\(\left(x+1\right)\left(2x-3\right)-x^2=\left(x-2\right)^2\)
\(\Leftrightarrow2x^2-3x+2x-3-x^2=x^2-4x+4\)
\(\Leftrightarrow x^2-x-3-x^2+4x-4=0\)
\(\Leftrightarrow3x-7=0\)
\(\Leftrightarrow3x=7\)
\(\Leftrightarrow x=\frac{7}{3}\)
Bài 1 :
\(\left(x+1\right)\left(2x-3\right)-x^2=\left(x-2\right)^2\)
\(\Leftrightarrow2x^2-3x+2x-3-x^2=x^2-4x+4\)
\(\Leftrightarrow2x^2-x^2-x^2-3x+2x+4x=4+3\)
\(\Leftrightarrow3x=7\)
\(\Leftrightarrow x=7:3\)
\(\Leftrightarrow x=\frac{7}{3}\)
Vậy \(S=\left\{\frac{7}{3}\right\}\) .
Học tốt
( x + 1 )( 2x - 3 ) - x2 = ( x - 2 )2
<=> 2x2 - x - 3 - x2 = x2 - 4x + 4
<=> 2x2 - x - x2 - x2 + 4x = 4 + 3
<=> 3x = 7
<=> x = 7/3
Vậy S = { 7/3 }