Giải bài 80-81 trang 129 SGK giải tích nâng cao 12
Giải các bất phương trình sau:
80.a) \({{2}^{3-6x}}>1\); b) \({{16}^{x}}>0,125 \)
81.a) \({{\log }_{5}}\left( 3x-1 \right)<1;\) b) \({{\log }_{\frac{1}{3}}}\left( 5x-1 \right)>0\)
c) \({{\log }_{0,5}}\left( {{x}^{2}}-5x+6 \right)\ge -1;\) d) \({{\log }_{3}}\dfrac{1-2x}{x}\le 0\).
80.a) \({{2}^{3-6x}}>1\Leftrightarrow 3-6x>0\Leftrightarrow x<\dfrac{1}{2} \)
\(b)\,{{16}^{x}}>0,125\Leftrightarrow {{2}^{4x}}>{{2}^{-3}} \Leftrightarrow 4x>-3 \Leftrightarrow x>\dfrac{-3}{4}\)
81.a) Điều kiện: \(x>\dfrac{1}{3}\)
\(\,{{\log }_{5}}\left( 3x-1 \right)<1\Leftrightarrow 3x-1<5\Leftrightarrow x<2\)
Vậy \(S=\left( \dfrac{1}{3};2 \right)\)
b) Điều kiện: \(x>\dfrac{1}{5} \)
\(\,{{\log }_{\frac{1}{3}}}\left( 5x-1 \right)>0\Leftrightarrow 5x-1<1\Leftrightarrow x<\dfrac{2}{5}\)
Vậy \(S=\left( \dfrac{1}{5};\dfrac{2}{5} \right)\)
c) Điều kiện: \(x\in \left( -\infty ;2 \right)\cup \left( 3;+\infty \right)\)
\({{\log }_{0,5}}\left( {{x}^{2}}-5x+6 \right)\ge -1\Leftrightarrow {{x}^{2}}-5x+6\le 2 \\ \Leftrightarrow {{x}^{2}}-5x+4\le 0\Leftrightarrow x\in \left( 1;4 \right) \)
Vậy \(S=\left( 1;2 \right)\cup \left( 3;4 \right)\)
d) Điều kiện: \(x\in \left( 0;\dfrac{1}{2} \right)\)
\({{\log }_{3}}\dfrac{1-2x}{x}\le 0\Leftrightarrow \dfrac{1-2x}{x}\le 1 \\ \Leftrightarrow \dfrac{1-3x}{x}\le 0\Leftrightarrow x\in \left( 0;\dfrac{1}{3} \right] \)
Vậy \(S=\left( 0;\dfrac{1}{3} \right]\)