Question

2x+1/x +3≥ 3-5x/5 +4x+1/4

giải bpt

Answer 1

ĐKXĐ x<>0

Ta có: \(\frac{2x+1}{x}+3\ge\frac{3-5x}{5}+\frac{4x+1}{4}\)

=>\(\frac{2x+1+3x}{x}\ge\frac{4\left(3-5x\right)+5\left(4x+1\right)}{20}=\frac{17}{20}\)

=>\(\frac{5x+1}{x}\ge\frac{17}{20}\)

=>\(\frac{5x+1}{x}-\frac{17}{20}\ge0\)

=>\(\frac{20\cdot\left(5x+1\right)-17x}{20x}\ge0\)

=>\(\frac{83x+20}{x}\ge0\)

TH1: \(\begin{cases}83x+20\ge0\\ x>0\end{cases}\Rightarrow\begin{cases}x\ge-\frac{20}{83}\\ x>0\end{cases}\Rightarrow x>0\)

TH2: \(\begin{cases}83x+20\le0\\ x<0\end{cases}\Rightarrow\begin{cases}83x\le-20\\ x<0\end{cases}\Rightarrow x\le-\frac{20}{83}\)