Question

\(\frac{x\left(3x-1\right)}{2-\sqrt{4+x-3x^2}}\le\frac{1}{2}\)

Answer 1

ĐKXĐ: \(\left\{{}\begin{matrix}-1\le x\le\frac{4}{3}\\x\ne\left\{0;\frac{1}{3}\right\}\end{matrix}\right.\)

\(\Leftrightarrow\frac{x\left(3x-1\right)\left(2+\sqrt{4+x-3x^2}\right)}{\left(2-\sqrt{4+x-3x^2}\right)\left(2+\sqrt{4+x-3x^2}\right)}\le\frac{1}{2}\)

\(\Leftrightarrow\frac{x\left(3x-1\right)\left(2+\sqrt{4+x-3x^2}\right)}{3x^2-x}\le\frac{1}{2}\)

\(\Leftrightarrow2+\sqrt{4+x-3x^2}\le\frac{1}{2}\)

\(\Leftrightarrow\sqrt{4+x-3x^2}\le-\frac{3}{2}\)

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