Google
Câu hỏi: Có bao nhiêu cặp số nguyên $\left( x,y \right)$ thỏa mãn.
B. $20$.
C. $23$.
D. $22$.
${{\log }_{3}}{{\left( {{x}^{2}}+{{y}^{2}}+7x+14y \right)}^{2}}+{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)\le {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+30x+60y \right)+2{{\log }_{3}}\left( x+2y \right)$ ?
A. $21$.B. $20$.
C. $23$.
D. $22$.
Điều kiện $\left\{ \begin{matrix}
{{x}^{2}}+{{y}^{2}}+7x+14y\ne 0 \\
{{x}^{2}}+{{y}^{2}}+30x+60y>0 \\
x+2y>0 \\
\end{matrix} \right.\Leftrightarrow x+2y>0$.
${{\log }_{3}}{{\left( {{x}^{2}}+{{y}^{2}}+7x+14y \right)}^{2}}+{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)\le {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+30x+60y \right)+2{{\log }_{3}}\left( x+2y \right)$
$\Leftrightarrow 2{{\log }_{3}}\left( {{x}^{2}}+{{y}^{2}}+7x+14y \right)+{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)\le {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+30x+60y \right)+2{{\log }_{3}}\left( x+2y \right)$
$\Leftrightarrow 2{{\log }_{3}}\left( {{x}^{2}}+{{y}^{2}}+7x+14y \right)-2{{\log }_{3}}\left( x+2y \right)\le {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+30x+60y \right)-{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)$
$\Leftrightarrow 2{{\log }_{3}}\dfrac{{{x}^{2}}+{{y}^{2}}+7x+14y}{x+2y}\le {{\log }_{2}}\dfrac{{{x}^{2}}+{{y}^{2}}+30x+60y}{{{x}^{2}}+{{y}^{2}}}$
$\Leftrightarrow 2{{\log }_{3}}\left( \dfrac{{{x}^{2}}+{{y}^{2}}}{x+2y}+7 \right)\le {{\log }_{2}}\left( 1+30\dfrac{x+2y}{{{x}^{2}}+{{y}^{2}}} \right)$.
$\Leftrightarrow 2{{\log }_{3}}\left( \dfrac{{{x}^{2}}+{{y}^{2}}}{x+2y}+7 \right)-{{\log }_{2}}\left( 1+30\dfrac{x+2y}{{{x}^{2}}+{{y}^{2}}} \right)\le 0$
Đặt $t=\dfrac{{{x}^{2}}+{{y}^{2}}}{x+2y}$ với $t>0$, khi đó, bất phương trình tương đương $2{{\log }_{3}}\left( t+7 \right)-{{\log }_{2}}\left( 1+\dfrac{30}{t} \right)\le 0$
Xét hàm số $f\left( t \right)=2{{\log }_{3}}\left( t+7 \right)-{{\log }_{2}}\left( 1+\dfrac{30}{t} \right)$.
Ta có ${f}'\left( t \right)=\dfrac{2}{\left( t+7 \right)\ln 3}+\dfrac{\dfrac{30}{{{t}^{2}}}}{\left( 1+\dfrac{30}{t} \right)\ln 2}>0,\forall t>0$.
Nên $f\left( t \right)$ đồng biến trên $\left( 0;+\infty \right)$.
Mặt khác $f\left( t \right)=0\Leftrightarrow t=2$ nên $2{{\log }_{3}}\left( t+7 \right)-{{\log }_{2}}\left( 1+\dfrac{30}{t} \right)\le 0\Leftrightarrow t\le 2$
$\Leftrightarrow \dfrac{{{x}^{2}}+{{y}^{2}}}{x+2y}\le 2\Leftrightarrow {{x}^{2}}+{{y}^{2}}\le 2x+4y\Leftrightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}\le 5$.
Mà $x+2y>0$ nên ${{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}\le 5$. Có tất cả $20$ cặp $\left( x,y \right)$ thỏa mãn.
{{x}^{2}}+{{y}^{2}}+7x+14y\ne 0 \\
{{x}^{2}}+{{y}^{2}}+30x+60y>0 \\
x+2y>0 \\
\end{matrix} \right.\Leftrightarrow x+2y>0$.
${{\log }_{3}}{{\left( {{x}^{2}}+{{y}^{2}}+7x+14y \right)}^{2}}+{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)\le {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+30x+60y \right)+2{{\log }_{3}}\left( x+2y \right)$
$\Leftrightarrow 2{{\log }_{3}}\left( {{x}^{2}}+{{y}^{2}}+7x+14y \right)+{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)\le {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+30x+60y \right)+2{{\log }_{3}}\left( x+2y \right)$
$\Leftrightarrow 2{{\log }_{3}}\left( {{x}^{2}}+{{y}^{2}}+7x+14y \right)-2{{\log }_{3}}\left( x+2y \right)\le {{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}}+30x+60y \right)-{{\log }_{2}}\left( {{x}^{2}}+{{y}^{2}} \right)$
$\Leftrightarrow 2{{\log }_{3}}\dfrac{{{x}^{2}}+{{y}^{2}}+7x+14y}{x+2y}\le {{\log }_{2}}\dfrac{{{x}^{2}}+{{y}^{2}}+30x+60y}{{{x}^{2}}+{{y}^{2}}}$
$\Leftrightarrow 2{{\log }_{3}}\left( \dfrac{{{x}^{2}}+{{y}^{2}}}{x+2y}+7 \right)\le {{\log }_{2}}\left( 1+30\dfrac{x+2y}{{{x}^{2}}+{{y}^{2}}} \right)$.
$\Leftrightarrow 2{{\log }_{3}}\left( \dfrac{{{x}^{2}}+{{y}^{2}}}{x+2y}+7 \right)-{{\log }_{2}}\left( 1+30\dfrac{x+2y}{{{x}^{2}}+{{y}^{2}}} \right)\le 0$
Đặt $t=\dfrac{{{x}^{2}}+{{y}^{2}}}{x+2y}$ với $t>0$, khi đó, bất phương trình tương đương $2{{\log }_{3}}\left( t+7 \right)-{{\log }_{2}}\left( 1+\dfrac{30}{t} \right)\le 0$
Xét hàm số $f\left( t \right)=2{{\log }_{3}}\left( t+7 \right)-{{\log }_{2}}\left( 1+\dfrac{30}{t} \right)$.
Ta có ${f}'\left( t \right)=\dfrac{2}{\left( t+7 \right)\ln 3}+\dfrac{\dfrac{30}{{{t}^{2}}}}{\left( 1+\dfrac{30}{t} \right)\ln 2}>0,\forall t>0$.
Nên $f\left( t \right)$ đồng biến trên $\left( 0;+\infty \right)$.
Mặt khác $f\left( t \right)=0\Leftrightarrow t=2$ nên $2{{\log }_{3}}\left( t+7 \right)-{{\log }_{2}}\left( 1+\dfrac{30}{t} \right)\le 0\Leftrightarrow t\le 2$
$\Leftrightarrow \dfrac{{{x}^{2}}+{{y}^{2}}}{x+2y}\le 2\Leftrightarrow {{x}^{2}}+{{y}^{2}}\le 2x+4y\Leftrightarrow {{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}\le 5$.
Mà $x+2y>0$ nên ${{\left( x-1 \right)}^{2}}+{{\left( y-2 \right)}^{2}}\le 5$. Có tất cả $20$ cặp $\left( x,y \right)$ thỏa mãn.
Đáp án B.
