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Câu hỏi: Tập nghiệm của phương trình ${{4}^{{{x}^{2}}}}={{2}^{x+1}}$ là
A. $S=\left\{ 0;1 \right\}$
B. $S=\left\{ \dfrac{1-\sqrt{5}}{2};\dfrac{1+\sqrt{5}}{2} \right\}$
C. $S=\left\{ -1;\dfrac{1}{2} \right\}$
D. $S=\left\{ -\dfrac{1}{2};1 \right\}$
A. $S=\left\{ 0;1 \right\}$
B. $S=\left\{ \dfrac{1-\sqrt{5}}{2};\dfrac{1+\sqrt{5}}{2} \right\}$
C. $S=\left\{ -1;\dfrac{1}{2} \right\}$
D. $S=\left\{ -\dfrac{1}{2};1 \right\}$
${{4}^{{{x}^{2}}}}={{2}^{x+1}}\Leftrightarrow {{2}^{2{{\text{x}}^{2}}}}={{2}^{x+1}}\Leftrightarrow 2{{\text{x}}^{2}}=x+1\Leftrightarrow 2{{x}^{2}}-x-1=0\Leftrightarrow \left( 2\text{x}+1 \right)\left( x-1 \right)=0$
$\Leftrightarrow \left[ \begin{aligned}
& 2\text{x}+1=0 \\
& x-1=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=-\dfrac{1}{2} \\
& x=1 \\
\end{aligned} \right.$.
$\Leftrightarrow \left[ \begin{aligned}
& 2\text{x}+1=0 \\
& x-1=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=-\dfrac{1}{2} \\
& x=1 \\
\end{aligned} \right.$.
Đáp án D.