Google
Câu hỏi: Cho biết ${{9}^{x}}-{{12}^{2}}=0,$ tính giá trị biểu thức $P=\dfrac{1}{{{3}^{-x-1}}}-{{8.9}^{\dfrac{x-1}{2}}}+19$
A. 15.
B. 31.
C. 23.
D. 22.
A. 15.
B. 31.
C. 23.
D. 22.
Ta có:
$\begin{aligned}
& {{9}^{x}}-{{12}^{2}}=0\Leftrightarrow {{9}^{x}}={{12}^{2}}\Leftrightarrow x={{\log }_{{{3}^{2}}}}{{12}^{2}}={{\log }_{3}}12\Rightarrow {{3}^{x}}={{3}^{{{\log }_{3}}12}}=12 \\
& P=\dfrac{1}{{{3}^{-x-1}}}-{{8.9}^{\dfrac{x-1}{2}}}+19={{3}^{x+1}}-{{8.3}^{x-1}}+19={{3.3}^{x}}-\dfrac{8}{3}{{.3}^{x}}+19=3.12-\dfrac{8}{3}.12+19=23 \\
\end{aligned}$
$\begin{aligned}
& {{9}^{x}}-{{12}^{2}}=0\Leftrightarrow {{9}^{x}}={{12}^{2}}\Leftrightarrow x={{\log }_{{{3}^{2}}}}{{12}^{2}}={{\log }_{3}}12\Rightarrow {{3}^{x}}={{3}^{{{\log }_{3}}12}}=12 \\
& P=\dfrac{1}{{{3}^{-x-1}}}-{{8.9}^{\dfrac{x-1}{2}}}+19={{3}^{x+1}}-{{8.3}^{x-1}}+19={{3.3}^{x}}-\dfrac{8}{3}{{.3}^{x}}+19=3.12-\dfrac{8}{3}.12+19=23 \\
\end{aligned}$
Đáp án C.