Giải bài 63 trang 51 sách bài tập toán 11 – Cánh diều

Đề bài

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

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

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