\(2x-2x^2-5\)

=\(-2\left(x^2-x+\dfrac{5}{2}\right)\)

=\(-2\left(x^2-x+\dfrac{1}{4}+\dfrac{9}{4}\right)\)

\(=-2\left(x-\dfrac{1}{2}\right)^2-\dfrac{9}{2}\)

Với mọi x thì \(-\left(x-\dfrac{1}{2}\right)-\dfrac{9}{2}>=-\dfrac{9}{2}\)

Để \(-2\left(x-\dfrac{1}{2}\right)-\dfrac{9}{2}=-\dfrac{9}{2}\)thì

\(\left(x-\dfrac{1}{2}\right)^2=0\)=>\(x-\dfrac{1}{2}=0\)=>\(x=\dfrac{1}{2}\)

Vậy ...

x³ - 2x² + 6x = 12

x³ - 2x² + 6x - 12 = 0

x²(x - 2) + 6(x - 2)=0

(x - 2)(x² + 6) = 0

\(\Leftrightarrow \begin{bmatrix} x - 2 = 0 & & \\ x^{2} + 6 = 0& & \end{bmatrix}\) bỏ dấu ngoặc bên phải nha pn

\(\Leftrightarrow \begin{bmatrix} x = 2 & & \\ x^{2} = - 6 & & \end{bmatrix}\) không tìm được giá trị của x (pn ghi cái này kế pn chỗ x² = - 6 nhé

Vậy x = 2

\(x^3-2x^2+6x=12\)

\(\Rightarrow\) \(x^3-2x^2+6x-12=0\)

\(\Rightarrow x^2\left(x-2\right)+6\left(x-2\right)=0\)

\(\Rightarrow\left(x-2\right)\left(x^2+6\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x-2=0\\x^2+6=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=2\\x\in\varnothing\end{matrix}\right.\)

Vậy $x=2$

\(x^3-2x^2+6x=12\)

\(\Rightarrow x^3-2x^2+6x-12=0\)

\(\Rightarrow x^2\left(x-2\right)+6\left(x-2\right)-0\)

\(\Rightarrow\left(x^2+6\right)\left(x-2\right)=0\)

\(\Rightarrow\left\{{}\begin{matrix}x^2+6=0\\x-2=0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x\in\varnothing\\x=2\end{matrix}\right.\)

Vậy \(x=2\)