cho x biết |x^2+2.|x-1/2||=x^2+2

Đáp án:$\,x = \dfrac{3}{2};x =  - \dfrac{1}{2}$

Giải thích các bước giải:

$\begin{array}{l}
\left| {{x^2} + 2} \right|.\left| {x - \dfrac{1}{2}} \right| = {x^2} + 2\\
 \Leftrightarrow \left( {{x^2} + 2} \right).\left| {x - \dfrac{1}{2}} \right| = {x^2} + 2\\
 \Leftrightarrow \left( {{x^2} + 2} \right).\left| {x - \dfrac{1}{2}} \right| - \left( {{x^2} + 2} \right) = 0\\
 \Leftrightarrow \left( {{x^2} + 2} \right).\left( {\left| {x - \dfrac{1}{2}} \right| - 1} \right) = 0\\
 \Leftrightarrow \left| {x - \dfrac{1}{2}} \right| - 1 = 0\left( {do:{x^2} + 2 > 0} \right)\\
 \Leftrightarrow \left| {x - \dfrac{1}{2}} \right| = 1\\
 \Leftrightarrow \left[ \begin{array}{l}
x - \dfrac{1}{2} = 1\\
x - \dfrac{1}{2} =  - 1
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x = 1 + \dfrac{1}{2} = \dfrac{3}{2}\\
x =  - 1 + \dfrac{1}{2} =  - \dfrac{1}{2}
\end{array} \right.\\
Vậy\,x = \dfrac{3}{2};x =  - \dfrac{1}{2}
\end{array}$