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NV
9 tháng 9 2021

ĐKXĐ:

a.

\(sin5x\ne0\Leftrightarrow5x\ne k\pi\Rightarrow x\ne\dfrac{k\pi}{5}\)

b.

\(cos6x\ne0\Leftrightarrow6x\ne\dfrac{\pi}{2}+k\pi\Rightarrow x\ne\dfrac{\pi}{12}+\dfrac{k\pi}{6}\)

d.

\(sin\left(3x-\dfrac{\pi}{6}\right)\ne0\Leftrightarrow3x-\dfrac{\pi}{6}\ne k\pi\Rightarrow x\ne\dfrac{\pi}{18}+\dfrac{k\pi}{3}\)

e.

\(sin\left(4x-\dfrac{\pi}{3}\right)\ne0\Leftrightarrow4x-\dfrac{\pi}{3}\ne k\pi\Rightarrow x\ne\dfrac{\pi}{12}+\dfrac{k\pi}{4}\)

NV
15 tháng 7 2021

a.

\(\left\{{}\begin{matrix}sin\left(3x+\dfrac{\pi}{6}\right)\ne0\\cos2x\ne0\\sinx\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne-\dfrac{\pi}{18}+\dfrac{k\pi}{3}\\x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\\x\ne-\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)

b.

Do \(5+2cot^2x-sinx=4+2cot^2x+\left(1-sinx\right)>0\) nên hàm xác định khi:

\(\left\{{}\begin{matrix}sinx\ne0\\sin\left(x+\dfrac{\pi}{2}\right)\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\end{matrix}\right.\) \(\Leftrightarrow sin2x\ne0\)

\(\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

NV
10 tháng 9 2021

a.

\(y'=-\dfrac{3}{2}x^3+\dfrac{6}{5}x^2-x+5\)

b.

\(y'=\dfrac{\left(x^2+4x+5\right)'}{2\sqrt{x^2+4x+5}}=\dfrac{2x+4}{2\sqrt{x^2+4x+5}}=\dfrac{x+2}{\sqrt{x^2+4x+5}}\)

c.

\(y=\left(3x-2\right)^{\dfrac{1}{3}}\Rightarrow y'=\dfrac{1}{3}\left(3x-2\right)^{-\dfrac{2}{3}}=\dfrac{1}{3\sqrt[3]{\left(3x-2\right)^2}}\)

d.

\(y'=2\sqrt{x+2}+\dfrac{2x-1}{2\sqrt{x+2}}=\dfrac{6x+7}{2\sqrt{x+2}}\)

e.

\(y'=3sin^2\left(\dfrac{\pi}{3}-5x\right).\left[sin\left(\dfrac{\pi}{3}-5x\right)\right]'=-15sin^2\left(\dfrac{\pi}{3}-5x\right).cos\left(\dfrac{\pi}{3}-5x\right)\)

g.

\(y'=4cot^3\left(\dfrac{\pi}{6}-3x\right)\left[cot\left(\dfrac{\pi}{3}-3x\right)\right]'=12cot^3\left(\dfrac{\pi}{6}-3x\right).\dfrac{1}{sin^2\left(\dfrac{\pi}{3}-3x\right)}\)

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NV
22 tháng 3 2021

a.

\(y'=\dfrac{3}{cos^2\left(3x-\dfrac{\pi}{4}\right)}-\dfrac{2}{sin^2\left(2x-\dfrac{\pi}{3}\right)}-sin\left(x+\dfrac{\pi}{6}\right)\)

b.

\(y'=\dfrac{\dfrac{\left(2x+1\right)cosx}{2\sqrt{sinx+2}}-2\sqrt{sinx+2}}{\left(2x+1\right)^2}=\dfrac{\left(2x+1\right)cosx-4\left(sinx+2\right)}{\left(2x+1\right)^2}\)

c.

\(y'=-3sin\left(3x+\dfrac{\pi}{3}\right)-2cos\left(2x+\dfrac{\pi}{6}\right)-\dfrac{1}{sin^2\left(x+\dfrac{\pi}{4}\right)}\)

NV
6 tháng 6 2021

1.

ĐKXĐ: \(\left\{{}\begin{matrix}cosx\ne0\\tanx-sinx\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\\dfrac{sinx}{cosx}-sinx\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}cosx\ne0\\sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

2.

ĐKXĐ: \(sin2x\ne0\Leftrightarrow x\ne\dfrac{k\pi}{2}\)

3. 

ĐKXĐ: \(\left\{{}\begin{matrix}sin\left(x-\dfrac{\pi}{4}\right)\ne0\\cos\left(x-\dfrac{\pi}{4}\right)\ne0\end{matrix}\right.\)

\(\Leftrightarrow sin\left(2x-\dfrac{\pi}{2}\right)\ne0\Leftrightarrow cos2x\ne0\)

\(\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

6 tháng 6 2021

cho hỏi cái này tí nha    \(sin\alpha\)=1/2  và \(cos\alpha\)=\(\dfrac{-\sqrt{3}}{2}\)

thì góc đó là \(\alpha=?\pi\)

NV
30 tháng 6 2021

ĐKXĐ:

a. \(cosx\ne0\Leftrightarrow x\ne\dfrac{\pi}{2}+k\pi\)

b. \(cos2x\ne0\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

c. \(sin\left(x+\dfrac{\pi}{4}\right)\ne0\Leftrightarrow x+\dfrac{\pi}{4}\ne k\pi\Leftrightarrow x\ne-\dfrac{\pi}{4}+k\pi\)

25 tháng 6 2021

1. \(sin\left(\dfrac{\pi}{3}-x\right)\ne0\Leftrightarrow\dfrac{\pi}{3}-x\ne k\pi\Leftrightarrow x\ne\dfrac{\pi}{3}-k\pi\)

2. \(cos2x\ne0\Leftrightarrow2x\ne\dfrac{\pi}{2}+k\pi\Leftrightarrow x\ne\dfrac{\pi}{4}+\dfrac{k\pi}{2}\)

3. \(\sqrt{1+sinx}-\sqrt{2}\ge0\Leftrightarrow1+sinx\ge2\Leftrightarrow sinx\ge1\Leftrightarrow sinx=1\Leftrightarrow x=\dfrac{\pi}{2}+k2\pi\)

4. \(\sqrt{2-2cosx}-2\ne0\Leftrightarrow2-2cosx\ne4\Leftrightarrow cosx\ne-1\Leftrightarrow x\ne\pi+k2\pi\)

5. \(1-\sqrt{1+sin3x}\ne0\Leftrightarrow sin3x\ne0\Leftrightarrow3x\ne k\pi\Leftrightarrow x\ne\dfrac{k\pi}{3}\)

26 tháng 6 2021

câu 4 sao ra 2-2cosx\(\ne\)4 ạ