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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. \(y=x^{\dfrac{1}{3}}+\left(x+1\right)^{\dfrac{2}{3}}\Rightarrow y'=\dfrac{1}{3}x^{-\dfrac{2}{3}}+\dfrac{2}{3}\left(x+1\right)^{-\dfrac{1}{3}}=\dfrac{1}{3\sqrt[3]{x^2}}+\dfrac{2}{3\sqrt[3]{x+1}}\)
1, \(y=2-sin\left(\dfrac{3x}{2}+x\right).cos\left(x+\dfrac{\pi}{2}\right)\)
\(y=2-\left(-cosx\right).\left(-sinx\right)\)
y = 2 - sinx.cosx
y = \(2-\dfrac{1}{2}sin2x\)
Max = 2 + \(\dfrac{1}{2}\) = 2,5
Min = \(2-\dfrac{1}{2}\) = 1,5
2, y = \(\sqrt{5-\dfrac{1}{2}sin^22x}\)
Min = \(\sqrt{5-\dfrac{1}{2}}=\dfrac{3\sqrt{2}}{2}\)
Max = \(\sqrt{5}\)
Hàm số xác định khi: \(\left\{{}\begin{matrix}tanx\ne\pm1;cosx\ne0\\cosx\ne-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\pm\dfrac{\pi}{4}+k\pi\\x\ne\dfrac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)
a: ĐKXĐ: \(x\in R\)
=>TXĐ: D=R
b; ĐKXĐ: 2x-4>=0
=>x>=2
TXĐ: D=[2;+\(\infty\))
c: ĐKXĐ: 1-cos^2x>=0
=>sin^2x>=0(luôn đúng)
ĐK: Biểu thức xác định với mọi `x`.
`y_(min) <=> (\sqrt(2-cos(x-π/6))+3)_(max) <=> (cos(x-π/6))_(max)`
`<=> cos(x-π/6)=1 <=> x-π/6=k2π <=> x = π/6+k2π ( k \in ZZ)`.
`=> y_(min) = 1`
`y_(max) <=> (\sqrt(2-cos(x-π/6))+3)_(min) <=> (cos(x-π/6))_(min)`
`<=> cos(x-π/6)=-1 <=> x -π/6= π+k2π <=> x = (7π)/6+k2π (k \in ZZ)`
`=> y_(max) = (6-2\sqrt3)/3`.
a, Hàm số xác định khi: \(\left\{{}\begin{matrix}cos\dfrac{x}{2}\ne3\\tanx\ne\sqrt{3}\\cosx\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne\dfrac{\pi}{3}+k\pi\\x\ne\dfrac{\pi}{2}+k\pi\end{matrix}\right.\)
a, \(cos3x^2\in\left[-1;1\right]\)
\(\Rightarrow1-cos3x^2\in\left[0;2\right]\)
\(\Rightarrow\sqrt{1-cos3x^2}\in\left[0;\sqrt{2}\right]\)
\(\Rightarrow y=\sqrt{1-cos3x^2}-2\in\left[-2;\sqrt{2}-2\right]\)
\(\Rightarrow y_{min}=-2\Leftrightarrow cos3x^2=1\Leftrightarrow3x^2=k2\pi\Leftrightarrow x=\pm\sqrt{\dfrac{k2\pi}{3}}\)
b, ĐK: \(x\ge1\)
\(cos\sqrt{x-1}\in\left[-1;1\right]\)
\(\Rightarrow y=2008cos\sqrt{x-1}\in\left[-2008;2008\right]\)
\(\Rightarrow y_{min}=-2008\Leftrightarrow cos\sqrt{x-1}=-1\Leftrightarrow\sqrt{x-1}=\pi+k2\pi\Leftrightarrow x=1+\left(\pi+k2\pi\right)^2\)
\(y_{max}=2008\Leftrightarrow cos\sqrt{x-1}=1\Leftrightarrow\sqrt{x-1}=k2\pi\Leftrightarrow x=1+4k^2\pi^2\)