Question

Giải phương trình sau

a) \(3^{2x}-30.3^x+81=0\)

b) \(\left(16x^2-1\right)\left(4x+3\right)=64x^3-1\)

Answer 1

\(3^{2x}-30.3^x+81=0\\ \Leftrightarrow\left(3^x\right)^2-2.3^x.15+15^2=144\\ \Leftrightarrow\left(3^x-15\right)^2=144\\ \Leftrightarrow\left(3^x-15\right)^2-144=0\\ \Leftrightarrow\left(3^x-15-12\right)\left(3^x-15+12\right)=0\\ \Leftrightarrow\left(3^x-27\right)\left(3^x-3\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}3^x-27=0\\3^x-3=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)