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Câu hỏi: Nếu ${{\left( \sqrt{3}-\sqrt{2} \right)}^{x}}>\sqrt{3}+\sqrt{2}$ thì
A. $x>-1.$
B. $\forall x\in \mathbb{R}.$
C. $x<1.$
D. $x<-1.$
A. $x>-1.$
B. $\forall x\in \mathbb{R}.$
C. $x<1.$
D. $x<-1.$
Ta có $\left( \sqrt{3}+\sqrt{2} \right)\left( \sqrt{3}-\sqrt{2} \right)=1\Leftrightarrow \sqrt{3}+\sqrt{2}=\dfrac{1}{\sqrt{3}-\sqrt{2}}={{\left( \sqrt{3}-\sqrt{2} \right)}^{-1}}.$
Nên ${{\left( \sqrt{3}-\sqrt{2} \right)}^{x}}>\sqrt{3}+\sqrt{2}\Leftrightarrow {{\left( \sqrt{3}-\sqrt{2} \right)}^{x}}>{{\left( \sqrt{3}-\sqrt{2} \right)}^{-1}}\Leftrightarrow x<-1$
(vì $0<\sqrt{3}-\sqrt{2}<1$ )
Nên ${{\left( \sqrt{3}-\sqrt{2} \right)}^{x}}>\sqrt{3}+\sqrt{2}\Leftrightarrow {{\left( \sqrt{3}-\sqrt{2} \right)}^{x}}>{{\left( \sqrt{3}-\sqrt{2} \right)}^{-1}}\Leftrightarrow x<-1$
(vì $0<\sqrt{3}-\sqrt{2}<1$ )
Đáp án D.