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Cho \(0< x< y\le z\le1\) và \(3x+2y+z\le4\). Tìm Max \(S=3x^2+2y^2+z^2\) - Hoc24
Tham khảo
Khai triển Abel ta có:
\(S=\left(z-y\right)z+\left(y-x\right)\left(z+2y\right)+x\left(3x+2y+z\right)\)
\(\le\left(z-y\right).1+\left(y-x\right).3+4x=x+2y+z\)
\(=\left(1-1\right)z+\left(1-\dfrac{1}{3}\right)\left(2y+z\right)+\dfrac{1}{3}\left(3x+2y+z\right)\)
\(\le\dfrac{2}{3}.3+\dfrac{1}{3}.4=\dfrac{10}{3}\)
Dấu = xảy ra khi \(x=\dfrac{1}{3},y=z=1\)
Với \(cosx=0\) không phải nghiệm
Với \(cosx\ne0\) , chia 2 vế cho \(cos^3x\):
\(4tan^3x+3tan^2x-tanx.\left(1+tan^2x\right)-1=0\)
\(\Leftrightarrow3tan^3x+3tan^2x-tanx-1=0\)
\(\Leftrightarrow\left(tanx+1\right)\left(3tan^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=\dfrac{1}{\sqrt{3}}\\tanx=-\dfrac{1}{\sqrt{3}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=\pm\dfrac{\pi}{6}+k\pi\end{matrix}\right.\)
a. \(y'=3sin^2x.\left(sinx\right)'=3sin^2x.cosx\)
b. \(y'=3cos^2x.\left(cosx\right)'=-3cos^2x.sinx\)
c. \(y'=cosx.cos^2x+2cosx.\left(-sinx\right).sinx=cos^3x-2cosx.sin^2x\)
d. \(y=x^{\dfrac{1}{3}}+\left(x+1\right)^{\dfrac{2}{3}}\Rightarrow y'=\dfrac{1}{3}x^{-\dfrac{2}{3}}+\dfrac{2}{3}\left(x+1\right)^{-\dfrac{1}{3}}=\dfrac{1}{3\sqrt[3]{x^2}}+\dfrac{2}{3\sqrt[3]{x+1}}\)
\(\Leftrightarrow sin^3x+3sin^2x+3sinx+1-cos^3x+sinx-cosx+1=0\)
\(\Leftrightarrow\left(sinx+1\right)^3-cos^3x+sinx-cosx+1=0\)
\(\Leftrightarrow\left(sinx-cosx+1\right)\left[\left(sinx+1\right)^2+cosx\left(sinx+1\right)+cos^2x\right]+sinx-cosx+1=0\)
\(\Leftrightarrow\left(sinx-cosx+1\right)\left(2sinx+sinx.cosx+cosx+2\right)+sinx-cosx+1=0\)
\(\Leftrightarrow\left(sinx-cosx+1\right)\left(2sinx+cosx+sinx.cosx+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sinx-cosx=-1\Leftrightarrow sin\left(x-\frac{\pi}{4}\right)=-\frac{\sqrt{2}}{2}\Leftrightarrow...\\2sinx+cosx+sinx.cosx+3=0\left(1\right)\end{matrix}\right.\)
Xét (1):
\(\Leftrightarrow2\left(sinx+1\right)+cosx\left(sinx+1\right)+1=0\)
\(\Leftrightarrow\left(cosx+2\right)\left(sinx+1\right)+1=0\)
Do \(sinx;cosx\ge-1\Rightarrow\left(cosx+2\right)\left(sinx+1\right)\ge0\)
\(\Rightarrow\left(cosx+2\right)\left(sinx+1\right)+1=0\) vô nghiệm
a, (sinx + cosx)(1 - sinx . cosx) = (cosx - sinx)(cosx + sinx)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx-sinx=1-sinx.cosx\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\cosx+sinx.cosx-1-sinx=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sinx+cosx=0\\\left(cosx-1\right)\left(sinx+1\right)=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x+\dfrac{\pi}{4}\right)=0\\cosx=1\\sinx=-1\end{matrix}\right.\)
b, (sinx + cosx)(1 - sinx . cosx) = 2sin2x + sinx + cosx
⇔ (sinx + cosx)(1 - sinx.cosx - 1) = 2sin2x
⇔ (sinx + cosx).(- sinx . cosx) = 2sin2x
⇔ 4sin2x + (sinx + cosx) . sin2x = 0
⇔ \(\left[{}\begin{matrix}sin2x=0\\\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)+4=0\end{matrix}\right.\)
⇔ sin2x = 0
c, 2cos3x = sin3x
⇔ 2cos3x = 3sinx - 4sin3x
⇔ 4sin3x + 2cos3x - 3sinx(sin2x + cos2x) = 0
⇔ sin3x + 2cos3x - 3sinx.cos2x = 0
Xét cosx = 0 : thay vào phương trình ta được sinx = 0. Không có cung x nào có cả cos và sin = 0 nên cosx = 0 không thỏa mãn phương trình
Xét cosx ≠ 0 chia cả 2 vế cho cos3x ta được :
tan3x + 2 - 3tanx = 0
⇔ \(\left[{}\begin{matrix}tanx=1\\tanx=-2\end{matrix}\right.\)
d, cos2x - \(\sqrt{3}sin2x\) = 1 + sin2x
⇔ cos2x - sin2x - \(\sqrt{3}sin2x\) = 1
⇔ cos2x - \(\sqrt{3}sin2x\) = 1
⇔ \(2cos\left(2x+\dfrac{\pi}{3}\right)=1\)
⇔ \(cos\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}=cos\dfrac{\pi}{3}\)
e, cos3x + sin3x = 2cos5x + 2sin5x
⇔ cos3x (1 - 2cos2x) + sin3x (1 - 2sin2x) = 0
⇔ cos3x . (- cos2x) + sin3x . cos2x = 0
⇔ \(\left[{}\begin{matrix}sin^3x=cos^3x\\cos2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}sinx=cosx\\cos2x=0\end{matrix}\right.\)
⇔ \(\left[{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)=0\\cos2x=0\end{matrix}\right.\)
a, Đặt \(t=cos3x\left(t\in\left[-1;1\right]\right)\)
\(y=9-sin^23x-\sqrt{2}cos3x\)
\(=cos^23x-\sqrt{2}cos3x+8\)
\(\Leftrightarrow y=f\left(t\right)=t^2-\sqrt{2}t+8\)
\(\Rightarrow minf\left(t\right)\le y\le maxf\left(x\right)\)
\(\Rightarrow min\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{\sqrt{2}}{2}\right)\right\}\le y\le max\left\{f\left(-1\right);f\left(1\right);f\left(\dfrac{\sqrt{2}}{2}\right)\right\}\)
\(\Rightarrow\dfrac{15}{2}\le y\le9+\sqrt{2}\)
\(\Rightarrow y_{max}=9+\sqrt{2}\)
b, Đặt \(t=sin3x\left(t\in\left[-1;1\right]\right)\)
\(y=3sin3x-8cos^23x+4\)
\(=3sin3x+8-8cos^23x-4\)
\(=8sin^23x+3sin3x-4\)
\(\Leftrightarrow y=f\left(t\right)=8t^2+3t-4\)
\(\Rightarrow minf\left(x\right)\le y\le maxf\left(t\right)\)
\(\Rightarrow min\left\{f\left(-1\right);f\left(1\right);f\left(-\dfrac{3}{16}\right)\right\}\le y\le max\left\{f\left(-1\right);f\left(1\right);f\left(-\dfrac{3}{16}\right)\right\}\)
\(\Rightarrow-\dfrac{137}{32}\le y\le7\)
\(\Rightarrow y_{max}=7\)