Question

\(p=2x+\sqrt{1-x}\)voi 0<=x<=1. Tim gia tri lon nhat cua P

Answer 1

\(P=-2\left[\left(1-x\right)-2.\frac{\sqrt{1-x}}{4}+\frac{1}{16}\right]+2+\frac{1}{8}=-2\left(\sqrt{1-x}-\frac{1}{4}\right)^2+\frac{17}{8}\le\frac{17}{8}\)

Max P=17/8  khi 1-x =1/16  hay x = 15/16

Answer 2

\(P=-2\left[\left(1-x\right)-\frac{2\sqrt{1-x}}{4}+\frac{1}{16}\right]+\frac{1}{8}=-2\left(\sqrt{1-x}-\frac{1}{4}\right)^2+\frac{1}{8}\le\frac{1}{8}\)

Max P = 1/8 khi 1- x =1/16  => x =1-1/16 =15/16