Trả lời câu hỏi 63 - Mục câu hỏi trắc nghiệm trang 51

1. Nội dung câu hỏi

Giải mỗi bất phương trình sau:

a) \({\left( {0,2} \right)^{2x + 1}} > 1;\)

b) \({27^{2x}} \le \frac{1}{9};\)

c) \({\left( {\frac{1}{2}} \right)^{{x^2} - 5x + 4}} \ge 4;\)

d) \({\left( {\frac{1}{{25}}} \right)^{x + 1}} < {125^{2x}};\)

e) \({\left( {\sqrt 2  - 1} \right)^{3x - 2}} < {\left( {\sqrt 2  + 1} \right)^{4 - x}};\)

g) \({\left( {0,5} \right)^{2{x^2} - x}} > {\left( {\sqrt 2 } \right)^{4x - 12}}.\)

3. Lời giải chi tiết

a) \({\left( {0,2} \right)^{2x + 1}} > 1 \Leftrightarrow 2x + 1 < {\log _{0,2}}1 \Leftrightarrow 2x + 1 < 0 \Leftrightarrow x <  - \frac{1}{2}.\)

b) \({27^{2x}} \le \frac{1}{9} \Leftrightarrow {3^{6x}} \le {3^{ - 2}} \Leftrightarrow 6x \le  - 2 \Leftrightarrow x \le  - \frac{1}{3}.\)

c) \({\left( {\frac{1}{2}} \right)^{{x^2} - 5x + 4}} \ge 4 \Leftrightarrow {\left( {\frac{1}{2}} \right)^{{x^2} - 5x + 4}} \ge {\left( {\frac{1}{2}} \right)^{ - 2}} \Leftrightarrow {x^2} - 5x + 4 \le  - 2 \Leftrightarrow {x^2} - 5x + 6 \le 0\)

\( \Leftrightarrow \left( {x - 2} \right)\left( {x - 3} \right) \le 0 \Leftrightarrow 2 \le x \le 3.\)

d) \({\left( {\frac{1}{{25}}} \right)^{x + 1}} < {125^{2x}} \Leftrightarrow {\left( {{5^{ - 2}}} \right)^{x + 1}} < {\left( {{5^3}} \right)^{2x}} \Leftrightarrow {5^{ - 2x - 2}} < {5^{6x}} \Leftrightarrow  - 2x - 2 < 6x \Leftrightarrow x >  - \frac{1}{4}.\)

e) Ta có:

\(\begin{array}{l}{\left( {\sqrt 2  - 1} \right)^{3x - 2}} < {\left( {\sqrt 2  + 1} \right)^{4 - x}} \Leftrightarrow {\left( {{{\left( {\sqrt 2  + 1} \right)}^{ - 1}}} \right)^{3x - 2}} < {\left( {\sqrt 2  + 1} \right)^{4 - x}}\\ \Leftrightarrow {\left( {\sqrt 2  + 1} \right)^{2 - 3x}} < {\left( {\sqrt 2  + 1} \right)^{4 - x}} \Leftrightarrow 2 - 3x < 4 - x \Leftrightarrow 2x >  - 2 \Leftrightarrow x >  - 1.\end{array}\)

g) \({\left( {0,5} \right)^{2{x^2} - x}} > {\left( {\sqrt 2 } \right)^{^{4x - 12}}} \Leftrightarrow {\left( {{2^{ - 1}}} \right)^{2{x^2} - x}} > {\left( {{2^{\frac{1}{2}}}} \right)^{4x - 12}} \Leftrightarrow {2^{x - 2{x^2}}} > {2^{2x - 6}}\)

\( \Leftrightarrow x - 2{x^2} > 2x - 6 \Leftrightarrow 2{x^2} + x - 6 < 0 \Leftrightarrow \left( {2x - 3} \right)\left( {x + 2} \right) < 0 \Leftrightarrow  - 2 < x < \frac{3}{2}.\)