Question

cho \(P=\sqrt{\left(1-x\right)+\left(1-x\right)\sqrt{1-x^2}}+\sqrt{\left(1-x\right)-\left(1-x\right)\sqrt{1-x^2}}\) với \(-1\le x\le1\)

tính P khi x= \(-\dfrac{1}{2017}\)

Answer 1

Lời giải:
Đặt \(\sqrt{(1-x)+(1-x)\sqrt{1-x^2}}=a; \sqrt{(1-x)-(1-x)\sqrt{1-x^2}}=b\)

Khi đó: \(P=a+b\geq 0\)

Ta có:
\(a^2+b^2=(1-x)+(1-x)\sqrt{1-x^2}+(1-x)-(1-x)\sqrt{1-x^2}=2(1-x)\)

Và:

\(ab=(1-x)\sqrt{(1+\sqrt{1-x^2})(1-\sqrt{1-x^2})}\)

\(=(1-x)\sqrt{1-(1-x^2)}=(1-x)\sqrt{x^2}=|x|(1-x)\)

\(\Rightarrow P^2=(a+b)^2=a^2+b^2+2ab=2(1-x)+2|x|(1-x)\)

\(=2(1-x)(1+|x|)=\frac{4036}{2017}.\frac{2018}{2017}\)

\(\Rightarrow P=\sqrt{\frac{4036.2018}{2017^2}}=\frac{\sqrt{4036.2018}}{2017}\)