Câu hỏi: Có bao nhiêu số nguyên $a$ sao cho ứng với mỗi $a$ tồn tại số thực $b$ thỏa mãn ${{3}^{b}}+4{{a}^{2}}{{.3}^{-b}}-{{\left( \dfrac{5}{3} \right)}^{b}}=2\sqrt{3}a$
A. $13$.
B. $6$.
C. $7$.
D. $11$.
A. $13$.
B. $6$.
C. $7$.
D. $11$.
Biến đổi giả thiết thành
${{3}^{b}}+4{{a}^{2}}{{.3}^{-b}}-{{\left( \dfrac{5}{3} \right)}^{b}}=$ $2\sqrt{3}a\Leftrightarrow {{9}^{b}}+4{{a}^{2}}-{{5}^{b}}=2\sqrt{3}a{{.3}^{b}}$
$\Leftrightarrow {{\left( {{3}^{b}}-\sqrt{3}a \right)}^{2}}={{5}^{b}}-{{a}^{2}}$ $\Leftrightarrow {{3}^{b}}-\sqrt{3}a=\pm \sqrt{{{5}^{b}}-{{a}^{2}}}=\pm t,\left( t\ge 0 \right)$
$\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{3}a\pm t={{3}^{b}} \\
& {{a}^{2}}+{{t}^{2}}={{5}^{b}} \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& a=\sqrt{{{5}^{b}}}\sin x \\
& t=\sqrt{{{5}^{b}}}\cos x \\
& \sqrt{3}\sin x\pm \cos x={{\left( \dfrac{3}{\sqrt{5}} \right)}^{b}}\Rightarrow a= \\
& x\in \left[ -\dfrac{\pi }{2};\dfrac{\pi }{2} \right] \\
\end{aligned} \right. $ $ {{\left( \sqrt{5} \right)}^{\log \dfrac{3}{\sqrt{5}}\left( \sqrt{3}\sin x\pm \cos x \right)}}\sin x\Rightarrow a\in \left\{ 0,...6 \right\}$
${{3}^{b}}+4{{a}^{2}}{{.3}^{-b}}-{{\left( \dfrac{5}{3} \right)}^{b}}=$ $2\sqrt{3}a\Leftrightarrow {{9}^{b}}+4{{a}^{2}}-{{5}^{b}}=2\sqrt{3}a{{.3}^{b}}$
$\Leftrightarrow {{\left( {{3}^{b}}-\sqrt{3}a \right)}^{2}}={{5}^{b}}-{{a}^{2}}$ $\Leftrightarrow {{3}^{b}}-\sqrt{3}a=\pm \sqrt{{{5}^{b}}-{{a}^{2}}}=\pm t,\left( t\ge 0 \right)$
$\Leftrightarrow \left\{ \begin{aligned}
& \sqrt{3}a\pm t={{3}^{b}} \\
& {{a}^{2}}+{{t}^{2}}={{5}^{b}} \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& a=\sqrt{{{5}^{b}}}\sin x \\
& t=\sqrt{{{5}^{b}}}\cos x \\
& \sqrt{3}\sin x\pm \cos x={{\left( \dfrac{3}{\sqrt{5}} \right)}^{b}}\Rightarrow a= \\
& x\in \left[ -\dfrac{\pi }{2};\dfrac{\pi }{2} \right] \\
\end{aligned} \right. $ $ {{\left( \sqrt{5} \right)}^{\log \dfrac{3}{\sqrt{5}}\left( \sqrt{3}\sin x\pm \cos x \right)}}\sin x\Rightarrow a\in \left\{ 0,...6 \right\}$
Đáp án B.