\(4^x-10\times2^x+16=0\)
\(\Leftrightarrow2^{2x}-2\times5\times2^x+16=0\)
\(\Leftrightarrow\left[\left(2^x\right)^2-2\times2^x\times5+25\right]-9=0\)
\(\Leftrightarrow\left(2^x-5\right)^2-3^2=0\)
\(\Leftrightarrow\left(2^x-5-3\right)\left(2^x-5+3\right)=0\)
\(\Leftrightarrow\left(2^x-8\right)\left(2^x-2\right)=0\)
\(\Leftrightarrow2^x-8=0\) hoặc \(2^x-2=0\)
\(\cdot2^x-8=0\Leftrightarrow2^x=8\Leftrightarrow x=3\)
\(\cdot2^x-2=0\Leftrightarrow2^x=2\Leftrightarrow x=1\)
Vậy \(S=\left\{3;1\right\}\)
\(4^x-10\cdot2^x+16=0\)
\(=\left(2^x\right)^2-10\cdot2^x+16=0\)
Đặt \(t=2^x\). Ta có:
\(t^2-10t+16=0\)
\(\Rightarrow t^2-2\cdot t\cdot5+25-9=0\)
\(\Rightarrow\left(t-5\right)^2-3^2=0\)
\(\Rightarrow\left(t-8\right)\left(t-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}t=8\\t=2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)
Vậy S = {1,3}
\(4^x-10.2^x+16=0\)
\(\Leftrightarrow\left(2^x\right)^2-10.2^x=-16\)
\(\Leftrightarrow\left[{}\begin{matrix}2^x\left(2^x-10\right)=2\left(2-10\right)\Leftrightarrow2^x=1\Leftrightarrow2^x=2^0\Leftrightarrow x=0\\2^x\left(2^x-10\right)=8\left(8-10\right)\Leftrightarrow2^x=8\Leftrightarrow2^x=2^3\Leftrightarrow x=3\end{matrix}\right.\)
Vậy \(x\in\left\{0;3\right\}\)
