Đáp án:7)S=\((-\infty ;0]∩(\frac{1}{3};+\infty )\)
9)th1:S=\((-2;-1)∩(1;+\infty)\)
th2:S=\((1;2)∩(\frac{5}{3};+\infty)\)
Giải thích các bước giải:
7) \(\frac{x+2}{3x-1}≥-2 (đk: x\neq \frac{1}{3})\)
⇒\(\frac{x+2}{3x-1}+2≥0\)
⇔\(\frac{x+2+2(3x-1)}{3x-1}≥0\)
⇔\(\frac{7x}{3x-1}≥0\)
Cho 7x=0 ⇒ x=0
3x-1=0⇒ x=\(\frac{1}{3}\)
x \(-\infty\) 0 \(\frac{1}{3}\) \(+\infty \)
7x - 0 + | +
3x-1 - | - 0 +
f(x) + 0 - || +
⇒ S=\((-\infty ;0]∩(\frac{1}{3};+\infty )\)
9)\(|\frac{-5}{x+2}|<|\frac{10}{x-1}|\)
th1:\(\frac{-5}{x+2}<\frac{10}{x-1}\)
⇒\(\frac{-5}{x+2}-\frac{10}{x-1}<0\)
⇔\(\frac{-5(x-1)-10(x+2)}{(x+2)(x-1)}<0\)
⇔\(\frac{-5x+5-10x-20}{(x+2)(x-1)}<0\)
⇔\(\frac{-15x-15}{(x+2)(x-1)}<0\)
Cho -15x-15=0⇒ x=-1
x+2=0⇒ x=-2
x-1=0⇒ x=1
x \(-\infty \) -2 -1 1 \(+\infty\)
-15x-15 + | + 0 - | -
x+2 - 0 + | + | +
x-1 - | - | - 0 +
f(x) + || - || + || -
⇒S=\((-2;-1)∩(1;+\infty)\)
th2: tương tự
⇒S=\((-2;1)∩(\frac{5}{3};+\infty)\)
