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Câu hỏi: Khẳng định nào sau đây là sai?
A. ${{\left( 2-\sqrt{3} \right)}^{2019}}>{{\left( 2-\sqrt{3} \right)}^{2020}}.$
B. ${{\left( 1-\dfrac{\sqrt{3}}{2} \right)}^{2019}}>{{\left( 1-\dfrac{\sqrt{3}}{2} \right)}^{2020}}.$
C. ${{\left( 1+\sqrt{2} \right)}^{2019}}>{{\left( 1+\sqrt{2} \right)}^{2020}}.$
D. ${{\pi }^{2019}}>{{e}^{2019}}.$
A. ${{\left( 2-\sqrt{3} \right)}^{2019}}>{{\left( 2-\sqrt{3} \right)}^{2020}}.$
B. ${{\left( 1-\dfrac{\sqrt{3}}{2} \right)}^{2019}}>{{\left( 1-\dfrac{\sqrt{3}}{2} \right)}^{2020}}.$
C. ${{\left( 1+\sqrt{2} \right)}^{2019}}>{{\left( 1+\sqrt{2} \right)}^{2020}}.$
D. ${{\pi }^{2019}}>{{e}^{2019}}.$
Hàm số $y={{\left( 1+\sqrt{2} \right)}^{x}}$ đồng biến do cơ số $a=1+\sqrt{2}>1$
$\Rightarrow {{\left( 1+\sqrt{2} \right)}^{2019}}<{{\left( 1+\sqrt{2} \right)}^{2020}}$ nên C sai.
$\Rightarrow {{\left( 1+\sqrt{2} \right)}^{2019}}<{{\left( 1+\sqrt{2} \right)}^{2020}}$ nên C sai.
Đáp án C.