Question

tìm Min P = căn(x+2) + căn(2-x) - căn(4-x^2)

Answer 1

ĐK: \(-2\le x\le2\)

Đặt: \(\sqrt{x+2}+\sqrt{2-x}=t>0\)

=> \(t^2=\left(\sqrt{x+2}+\sqrt{2-x}\right)^2\le2\left(x+2+2-x\right)=8\)

=> \(0< t\le2\sqrt{2}\)

Ta có: \(t^2=\left(\sqrt{x+2}+\sqrt{2-x}\right)^2=x+2+2-x+2\sqrt{4-x^2}\)

=> \(\sqrt{4-x^2}=\frac{t^2-4}{2}\)

Ta có: \(P=t-\frac{t^2-4}{2}=\frac{\left(t+2\sqrt{2}-2\right)\left(2\sqrt{2}-t\right)}{2}+2\sqrt{2}-2\ge2\sqrt{2}-2\)

=> min P = \(2\sqrt{2}-2\) tại  \(t=2\sqrt{2}\)khi đó x = 0 

Vậy:...

Answer 2

em cảm ơn ạ