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21 tháng 7 2022
a: \(\Leftrightarrow\dfrac{2}{\sin2x}=2\)
\(\Leftrightarrow\sin2x=1\)
\(\Leftrightarrow2x=\dfrac{\Pi}{2}+k2\Pi\)
hay
b: \(\Leftrightarrow3\cdot tan^4x+3tan^2x-tan^2x-1=0\)
\(\Leftrightarrow3tan^2x-1=0\)
\(\Leftrightarrow tan^2x=\dfrac{1}{3}\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(\dfrac{1}{\sqrt{3}}\right)+k\Pi=\dfrac{\Pi}{6}+k\Pi\\x=-\dfrac{\Pi}{6}+k\Pi\end{matrix}\right.\)
TT
1
21 tháng 7 2022
a: \(\Leftrightarrow1-cos^4x-cos^2x=1\)
\(\Leftrightarrow cos^2x\left(cos^2x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\cosx=1\\cosx=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{2}+k\Pi\\x=k2\Pi\\x=\Pi+k2\Pi\end{matrix}\right.\)
b: \(\Leftrightarrow3\left(1+\tan^2x\right)+2\sqrt{3}tanx-6=0\)
\(\Leftrightarrow3\cdot tan^2x+2\sqrt{3}\cdot tanx-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=\dfrac{\sqrt{3}}{3}\\tanx=-\sqrt{3}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\Pi}{6}+k\Pi\\x=-\dfrac{\Pi}{3}+k\Pi\end{matrix}\right.\)
AH
Akai Haruma
Giáo viên
21 tháng 1 2020
Câu 2:
\(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)}=\frac{2-1}{1.2}+\frac{3-2}{2.3}+...+\frac{(n+1)-n}{n(n+1)}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...\frac{1}{n}-\frac{1}{n+1}\)
\(=1-\frac{1}{n+1}\)
\(\Rightarrow \lim_{n\to \infty}(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{n(n+1)})=\lim_{n\to \infty}(1-\frac{1}{n+1})=1-\lim_{n\to \infty}\frac{1}{n+1}=1-0=1\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 10 2020
d.
\(\sqrt{2}sin\left(x+\frac{\pi}{4}\right)=\sqrt{2}\)
\(\Leftrightarrow sin\left(x+\frac{\pi}{4}\right)=1\)
\(\Leftrightarrow x+\frac{\pi}{4}=\frac{\pi}{2}+k2\pi\)
\(\Leftrightarrow x=\frac{\pi}{4}+k2\pi\)
e.
\(\Leftrightarrow cosx.cos\left(\frac{\pi}{12}\right)-sinx.sin\left(\frac{\pi}{12}\right)=\frac{1}{2}\)
\(\Leftrightarrow cos\left(x+\frac{\pi}{12}\right)=\frac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\frac{\pi}{12}=\frac{\pi}{3}+k2\pi\\x+\frac{\pi}{12}=-\frac{\pi}{3}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
28 tháng 10 2020
2.a.
ĐKXĐ: ...
\(\sqrt{3}tanx-\frac{6}{tanx}+2\sqrt{3}-3=0\)
\(\Leftrightarrow\sqrt{3}tan^2x+\left(2\sqrt{3}-3\right)tanx-6=0\)
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-2\\tanx=\sqrt{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=arctan\left(-2\right)+k\pi\\x=\frac{\pi}{3}+k\pi\end{matrix}\right.\)
b.
ĐKXĐ: \(x\ne k\pi\)
\(1-sin2x=2sin^2x\)
\(\Leftrightarrow1-2sin^2x-sin2x=0\)
\(\Leftrightarrow cos2x-sin2x=0\)
\(\Leftrightarrow cos\left(2x+\frac{\pi}{4}\right)=0\)
\(\Leftrightarrow...\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
a/
\(\Leftrightarrow3\left(1-sin^22x\right)+4sin2x-4=0\)
\(\Leftrightarrow-3sin^22x+4sin2x-1=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=1\\sin2x=\frac{1}{3}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{4}+k\pi\\x=\frac{1}{2}arcsin\left(\frac{1}{3}\right)+k\pi\\x=\frac{\pi}{2}-\frac{1}{2}arcsin\left(\frac{1}{3}\right)+k\pi\end{matrix}\right.\)
b/
\(\Leftrightarrow\left[{}\begin{matrix}cos2x=-1\\cos2x=\frac{\sqrt{2}}{2}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\frac{\pi}{2}+k\pi\\x=\frac{\pi}{8}+k\pi\\x=-\frac{\pi}{8}+k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 10 2020
f/
\(\Leftrightarrow4\left(1-2sin^2\frac{x}{2}\right)-5sin\frac{x}{2}=1\)
\(\Leftrightarrow8sin^2\frac{x}{2}+5sin\frac{x}{2}-3=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin\frac{x}{2}=-1\\sin\frac{x}{2}=\frac{3}{8}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\pi+k4\pi\\x=2arcsin\left(\frac{3}{8}\right)+k4\pi\\x=2\pi-2arcsin\left(\frac{3}{8}\right)+k4\pi\end{matrix}\right.\)
3 tháng 10 2021
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NV
Nguyễn Việt Lâm
Giáo viên
13 tháng 4 2020
1/ \(y=x^{-1}+\frac{2}{3}x^{-2}-\frac{2}{3}\Rightarrow y'=-\frac{1}{x^2}-\frac{4}{3x^3}\)
\(3x^3y'+3x+4=3x^3\left(-\frac{1}{x^2}-\frac{4}{3x^3}\right)+3x+4\)
\(=-3x-4+3x+4=0\) (đpcm)
2/ \(y'\le0\)
\(\Leftrightarrow3x^2-10x+7\le0\)
\(\Leftrightarrow1\le x\le\frac{7}{3}\)
PH
25 tháng 2 2020
lim\(\frac{3n^2+n-5}{2n^2+1}\)=lim\(\frac{n^2\left(3+\frac{1}{n}-\frac{5}{n^2}\right)}{n^2\left(2+\frac{1}{n}\right)}\)=\(\frac{3}{2}\)
PH
25 tháng 2 2020
lim\(\frac{\sqrt{9n^2-n}+1}{4n-2}\)=lim\(\frac{n\sqrt{9-\frac{1}{n}+\frac{1}{n^2}}}{n\left(4-\frac{2}{n}\right)}\)=lim\(\frac{\sqrt{9}}{4}\)=\(\frac{3}{2}\)
\(\Leftrightarrow\dfrac{2}{\sin2x}=2\)
\(\Leftrightarrow\sin2x=1\)
\(\Leftrightarrow2x=\dfrac{\Pi}{2}+k2\Pi\)
hay \(x=\dfrac{\Pi}{4}+k\Pi\)