Question

Tìm x:

³\(\sqrt{1-2x^2\:}\) + 3 = 0

Answer 1

\(\sqrt[3]{1-2x^2}+3=0\)

\(\Leftrightarrow1-2x^2=-27\)

\(\Leftrightarrow2x^2-28=0\)

\(\Leftrightarrow2\left(x^2-14\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)

Answer 2

ĐK : \(x\in R\)

Ta có : \(\sqrt[3]{1-2x^2}+3=0\)

\(\Leftrightarrow\sqrt[3]{1-2x^2}=-3\)

\(\Leftrightarrow1-2x^2=-27\)

\(\Leftrightarrow-2x^2=-28\)

\(\Leftrightarrow x^2=14\)

\(\Leftrightarrow\) \(\left[{}\begin{matrix}x=\sqrt{14}\\x=-\sqrt{14}\end{matrix}\right.\)

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