P= \(1-cos^2x+2cos^2x=1+cos^2x\)

Ta có:

\(0\le cos^2x\le1\)

=> \(1\le P\le2\)

min P=1 <=> \(cos^2x=0\Leftrightarrow cosx=0\Leftrightarrow x=\frac{\pi}{2}+k\pi\)

ta có:

\(-1\le\sin x\le1\)

=> \(3.\left(-1\right)-2\le P\le3.1-2\)

suy ra: \(-5\le P\le1\)

\(maxP=1\)<=> sin x=1<=> \(x=\frac{\pi}{2}+k2\pi\)

\(P=\sqrt{c\left(a+b\right)}+\frac{\sqrt[3]{8.8.\left(2b+3c\right)}}{4}\)

\(\le\frac{c+a+b}{2}+\frac{8+8+2b+3c}{12}=\frac{6a+8b+9c+16}{12}\le\frac{32+16}{12}=4\)

Dự đoán xảy ra cực trị tại y = 2 và x = 1

Ta biến đổi nhưng sau: \(P=\left(8x^3+8+8\right)+\left(y^3+8+8\right)-32\)

\(\ge3\sqrt[3]{8x^3.8.8}+3\sqrt[3]{y^3.8.8}-32\)

\(=24x+12y-32=12\left(2x+y-\frac{8}{3}\right)\)

\(=12\left(6-\frac{8}{3}-xy\right)=12\left(\frac{10}{3}-xy\right)\)

\(=12\left(\frac{10}{3}-1x.2y\right)\ge12\left(\frac{10}{3}-\frac{\left(x+1\right)^2}{4}.\frac{\left(y+2\right)^2}{4}\right)\)

\(=12\left(\frac{10}{3}-\frac{\left[\left(x+1\right)\left(y+2\right)\right]^2}{4}\right)\)

\(=12\left(\frac{10}{3}-\frac{xy+2x+y+2}{4}\right)=12\left(\frac{10}{3}-\frac{6+2}{4}\right)=16\)

Vậy P min = 16 khi x = 1;y=2

Chết mọe,nhầm cmnr=(((