a/ \(P=sin^2x+cos^2x+cos^2x=1+cos^2x\)

Mà \(0\le cos^2x\le1\Rightarrow1\le P\le2\)

\(P_{min}=1\) khi \(cosx=0\)

\(P_{max}=2\) khi \(cosx=\pm1\)

b/ \(P=8sin^2x+3\left(1-2sin^2x\right)=3+2sin^2x\)

Mà \(0\le sin^2x\le1\Rightarrow3\le P\le5\)

\(P_{min}=3\) khi \(sinx=0\)

\(P_{max}=5\) khi \(sinx=\pm1\)

c/ \(P=\left(sin^2x-cos^2x\right)\left(sin^2x+cos^2x\right)=sin^2x-cos^2x=-cos2x\)

Mà \(-1\le cos2x\le1\Rightarrow-1\le P\le1\)

\(P_{min}=-1\) khi \(cos2x=1\)

\(P_{max}=1\) khi \(cos2x=-1\)

d/ \(P=\left(sin^2x+cos^2x\right)^3-3sin^2x.cos^2x\left(sin^2x+cos^2x\right)\)

\(=1-3sin^2x.cos^2x=1-\frac{3}{4}\left(2sinx.cosx\right)^2=1-\frac{3}{4}sin^22x\)

Mà \(0\le sin^22x\le1\Rightarrow\frac{1}{4}\le P\le1\)

\(P_{min}=\frac{1}{4}\) khi \(sin2x=\pm1\)

\(P_{max}=1\) khi \(sin2x=0\)

thanks bn nhiều

\(A=3sin^2x+6cos^2=3sin^2x+6\left(1-sin^2x\right)\)

\(=6-3sin^2x\)

Do : \(0\le sin^2x\le1\Rightarrow\left\{{}\begin{matrix}6-3sin^2x\ge3\\6-3sin^2x\le6\end{matrix}\right.\)

\(P=\frac{1-sin^2x.cos^2x}{cos^2x}-cos^2x=\frac{1}{cos^2x}-sin^2x-cos^2x\)

\(=1+tan^2x-\left(sin^2x+cos^2x\right)=1+tan^2x-1=tan^2x\)

\(M=\frac{2cos^2x-1}{sinx+cosx}=\frac{2cos^2x-\left(sin^2x+cos^2x\right)}{sinx+cosx}=\frac{cos^2x-sin^2x}{sinx+cosx}\)

\(\frac{\left(cosx-sinx\right)\left(cosx+sinx\right)}{sinx+cosx}=cosx-sinx\)

\(P=4sin^2x+\sqrt{2}\left(sin2x.cos\frac{\pi}{4}+cos2x.sin\frac{\pi}{4}\right)\)

\(P=4sin^2x+sin2x+cos2x\)

\(P=2\left(1-cos2x\right)+sin2x+cos2x\)

\(P=2+sin2x-cos2x\)

\(P=2+\sqrt{2}sin\left(2x-\frac{\pi}{4}\right)\)

Do \(sin\left(2x-\frac{\pi}{4}\right)\le1\Rightarrow P\le2+\sqrt{2}\)

\(\Rightarrow P_{max}=2+\sqrt{2}\) khi \(sin\left(2x-\frac{\pi}{4}\right)=1\Leftrightarrow x=\frac{3\pi}{8}+k\pi\)

\(P=sin^4x-cos^4x=\left(sin^2x+cos^2x\right)\left(sin^2x-cos^2x\right)\)

\(\Rightarrow P=-\left(cos^2x-sin^2x\right)=-cos2x\)

Do \(-1\le cos2x\le1\Rightarrow-1\le P\le1\)

\(\Rightarrow\left\{{}\begin{matrix}P_{min}=-1\Rightarrow x=k\pi\\P_{max}=1\Rightarrow x=\frac{\pi}{2}+k\pi\end{matrix}\right.\)

b/

\(P=sin^6x+cos^6x=\left(sin^2x+cos^2x\right)\left(sin^4x+cos^4x-sin^2x.cos^2x\right)\)

\(P=sin^4x+cos^4x+2sin^2x.cos^2x-3sin^2x.cos^2x\)

\(P=\left(sin^2x+cos^2x\right)^2-\frac{3}{4}\left(2sinx.cosx\right)^2\)

\(P=1-\frac{3}{4}sin^22x\)

Do \(0\le sin^22x\le1\Rightarrow\frac{1}{4}\le P\le1\)

\(\Rightarrow\left\{{}\begin{matrix}P_{min}=\frac{1}{4}\Rightarrow x=\frac{k\pi}{4}\\P_{max}=1\Rightarrow x=\frac{k\pi}{2}\end{matrix}\right.\)

c/

\(P=1-2\left|cos3x\right|\)

Do \(0\le\left|cos3x\right|\le1\Rightarrow-1\le P\le1\)

\(\Rightarrow\left\{{}\begin{matrix}P_{min}=-1\Rightarrow x=\frac{k\pi}{3}\\P_{max}=1\Rightarrow x=\frac{k\pi}{6}\end{matrix}\right.\)

câu 1) ta có : \(M=\left(x^2-x\right)^2+\left(2x-1\right)^2=x^4-2x^3+x^2+4x^2-4x+1\)

\(=\left(x^2-x+2\right)^2-3=\left(\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}\right)^2-3\)

\(\Rightarrow\dfrac{1}{16}\le M\le61\)

\(\Rightarrow M_{min}=\dfrac{1}{16}\)khi \(x=\dfrac{1}{2}\) ; \(M_{max}=61\) khi \(x=3\)

câu 2) điều kiện xác định : \(0\le x\le2\)
đặt \(\sqrt{2x-x^2}=t\left(t\ge0\right)\)

\(\Rightarrow M=-t^2+4t+3=-\left(t-2\right)^2+7\)

\(\Rightarrow3\le M\le7\)

câu 3) ta có : \(M=\left(x-2\right)^2+6\left|x-2\right|-6\ge-6\)

\(\Rightarrow M_{min}=-6\) khi \(x=2\)

4) điều kiện xác định \(-6\le x\le10\)

ta có : \(M=5\sqrt{x+6}+2\sqrt{10-x}-2\)

áp dụng bunhiacopxki dạng căn ta có :

\(-\sqrt{\left(5^2+2^2\right)\left(x+6+10-x\right)}\le5\sqrt{x+6}+2\sqrt{10-x}\le\sqrt{\left(5^2+2^2\right)\left(x+6+10-x\right)}\)

\(\Leftrightarrow-4\sqrt{29}\le5\sqrt{x+6}+2\sqrt{10-x}\le4\sqrt{29}\)

\(\Rightarrow-2-4\sqrt{29}\le B\le-2+4\sqrt{29}\)

\(\Rightarrow M_{max}=-2+4\sqrt{29}\) khi \(\dfrac{\sqrt{x+6}}{5}=\dfrac{\sqrt{10-x}}{2}\Leftrightarrow x=\dfrac{226}{29}\)