Cho hệ 9x^2-4y^2=5 và log m (3x+2y)-log 3 (3x-2y)=1 có nghiệm

* Hướng dẫn giải

Đáp án C

Ta có: $9x^2-4y^2=5\iff \left(3x+2y\right)\left(3x-2y\right)=5\iff 3x-2y=\frac{5}{3x+2y}$ 

Khi đó: ${\log }_m\left(3x+2y\right)={\log }_3\left(3x-2y\right)=1$

$\iff {\log }_m\left(3x+2y\right)-{\log }_3\left(\frac{5}{3x+2y}\right)=1$ 

$\begin{array}{l}\iff {\log }_m\left(3x+2y\right)+{\log }_3\left(3x+2y\right)-{\log }_{3}5=1\\ \iff {\log }_{m}3.{\log }_3\left(3x+2y\right)+{\log }_3\left(3x+2y\right)={\log }_{3}15\\ \iff {\log }_3\left(3x+2y\right)\left[1+{\log }_{m}3\right]={\log }_{3}15\end{array}$ 

Vì $3x+2y\le 5$ 

nên ${\log }_3\left(3x+2y\right)\le {\log }_{3}5\Rightarrow \frac{{\log }_{3}15}{1+{\log }_{m}3}\le {\log }_{3}5$

$\iff \frac{{\log }_{3}15}{{\log }_{3}5}\le 1+{\log }_{m}3$

$\iff {\log }_{m}3\ge {\log }_{5}15-1={\log }_{5}3\iff m\le 5.$