Đáp án:
Giải thích các bước giải:
a) $-6x-3=0 \Rightarrow x=\frac{-1}{2}\\
x+1=0 \Rightarrow x=-1\\
\Rightarrow -1<x<\frac{-1}{2}$
b) $ x+1=0 \Rightarrow x=-1\\
x-1=0 \Rightarrow x=1\\
x-2=0 \Rightarrow x=2\\
BXD: x -1 1 2\\
f(x) - 0 + 0 - 0+\\
\Rightarrow x \in (-1,1)\cup (2,+\infty )$
c) $\frac{-4}{3x+1}<\frac{3}{2x-1}\\
\Leftrightarrow \frac{-4(2x-1)}{(3x+1)(2x-1)}-\frac{3(3x+1)}{(3x+1)(2x-1)}<0\\
\Leftrightarrow \frac{-8x+4-9x-3}{(3x+1)(2x-1)}<0\\
\Leftrightarrow \frac{-17x+1}{(3x+1)(2x-1)}<0\\
-17x+1=0 \Rightarrow x=\frac{1}{17}\\
3x+1=0 \Rightarrow x=\frac{-1}{3}\\
2x-1=0 \Rightarrow x=\frac{1}{2}\\
BXD x \frac{-1}{3} \frac{1}{17} \frac{1}{2}\\
f(x) + 0 - 0 + 0 -\\
\Rightarrow x \in (\frac{-1}{3}, \frac{1}{17} )\cup ( \frac{1}{2},+\infty )$
d) $\frac{2x-5}{2-x} \geqslant -1\\
\Leftrightarrow \frac{2x-5}{2-x} +1 \geqslant 0\\
\Leftrightarrow \frac{2x-5+2-x}{2-x}\geqslant 0\\
\Leftrightarrow \frac{x-3}{2-x}\geqslant 0\\
x-3=0 \Rightarrow x=3\\
2-x=0 \Rightarrow x=2\\
BXD x 2 3\\
f(x) - 0 + 0 -\\
\Rightarrow x \in [2;3]$
