\(\left(x+2\right)\left(3-4x\right)-\left(x^2-4x+4\right)=0\)
\(-4x^2+3x-8x+6-x^2+4x-4=0\)
\(-5x^2-x+2=0\Leftrightarrow5x^2+x-2=0\)
\(\Leftrightarrow x^2+\dfrac{2}{10}x=\dfrac{2}{5}\Leftrightarrow\left(x+\dfrac{1}{10}\right)^2=\dfrac{2}{5}+\dfrac{1}{10^2}=\dfrac{41}{10^2}\)
\(\Rightarrow\left[\begin{matrix}x=\dfrac{-1-\sqrt{41}}{10}\\x=\dfrac{-1+\sqrt{41}}{10}\end{matrix}\right.\)
Thấy đáp số @thientuyetlinh có vẻ chưa đúng
--> mới làm thử %. chưa test lại bạn ktra xem ai đúng
\(\left(x+2\right)\left(3-4x\right)-\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)-\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x\right)+\left(x+2\right)^2=0\)
\(\Leftrightarrow\left(x+2\right)\left(3-4x+x+2\right)=0\)
\(\Leftrightarrow\left(x+2\right)\left(-3x+5\right)=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}x+2=0\\-3x+5=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=-2\\x=\dfrac{5}{3}\end{matrix}\right.\)
Vậy PT có nghiệm là x=-2;\(\dfrac{5}{3}\)
