a) Ta có: \(2x^2-6x+1=0\)
\(\Leftrightarrow2\left(x^2-3x+\frac{1}{2}\right)=0\)
Vì 2>0
nên \(x^2-3x+\frac{1}{2}=0\)
\(\Leftrightarrow x^2-2\cdot x\cdot\frac{3}{2}+\frac{9}{4}-\frac{7}{4}=0\)
\(\Leftrightarrow\left(x-\frac{3}{2}\right)^2=\frac{7}{4}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\frac{3}{2}=\sqrt{\frac{7}{4}}\\x-\frac{3}{2}=-\sqrt{\frac{7}{4}}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\frac{3+\sqrt{7}}{2}\\x=\frac{3-\sqrt{7}}{2}\end{matrix}\right.\)
Vậy: \(x=\frac{3\pm\sqrt{7}}{2}\)
b) Ta có: \(3x^2+4x-4=0\)
\(\Leftrightarrow3x^2-6x+2x-4=0\)
\(\Leftrightarrow3x\left(x-2\right)+2\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=\frac{-2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{2;\frac{-2}{3}\right\}\)
3x2 + 4x - 4 = 0
\(\Leftrightarrow\) 3x2 + 6x - 2x - 4 = 0
\(\Leftrightarrow\) 3x(x + 2) - 2(x + 2) = 0
\(\Leftrightarrow\) (x + 2)(3x - 2) = 0
\(\Leftrightarrow\) x + 2 = 0 hoặc 3x - 2 = 0
\(\Leftrightarrow\) x = -2 và x = \(\frac{2}{3}\)
Vậy S = {-2; \(\frac{2}{3}\)}
Mk giúp bn phần dưới thôi, còn phần trên mk làm sau
Chúc bn học tốt!!
