Question

Answer 1

\(a,\) Áp dụng tcdtsbn: 

\(\dfrac{a}{b}=\dfrac{1}{-3}\Rightarrow\dfrac{a}{1}=\dfrac{b}{-3}=\dfrac{a+2b}{1-6}=\dfrac{10}{-5}=-2\\ \Rightarrow\left\{{}\begin{matrix}a=-2\\b=6\end{matrix}\right.\\ b,\Rightarrow\left|2a-3\right|=\dfrac{12}{3}=4\Rightarrow\left[{}\begin{matrix}2a-3=4\\3-2a=4\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=\dfrac{7}{2}\\a=-\dfrac{1}{2}\end{matrix}\right.\\ c,\Rightarrow2\cdot3^{2a-1}-4\cdot3^{2a}=-30\\ \Rightarrow3^{2a-1}\left(2-4\cdot3\right)=-30\\ \Rightarrow3^{2a-1}=-30:\left(-10\right)=3=3^1\\ \Rightarrow2a-1=1\Rightarrow a=1\)