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Câu hỏi: Biết rằng nếu $x\in \mathbb{R}$ thỏa mãn ${{27}^{x}}+{{27}^{-x}}=4048$ thì ${{3}^{x}}+{{3}^{-x}}=6a+b$ trong đó $a,b\in \mathbb{N};1<a\le 9$. Tổng a + b bằng
A. 7.
B. 4.
C. 6.
D. 5.
A. 7.
B. 4.
C. 6.
D. 5.
$\begin{aligned}
& {{27}^{x}}+{{27}^{-x}}=4048\Leftrightarrow {{3}^{3x}}+{{3}^{-3x}}=4048. \\
& \Leftrightarrow {{\left( {{3}^{x}}+{{3}^{-x}} \right)}^{3}}-{{3.3}^{x}}{{.3}^{-x}}.\left( {{3}^{x}}+{{3}^{-x}} \right)=4048 \\
& \Leftrightarrow {{\left( {{3}^{x}}+{{3}^{-x}} \right)}^{3}}-3\left( {{3}^{x}}+{{3}^{-x}} \right)-4048=0\Leftrightarrow {{3}^{x}}+{{3}^{-x}}=16. \\
\end{aligned}$
Theo giả thiết ${{3}^{x}}+{{3}^{-x}}=6a+b\Rightarrow 6a+b=16$ trong đó $a,b\in \mathbb{N};1<a\le 9\Rightarrow \left\{ \begin{aligned}
& a=2 \\
& b=4 \\
\end{aligned} \right.$.
Vậy $a+b=6$.
& {{27}^{x}}+{{27}^{-x}}=4048\Leftrightarrow {{3}^{3x}}+{{3}^{-3x}}=4048. \\
& \Leftrightarrow {{\left( {{3}^{x}}+{{3}^{-x}} \right)}^{3}}-{{3.3}^{x}}{{.3}^{-x}}.\left( {{3}^{x}}+{{3}^{-x}} \right)=4048 \\
& \Leftrightarrow {{\left( {{3}^{x}}+{{3}^{-x}} \right)}^{3}}-3\left( {{3}^{x}}+{{3}^{-x}} \right)-4048=0\Leftrightarrow {{3}^{x}}+{{3}^{-x}}=16. \\
\end{aligned}$
Theo giả thiết ${{3}^{x}}+{{3}^{-x}}=6a+b\Rightarrow 6a+b=16$ trong đó $a,b\in \mathbb{N};1<a\le 9\Rightarrow \left\{ \begin{aligned}
& a=2 \\
& b=4 \\
\end{aligned} \right.$.
Vậy $a+b=6$.
Đáp án C.