Question

Đề : Tìm x

2x^2+5x-3=0

Answer 1

Sửa đề

\(2x^2-5x+3=0\)

<=> \(\left(2x^2-2x\right)-\left(3x-3\right)\)

<=> \(2x\left(x-1\right)-3\left(x-1\right)=0\)

<=> \(\left(x-1\right)\left(2x-3\right)=0\)

=>\(\left[{}\begin{matrix}x-1=0\\2x-3=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=1\\x=\frac{3}{2}\end{matrix}\right.\)

Answer 2

2x² + 5x - 3 = 0

\(\Leftrightarrow\) 2x² - x + 6x - 3 = 0

\(\Leftrightarrow\) (2x² - x) + (6x - 3) = 0

\(\Leftrightarrow\) x(2x - 1) + 3(2x - 1) = 0

\(\Leftrightarrow\) (2x - 1)(x + 3) = 0

\(\Leftrightarrow\) 2x - 1 = 0 hoặc x + 3 = 0

1) 2x - 1 = 0 \(\Leftrightarrow\) x = \(\frac{1}{2}\)

2) x + 3 = 0 \(\Leftrightarrow\) x = -3

Vậy............