Ta có: \(\left(2x-3\right)^2=x\left(4x-3\right)\)
\(\Leftrightarrow4x^2-12x+9=4x^2-3x\)
\(\Leftrightarrow4x^2-12x+9-4x^2+3x=0\)
\(\Leftrightarrow-9x+9=0\)
\(\Leftrightarrow-9\left(x-1\right)=0\)
mà -9≠0
nên x-1=0
hay x=1
Vậy: x=1
(2x - 3)2 = x(4x - 3)
\(\Leftrightarrow\) 4x2 - 12x + 9 = 4x2 - 3x
\(\Leftrightarrow\) 4x2 - 4x2 - 12x + 3x = -9
\(\Leftrightarrow\) -9x = -9
\(\Leftrightarrow\) x = 1
Vậy S = {1}
Chúc bn học tốt!!
\(\left(2x-3\right)^2=x\left(4x-3\right)\)
\(\Leftrightarrow4x^2-12x+9=4x^2-3x\)
\(\Leftrightarrow-9x=-9\)
\(\Leftrightarrow x=1\)
