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Question

$\sqrt{1+\sqrt{1-4x^{2}}}=x\left ( 1+\sqrt{1+\sqrt{1+2\sqrt{1-4x^{2}}}}\right )$

$\sqrt{1+\sqrt{1-x^{2}}}\left ( \sqrt{\left ( 1+x \right )^{3}}-\sqrt{\left ( 1-x \right )^{3}} \right )=\frac{2}{\sqrt{3}}+\sqrt{\frac{1-x^{2}}{3}}$

$\left ( 64x^{3}-112x^{2}+56x-7 \right )^{2}=4-4x$

lovesomebody121 · Answer 1 · 2/25/2016 12:55:43 PM

2, Đk x

$\epsilon$ [-1,1] cho nên đặt x=cost , t

$\epsilon$ [0;

$\pi$ ]

Khi đó pt trở thành:

$\sqrt{1+\sqrt{1-\cos^{2}t}}$

$($

$\sqrt{(1+cost)^{3}}$

$-$

$\sqrt{(1-cost)^{3}}$

$)$ =

$\frac{2}{\sqrt{3}}$ +

$\sqrt{\frac{1-cos^{2}t}{3}}$

$\Leftrightarrow$

$\sqrt{1+sint}$

$\left ( \sqrt{8\left ( \frac{1+cost}{2}^{3} \right )}- \sqrt{8\left ( \frac{1-cost}{2}^{3} \right )} \right )$ =

$\frac{2+sint}{\sqrt{3}}$

$\Leftrightarrow$ 2

$\sqrt{2}$

$\left ( sin\tfrac{t}{2} +cos\tfrac{t}{2} \right )$

$\left ( cos\tfrac{t}{2} -sin\tfrac{t}{2}\right )$

$\left ( 1+\frac{sint}{2} \right )$ =

$\frac{2+sint}{\sqrt{3}}$

Đến đây bạn tự làm nốt sẽ ra được cost =

$\frac{1}{\sqrt{6}}$ => x =

$\frac{1}{\sqrt{6}}$

1 Đáp án