\(y=-\dfrac{cosx}{3sin^2x}+\dfrac{4}{3}cotx\)

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2 tháng 5 2017

\(y'=-\left(\dfrac{-sinx.3sin^2x-cosx.3cos^2x}{9sin^4x}\right)-\dfrac{4}{3sin^2x}=\dfrac{3sin^3x+3cos^3x-12sin^2x}{9sin^4x}\)

19 tháng 12 2018

cotx=\(\dfrac{\cos x}{\sin x}\left(đk:\sin x\ne0\right)\)

\(\Rightarrow1+\dfrac{\cos x}{\sin x}=1-\dfrac{\cos x}{\sin^2x}\)

\(\Leftrightarrow\dfrac{\cos x}{\sin x}+\dfrac{\cos x}{\sin^2x}=0\)

\(\Leftrightarrow\cos x\sin x+\cos x=0\)

\(\Leftrightarrow\cos x\left(\sin x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\cos x=0\\\sin x+1=0\end{matrix}\right.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\sqcap}{2}+k\sqcap\\x=\dfrac{-\sqcap}{2}+k2\sqcap\end{matrix}\right.\)(tmđk)

9 tháng 4 2017

a) y' = 5x4 - 12x2 + 2.

b) y' = - + 2x - 2x3.

c) y' = 2x3 - 2x2 + .

d) y = 24x5 - 9x7 => y' = 120x4 - 63x6.

9 tháng 4 2017

a) = = .

b) = = .

c) = = .

d) y' =\(\dfrac{\left(x^2+7x+3\right)'\left(x^2-3x\right)-\left(x^2+7x+3\right)\left(x^2-3x\right)'}{\left(x^2-3x\right)^2}\)=\(\dfrac{\left(2x+7\right)\left(x^2-3x\right)-\left(x^2+7x+3\right)\left(2x-3\right)}{\left(x^2-3x\right)^2}\)=\(\dfrac{-2x^2-6x+9}{\left(x^2-3x\right)^2}\)

20 tháng 8 2018

a.\(\dfrac{sin2x+cosx-\sqrt{3}\left(cos2x+sinx\right)}{2sin2x-\sqrt{3}}=1\left(1\right)\)

ĐKXĐ: sin2x≠\(\dfrac{\sqrt{3}}{2}\)

(1) ⇔ sin2x + cosx - \(\sqrt{3}\) ( cos2x + sinx) = 2sin2x - \(\sqrt{3}\)

⇔cosx - \(\sqrt{3}\) sinx = \(\sqrt{3}\) cos2x + sin2x +\(\sqrt{3}\)

\(\dfrac{1}{2}cosx-\dfrac{\sqrt{3}}{2}sinx=\dfrac{\sqrt{3}}{2}cos2x+\dfrac{1}{2}sin2x+\dfrac{\sqrt{3}}{2}\)

\(sin\left(\dfrac{\Pi}{6}-x\right)=sin\left(2x+\dfrac{\Pi}{3}\right)-sin\dfrac{\Pi}{3}\)

\(sin\left(\dfrac{\Pi}{6}-x\right)=2cos\left(x+\dfrac{\Pi}{3}\right)sinx\)

\(sin\left(\dfrac{\Pi}{6}-x\right)=2sin\left(\dfrac{\Pi}{6}-x\right)sinx\)

\(sin\left(\dfrac{\Pi}{6}-x\right)\left(2sinx-1\right)=0\)

Đến đây tự giải tiếp nha nhớ đối chiếu đk.

20 tháng 8 2018

b.\(\left(2cosx-1\right)cotx=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\left(1\right)\)

ĐKXĐ: sinx≠0 và cosx≠1

(1)⇔\(\left(2cosx-1\right)\dfrac{cosx}{sinx}=\dfrac{3}{sinx}+\dfrac{2sinx}{cosx-1}\)

⇔cosx(2cosx-1)(cosx-1) = 3(cosx-1) + 2sin2x

⇔2cos3x - cos2x - 2cosx +1 = 0

⇔ (cosx-1)(cosx+1)(2cosx-1)=0

NV
18 tháng 9 2020

36.

\(sin^2x-cos^2x\ne0\Leftrightarrow cos2x\ne0\)

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

37.

\(cos3x\ne cosx\Leftrightarrow\left\{{}\begin{matrix}3x\ne x+k2\pi\\3x\ne-x+k2\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne k\pi\\x\ne\frac{k\pi}{2}\end{matrix}\right.\) \(\Leftrightarrow x\ne\frac{k\pi}{2}\)

38.

\(\left\{{}\begin{matrix}x\ge0\\sin\pi x\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\pi x\ne k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne k\end{matrix}\right.\)

39.

\(\left\{{}\begin{matrix}cos\left(x-\frac{\pi}{3}\right)\ne0\\tan\left(x-\frac{\pi}{3}\right)\ne-1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x-\frac{\pi}{3}\ne\frac{\pi}{2}+k\pi\\x-\frac{\pi}{3}\ne-\frac{\pi}{4}+k\pi\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{5\pi}{6}+k\pi\\x\ne-\frac{\pi}{12}+k\pi\end{matrix}\right.\)

NV
18 tháng 9 2020

33.

\(\left\{{}\begin{matrix}cosx\ne0\\cos\frac{x}{2}\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{\pi}{2}+k\pi\\x\ne\pi+k2\pi\end{matrix}\right.\)

34.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne0\\cotx\ne1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}sin2x\ne0\\cotx\ne1\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ne\frac{k\pi}{2}\\x\ne\frac{\pi}{4}+k\pi\end{matrix}\right.\)

35.

\(\left\{{}\begin{matrix}sinx\ne0\\cosx\ne1\end{matrix}\right.\) \(\Leftrightarrow sinx\ne0\)

\(\Leftrightarrow x\ne k\pi\)