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Chứng minh rằng: $\dfrac{2-\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}}{2-\sqrt{2+\sqrt{2+\sqrt{2}}}}<\dfrac{1}{3}$

Hỏi 08-05-13 02:07 PM

Xusint
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Bình chọn tăng 6 Bình chọn giảm

Đặt $a=\sqrt{2+\sqrt{2+\sqrt{2+\sqrt{2}}}}$

=> $a^2-2=\sqrt{2+\sqrt{2+\sqrt{2}}}$

=> $VT= \frac{2-a}{2-a^2+2}=\frac{2-a}{(2-a)(2+a)}=\frac{1}{a+2}$

Ta có $a>\sqrt{2}>1$ => $a+2>3$

=> $VT<\frac{1}{3}$

Trả lời 08-11-14 08:15 PM

๖ۣۜPXM๖ۣۜMinh4212♓
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