PT \(\Leftrightarrow9x^2-6x+1-9x+6=9x^2-18x-27\)
\(\Leftrightarrow9x^2-6x+1-9x+6-9x^2+18x+27=0\)
\(\Leftrightarrow3x+34=0\)
\(\Leftrightarrow x=-\dfrac{34}{3}\)
Vậy ...
Đúng 2
Bình luận (0)
Ta có: \(\left(3x-1\right)^2-3\left(3x-2\right)=9\left(x+1\right)\left(x-3\right)\)
\(\Leftrightarrow9x^2-6x+1-9x+6=9\left(x^2-3x+x-3\right)\)
\(\Leftrightarrow9x^2-15x+7=9x^2-18x-27\)
\(\Leftrightarrow9x^2-15x+7-9x^2+18x+27=0\)
\(\Leftrightarrow3x+34=0\)
\(\Leftrightarrow3x=-34\)
\(\Leftrightarrow x=-\dfrac{34}{3}\)
Vậy: \(S=\left\{-\dfrac{34}{3}\right\}\)
Đúng 2
Bình luận (0)