Có bao nhiêu cặp số nguyên (x,y) thỏa mãn

${\log }_3{\left(x^2+y^2+7x+14y\right)}^2+{\log }_2\left(x^2+y^2\right)\le {\log }_2\left(x^2+y^2+30x+60y\right)+2{\log }_3\left(x+2y\right)$ 

Đáp án đúng là: B

Điều kiện $\left\{\begin{array}{c}{x}^2+y^2+7x+14y\ne 0\\ x^2+y^2+30x+60y>0\\ x+2y>0\end{array}\right.\iff x+2y>0$.

${\log }_3{\left(x^2+y^2+7x+14y\right)}^2+{\log }_2\left(x^2+y^2\right)\le {\log }_2\left(x^2+y^2+30x+60y\right)+2{\log }_3\left(x+2y\right)$

$\iff 2{\log }_3\left(x^2+y^2+7x+14y\right)+{\log }_2\left(x^2+y^2\right)\le {\log }_2\left(x^2+y^2+30x+60y\right)+2{\log }_3\left(x+2y\right)\iff 2{\log }_3\left(x^2+y^2+7x+14y\right)+{\log }_2\left(x^2+y^2\right)\le {\log }_2\left(x^2+y^2+30x+60y\right)+2{\log }_3\left(x+2y\right)\iff 2{\log }_3\left(x^2+y^2+7x+14y\right)+{\log }_2\left(x^2+y^2\right)\le {\log }_2\left(x^2+y^2+30x+60y\right)+2{\log }_3\left(x+2y\right)$

$\iff 2{\log }_3\left(x^2+y^2+7x+14y\right)-2{\log }_3\left(x+2y\right)\le {\log }_2\left(x^2+y^2+30x+60y\right)-{\log }_2\left(x^2+y^2\right)$

$\iff 2{\log }_3\frac{x^2+y^2+7x+14y}{x+2y}\le {\log }_2\frac{x^2+y^2+30x+60y}{x^2+y^2}$

$\iff 2{\log }_3\left(\frac{x^2+y^2}{x+2y}+7\right)\le {\log }_2\left(1+30\frac{x+2y}{x^2+y^2}\right)$.

$\iff 2{\log }_3\left(\frac{x^2+y^2}{x+2y}+7\right)-{\log }_2\left(1+30\frac{x+2y}{x^2+y^2}\right)\le 0$

Đặt $t=\frac{x^2+y^2}{x+2y}$ với $t>0$.

Khi đó, bất phương trình tương đương $2{\log }_3\left(t+7\right)-{\log }_2\left(1+\frac{30}{t}\right)\le 0$

Xét hàm số $f\left(t\right)=2{\log }_3\left(t+7\right)-{\log }_2\left(1+\frac{30}{t}\right)$.

Ta có $f^{\text{'}}\left(t\right)=\frac{2}{\left(t+7\right)\ln 3}+\frac{\frac{30}{t^2}}{\left(1+\frac{30}{t}\right)\ln 2}>\mathrm{0,}\forall t>0$.

Nên f(t) đồng biến trên $\left(0;+\infty \right)$.

Mặt khác $f\left(t\right)=0\iff t=2$ nên $2{\log }_3\left(t+7\right)-{\log }_2\left(1+\frac{30}{t}\right)\le 0\iff t\le 2$

$\iff \frac{x^2+y^2}{x+2y}\le 2\iff x^2+y^2\le 2x+4y\iff {\left(x-1\right)}^2+{\left(y-2\right)}^2\le 5$.

Mà $x+2y>0$ nên ${\left(x-1\right)}^2+{\left(y-2\right)}^2\le 5$.

Vậy có tất cả 20 cặp (x,y) thỏa mãn.