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Câu hỏi: Cho ${{9}^{x}}+{{9}^{-x}}=14.$ Tính giá trị của biểu thức $P=\dfrac{6-3\left( {{3}^{x}}+{{3}^{-x}} \right)}{12+{{3}^{x+1}}+{{3}^{1-x}}}.$
A. $-\dfrac{1}{6}.$
B. $\dfrac{1}{6}.$
C. $-\dfrac{1}{4}.$
D. $\dfrac{1}{4}.$
A. $-\dfrac{1}{6}.$
B. $\dfrac{1}{6}.$
C. $-\dfrac{1}{4}.$
D. $\dfrac{1}{4}.$
Ta có ${{9}^{x}}+{{9}^{-x}}=14\Leftrightarrow {{\left( {{3}^{x}}+{{3}^{-x}} \right)}^{2}}=16\Leftrightarrow {{3}^{x}}+{{3}^{-x}}=4$
$\Rightarrow P=\dfrac{6-3\left( {{3}^{x}}+{{3}^{-x}} \right)}{12+{{3}^{x+1}}+{{3}^{1-x}}}=\dfrac{6-3\left( {{3}^{x}}+{{3}^{-x}} \right)}{12+3\left( {{3}^{x}}+{{3}^{-x}} \right)}=-\dfrac{1}{4}$.
$\Rightarrow P=\dfrac{6-3\left( {{3}^{x}}+{{3}^{-x}} \right)}{12+{{3}^{x+1}}+{{3}^{1-x}}}=\dfrac{6-3\left( {{3}^{x}}+{{3}^{-x}} \right)}{12+3\left( {{3}^{x}}+{{3}^{-x}} \right)}=-\dfrac{1}{4}$.
Đáp án C.