log }_{3}}\left( 5-\left| x \right|+2\left| y \right| \right)+2{{\log }_{2}}\left( 5-\left| x \right| \right)+3\ge {{\log }_{3}}\left| y \right|+{{\log }_{2}}{{\left( 5-\left| x \right|+3\left| y \right| \right)}^{2}}$
$\Leftrightarrow {{\log }_{3}}\left( \dfrac{5-\left| x \right|}{\left| y \right|}+2 \right)+2{{\log }_{2}}\left( \dfrac{5-\left| x \right|}{5-\left| x \right|+3\left| y \right|} \right)+3\ge 0$
$\Leftrightarrow {{\log }_{3}}\left( \dfrac{5-\left| x \right|}{\left| y \right|}+2 \right)-2{{\log }_{2}}\left( 1+3\dfrac{\left| y \right|}{5-\left| x \right|} \right)+3\ge 0$ (*).
Đặt $t=\dfrac{5-\left| x \right|}{\left| y \right|}>0$,
$\left( * \right)\Leftrightarrow {{\log }_{3}}\left( t+2 \right)-2{{\log }_{2}}\left( 1+\dfrac{3}{t} \right)+3\ge 0$ (**).
Xét hàm số $h\left( t \right)={{\log }_{3}}\left( t+2 \right)-2{{\log }_{2}}\left( 1+\dfrac{3}{t} \right)+3$.
${h}'\left( t \right)=\dfrac{1}{\ln 3\left( t+2 \right)}+\dfrac{6}{{{t}^{2}}.\left( 1+\dfrac{3}{t} \right).\ln 2}>0, \forall t>0$, mặt khác $h\left( 1 \right)=0$.
Do đó $\left( ** \right)\Leftrightarrow t\ge 1\Leftrightarrow \dfrac{5-\left| x \right|}{\left| y \right|}\ge 1\Leftrightarrow \left| y \right|\le 5-\left| x \right|$.
Với $x=0\Rightarrow \left| y \right|\le 5\Rightarrow y\in \left( -5,-4,-3,-2,-1,1,2,3,4,5 \right)$, có $10$ cặp $\left( x; y \right)$ thỏa mãn.
Với $x=\pm 1\Rightarrow \left| y \right|\le 4\Rightarrow y\in \left( -4,-3,-2,-1,1,2,3,4 \right)$, có $16$ cặp $\left( x; y \right)$ thỏa mãn.
Với $x=\pm 2\Rightarrow \left| y \right|\le 3\Rightarrow y\in \left( -3,-2,-1,1,2,3 \right)$, có $12$ cặp $\left( x; y \right)$ thỏa mãn.
Với $x=\pm 3\Rightarrow \left| y \right|\le 2\Rightarrow y\in \left( -2,-1,1,2 \right)$, có $8$ cặp $\left( x; y \right)$ thỏa mãn.
Với $x=\pm 4\Rightarrow \left| y \right|\le 1\Rightarrow y\in \left( -1,1 \right)$, có $4$ cặp $\left( x; y \right)$ thỏa mãn.
Vậy có $50$ cặp $\left( x; y \right)$ thỏa mãn.
Đáp án A.