Question

1, Tìm x a, 2*3^x=3^12*34+20*27^4 b, (2^x+1)^2+3*(2^2+1)=2^2*10  

Answer 1

1: Tìm x

a) Ta có: \(2\cdot3^x=3^{12}\cdot34+20\cdot27^4\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot34+20\cdot3^{12}\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot\left(34+20\right)\)

\(\Leftrightarrow2\cdot3^x-3^{12}\cdot54=0\)

\(\Leftrightarrow2\cdot3^x=3^{12}\cdot2\cdot27\)

\(\Leftrightarrow3^x=3^{12}\cdot3^3\)

\(\Leftrightarrow3^x=3^{15}\)

hay x=15

Vậy: x=15

b) Ta có: \(\left(2^x+1\right)^2+3\left(2^2+1\right)=2^2\cdot10\)

\(\Leftrightarrow\left(2^x+1\right)^2=40-3\cdot5=25\)

\(\Leftrightarrow\left[{}\begin{matrix}2^x+1=5\\2^x+1=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}2^x=4\\2^x=-6\left(loại\right)\end{matrix}\right.\Leftrightarrow x=2\)

Vậy: x=2