b: ta có: \(\sqrt{9a^2-6a+1}=a+1\)
\(\Leftrightarrow\left(3a-1\right)^2-\left(a+1\right)^2=0\)
\(\Leftrightarrow\left(3a-1-a-1\right)\left(3a-1+a+1\right)=0\)
\(\Leftrightarrow4a\left(2a-2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}a=0\\a=1\end{matrix}\right.\)
Đúng 0
Bình luận (0)
