a) 2x2 - 3x - 2 = 0.
<=> (2x + 1)(x - 2) = 0
<=> 2x + 1 = 0 hoặc x - 2 = 0
<=> x = -1/2 hoặc x = 2
b) 3x2 - 7x - 10 = 0.
<=> (x + 1)(3x - 10) = 0
<=> x = -1 hoặc x = 10/3
c) 2x2 - 5x + 3 = 0.
<=> (x - 1)(2x - 3) = 0
<=> x = 1 hoặc x = 3/2
a) 2x2-3x-2=0
⇔ \(\left(x-2\right)\left(x+\dfrac{1}{2}\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b)3x2-7x-10=0
\(\Leftrightarrow\left(x-\dfrac{10}{3}\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-1\end{matrix}\right.\)
c)2x2-5x+3=0
\(\Leftrightarrow\left(x-\dfrac{3}{2}\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
a. 2x2 - 3x - 2 = 0
<=> 2x2 - 4x + x - 2 = 0
<=> 2x(x - 2) + (x - 2) = 0
<=> (2x + 1)(x - 2)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{-1}{2}\\x=2\end{matrix}\right.\)
Vậy ................
b. 3x2 - 7x - 10
<=> 3x2 + 3x - 10x - 10 = 0
<=> 3x(x + 1) - 10(x + 1) = 0
<=> (3x - 10)(x + 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}3x-10=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{10}{3}\\x=-1\end{matrix}\right.\)
Vậy ..................
c. 2x2 - 5x + 3 = 0
<=> 2x2 - 2x - 3x + 3 = 0
<=> 2x(x - 1) - 3(x - 1) = 0
<=> (2x - 3)(x - 1) = 0
\(\Leftrightarrow\left[{}\begin{matrix}2x-3=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=1\end{matrix}\right.\)
Vậy ................
a: Ta có: \(2x^2-3x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left(2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-\dfrac{1}{2}\end{matrix}\right.\)
b: Ta có: \(3x^2-7x-10=0\)
\(\Leftrightarrow\left(x+1\right)\left(3x-10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=\dfrac{10}{3}\end{matrix}\right.\)
c: ta có: \(2x^2-5x+3=0\)
\(\Leftrightarrow\left(x-1\right)\left(2x-3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1\\x=\dfrac{3}{2}\end{matrix}\right.\)
