Question

Tìm x, biết:

2x² - x - 6 = 0

Answer 1

https://i.imgur.com/VImYD65.jpg

Answer 2

b, 2x2+x−6=02x2+x−6=0

⇒2x2+4x−3x−6=0⇒2x2+4x−3x−6=0

⇒(2x2+4x)−(3x+6)=0⇒(2x2+4x)−(3x+6)=0

⇒2x(x+2)−3(x+2)=0⇒2x(x+2)−3(x+2)=0

⇒(x+2)(2x−3)=0⇒(x+2)(2x−3)=0

⇒x+2=0⇒x+2=0 hoặc 2x−3=02x−3=0

⇒x=−2⇒x=−2 hoặc x=32x=32

Vậy x=−2x=−2 ; x=32

Answer 3

\(2x^2-x-6=0\)

\(\Leftrightarrow2x^2-4x+3x-6=0\)

\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)

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

\(\Leftrightarrow\left[{}\begin{matrix}2x+3=0\\x-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}2x=-3\\x=2\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\frac{3}{2}\\x=2\end{matrix}\right.\)

Vậy \(x\in\left\{-\frac{3}{2};2\right\}\)

Answer 4

\(2x^2-x-6=0\)

\(2x^2+3x-4x-6=0\)

\(x\left(2x+3\right)-2\left(2x+3\right)=0\)

\(\left(2x+3\right)\left(x-2\right)=0\)

⇒ 2x + 3 = 0 hoặc x - 2 = 0

2x =-3 hoặc x = 2

x = \(\frac{-3}{2}\) hoặc x = 2

Vậy x ∈ \(\left\{{}\begin{matrix}\frac{-3}{2}\\2\end{matrix}\right.\)

Answer 5

Answer 6

\(2x^2-x-6=0\) ( cái này là tam thức bậc 2 )

\(\Leftrightarrow2x^2-4x+3x-6=0\)

\(\Leftrightarrow2x\left(x-2\right)+3\left(x-2\right)=0\)

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

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

Vậy \(x=2;y=\frac{-3}{2}\)