Tìm x, biết: |x^2| x + 3/4|| = x^2

Ta có x² ≥ 0, ∀x ∈ ℝ.

Ta có \(\left| {{x^2}\left| {x + \frac{3}{4}} \right|} \right| = {x^2}\).

\( \Rightarrow {x^2}\left| {x + \frac{3}{4}} \right| = {x^2}\) hoặc \({x^2}\left| {x + \frac{3}{4}} \right| = - {x^2}\).

\( \Rightarrow {x^2}\left( {\left| {x + \frac{3}{4}} \right| - 1} \right) = 0\) hoặc \({x^2}\left( {\left| {x + \frac{3}{4}} \right| + 1} \right) = 0\).

\( \Rightarrow \left[ \begin{array}{l}{x^2} = 0\\\left| {x + \frac{3}{4}} \right| = 1\end{array} \right.\) hoặc \(\left[ \begin{array}{l}{x^2} = 0\\\left| {x + \frac{3}{4}} \right| = - 1\end{array} \right.\) (vô lí).

\( \Rightarrow \left[ \begin{array}{l}x = 0\\x + \frac{3}{4} = 1\\x + \frac{3}{4} = - 1\end{array} \right.\)\( \Rightarrow \left[ \begin{array}{l}x = 0\\x = \frac{1}{4}\\x = - \frac{7}{4}\end{array} \right.\)

Vậy \(\left[ \begin{array}{l}x = 0\\x = \frac{1}{4}\\x = - \frac{7}{4}\end{array} \right.\).