Question

\(\dfrac{\left(3x-2\right)^2}{3}-\dfrac{\left(2x+1\right)^2}{3}\le x\left(x+1\right)\)

Answer 1

\(\Leftrightarrow\left(3x-2\right)^2-\left(2x+1\right)^2< =3\left(x^2+x\right)\)

\(\Leftrightarrow9x^2-12x+4-4x^2-4x-1-3x^2-3x< =0\)

\(\Leftrightarrow2x^2-19x+3< =0\)

Đặt \(2x^2-19x+3=0\)

\(\Leftrightarrow\left\{{}\begin{matrix}x_1=\dfrac{19+\sqrt{337}}{4}\\x_2=\dfrac{19-\sqrt{337}}{4}\end{matrix}\right.\)

=>F(x)=2x²-19x+3<=0 khi \(x\in\left[\dfrac{19-\sqrt{337}}{4};\dfrac{19+\sqrt{337}}{4}\right]\)