Đáp án:
\(x > 1,639879256\)
Giải thích các bước giải:
\(\begin{array}{l}
\frac{{5x - 3}}{5} + \frac{{3x - 1}}{4} < \frac{{x\left( {2{x^2} + 3} \right)}}{2} - 5\\
\to \frac{{4\left( {5x - 3} \right) + 5\left( {3x - 1} \right)}}{{20}} - \frac{{10\left( {2{x^3} + 3x} \right) - 5.20}}{{20}} < 0\\
\to 20x - 12 + 15x - 5 - 20{x^3} - 30x + 100 < 0\\
\to - 20{x^3} + 5x + 80 < 0\\
Xét: - 20{x^3} + 5x + 80 = 0\\
\to x = 1,639879256
\end{array}\)
BXD:
x -∞ 1,6398 +∞
f(x) + 0 -
\(KL:x > 1,639879256\)
