Google
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.