Cho biểu thức $P=\sqrt{1-x+\left(1-x\right)\sqrt{1-x^2}}+\sqrt{1-x-\left(1-x\right)\sqrt{1-x^2}}$ với $-1\le x\le1$. Tính giá trị của biểu thức P khi $x=-\frac{1}{2017}$.

minhthu_ Đặt x=cos⁡α⇒1−x2=sin⁡αx = \cos \alpha \Rightarrow \sqrt{1 - x^2} = \sin \alphax=cosα⇒1−x2

​=sinα (vì −1≤x≤1-1 \le x \le 1−1≤x≤1).

Khi đó:

P=1−cos⁡α+(1−cos⁡α)sin⁡α+1−cos⁡α−(1−cos⁡α)sin⁡αP = \sqrt{1 - \cos \alpha + (1 - \cos \alpha)\sin \alpha} + \sqrt{1 - \cos \alpha - (1 - \cos \alpha)\sin \alpha}P=1−cosα+(1−cosα)sinα

​+1−cosα−(1−cosα)sinα

​Ta có:

1−cos⁡α=2sin⁡2α2,sin⁡α=2sin⁡α2cos⁡α21 - \cos \alpha = 2\sin^2 \dfrac{\alpha}{2}, \quad \sin \alpha = 2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}1−cosα=2sin22α​,sinα=2sin2α​cos2α​Thay vào:

P=2sin⁡2α2+4sin⁡3α2cos⁡α2+2sin⁡2α2−4sin⁡3α2cos⁡α2P = \sqrt{2\sin^2 \dfrac{\alpha}{2} + 4\sin^3 \dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}} + \sqrt{2\sin^2 \dfrac{\alpha}{2} - 4\sin^3 \dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}}P=2sin22α​+4sin32α​cos2α​

​+2sin22α​−4sin32α​cos2α​

​Rút 2sin⁡α2\sqrt{2}\sin\dfrac{\alpha}{2}2

​sin2α​ ra ngoài:

P=2sin⁡α2[1+2sin⁡α2cos⁡α2+1−2sin⁡α2cos⁡α2]P = \sqrt{2}\sin\dfrac{\alpha}{2} \left[\sqrt{1 + 2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}} + \sqrt{1 - 2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2}}\right]P=2

​sin2α​[1+2sin2α​cos2α​

​+1−2sin2α​cos2α​

​]Nhưng 2sin⁡α2cos⁡α2=sin⁡α2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} = \sin \alpha2sin2α​cos2α​=sinα.

Do đó:

P=2sin⁡α2[1+sin⁡α+1−sin⁡α]P = \sqrt{2}\sin\dfrac{\alpha}{2} \left[\sqrt{1 + \sin \alpha} + \sqrt{1 - \sin \alpha}\right]P=2

​sin2α​[1+sinα

​+1−sinα

​]Ta có:

1+sin⁡α+1−sin⁡α=2(cos⁡α2+sin⁡α2)\sqrt{1 + \sin \alpha} + \sqrt{1 - \sin \alpha} = \sqrt{2}\left(\cos\dfrac{\alpha}{2} + \sin\dfrac{\alpha}{2}\right)1+sinα

​+1−sinα

​=2

​(cos2α​+sin2α​)Suy ra:

P=2sin⁡α2(cos⁡α2+sin⁡α2)=2sin⁡α2cos⁡α2+2sin⁡2α2P = 2\sin\dfrac{\alpha}{2}(\cos\dfrac{\alpha}{2} + \sin\dfrac{\alpha}{2}) = 2\sin\dfrac{\alpha}{2}\cos\dfrac{\alpha}{2} + 2\sin^2\dfrac{\alpha}{2}P=2sin2α​(cos2α​+sin2α​)=2sin2α​cos2α​+2sin22α​ ⇒P=sin⁡α+1−cos⁡α\Rightarrow P = \sin \alpha + 1 - \cos \alpha⇒P=sinα+1−cosα Thay x=cos⁡α=−12017⇒sin⁡α=1−x2=1−120172x = \cos \alpha = -\dfrac{1}{2017} \Rightarrow \sin \alpha = \sqrt{1 - x^2} = \sqrt{1 - \dfrac{1}{2017^2}}x=cosα=−20171​⇒sinα=1−x2

​=1−201721​

​.

P=1−120172+1−(−12017)P = \sqrt{1 - \dfrac{1}{2017^2}} + 1 - \left(-\dfrac{1}{2017}\right)P=1−201721​

​+1−(−20171​) P=1+12017+1−120172P = 1 + \dfrac{1}{2017} + \sqrt{1 - \dfrac{1}{2017^2}}P=1+20171​+1−201721​

​ Đáp số:

P=1+12017+1−120172\boxed{P = 1 + \dfrac{1}{2017} + \sqrt{1 - \dfrac{1}{2017^2}}}P=1+20171​+1−201721​

​​