\(2x^2-5x+2=0\)
\(x^2-\frac{5}{2}x+1=0\)
\(x^2+2.\frac{5}{4}x+\frac{25}{16}-\frac{25}{16}+1=0\)
\(\left(x+\frac{5}{4}\right)^2-\frac{9}{16}=0\)
\(\left(x+\frac{5}{4}\right)^2-\left(\frac{3}{4}\right)^2=0\)
\(\left(x+\frac{5}{4}-\frac{3}{4}\right)\left(x+\frac{5}{4}+\frac{3}{4}\right)=0\)
\(\left(x+\frac{1}{2}\right)\left(x+2\right)=0\)
\(\Rightarrow\orbr{\begin{cases}x+\frac{1}{2}=0\\x+2=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=\frac{-1}{2}\\x=-2\end{cases}}\)
2x2-5x+2
\(\Leftrightarrow2x^2-4x-x+2\)
\(\Leftrightarrow\left(2x^2-4x\right)-\left(x-2\right)\)
\(\Leftrightarrow2x\left(x-2\right)-\left(x-2\right)\)
\(\Leftrightarrow\left(2x-1\right)\left(x-2\right)\)
\(\Leftrightarrow2x-1=0hoacx-2=0\)
Nếu 2x-1=0
\(\Leftrightarrow x=\frac{1}{2}\)
Nếu x-2=0 thì
\(\Leftrightarrow x=2\)
Vậy pt đãcho có tập nghiệm là:S=\(\hept{\begin{cases}\\\end{cases}2;\frac{1}{2}}\)
