Question

tìm các giá trị x\(\ge\)0 thỏa mãn bất phương trình

\(x^2-4x-6>\sqrt{x^3+3x^2+2x}\)

Answer 1

\(x^2-4x-6>\sqrt{x\left(x+1\right)\left(x+2\right)}\)

\(\Leftrightarrow x^2+2x-6\left(x+1\right)>\sqrt{\left(x^2+2x\right)\left(x+1\right)}\)

Đặt \(\left\{{}\begin{matrix}\sqrt{x^2+2x}=a\ge0\\\sqrt{x+1}=b>0\end{matrix}\right.\)

\(\Rightarrow a^2-6b^2>ab\Leftrightarrow a^2-ab-6b^2>0\)

\(\Leftrightarrow\left(a-3b\right)\left(a+2b\right)>0\)

\(\Leftrightarrow a-3b>0\Leftrightarrow a>3b\)

\(\Leftrightarrow\sqrt{x^2+2x}>3\sqrt{x+1}\)

\(\Leftrightarrow x^2+2x>9x+9\Leftrightarrow x^2-7x-9>0\)