tìm x 1/4x^2-(1/2x-4).1/2x=-14--3/2(x-8/3) 2x^2.(2x-3)-x^2.(4x^2-6x+2)

Đáp án:

$\begin{array}{l}
a)\\
\dfrac{1}{4}{x^2} - \left( {\dfrac{1}{2}x - 4} \right).\dfrac{1}{2}x =  - 14 - \dfrac{{ - 3}}{2}\left( {x - \dfrac{8}{3}} \right)\\
 \Leftrightarrow \dfrac{1}{4}{x^2} - \dfrac{1}{4}{x^2} + 2x =  - 14 + \dfrac{3}{2}x - \dfrac{3}{2}.\dfrac{8}{3}\\
 \Leftrightarrow 2x =  - 14 + \dfrac{3}{2}x - 4\\
 \Leftrightarrow 2x - \dfrac{3}{2}x =  - 18\\
 \Leftrightarrow \dfrac{1}{2}x =  - 18\\
 \Leftrightarrow x =  - 18.2\\
 \Leftrightarrow x =  - 36\\
Vậy\,x =  - 36\\
b)\\
2{x^2}\left( {2x - 3} \right) - x.\left( {4{x^2} - 6x + 2} \right) = 0\\
 \Leftrightarrow 4{x^3} - 6{x^2} - 4{x^3} + 6{x^2} - 2x = 0\\
 \Leftrightarrow  - 2x = 0\\
 \Leftrightarrow x = 0\\
Vậy\,x = 0
\end{array}$