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NV
Nguyễn Việt Lâm
Giáo viên
18 tháng 6 2021
\(\Leftrightarrow sin\left(2x+\dfrac{\pi}{3}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+\dfrac{\pi}{3}=\dfrac{\pi}{6}+k2\pi\\2x+\dfrac{\pi}{3}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{12}+k\pi\\x=\dfrac{\pi}{4}+k\pi\end{matrix}\right.\)
Nghiệm dương nhỏ nhất là \(x=\dfrac{\pi}{4}\approx0.79\)
Đáp án C
DN
2
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 7 2021
12.
\(y=\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\le\sqrt[]{2}\)
\(\Rightarrow M=\sqrt{2}\)
13.
Pt có nghiệm khi:
\(5^2+m^2\ge\left(m+1\right)^2\)
\(\Leftrightarrow2m\le24\)
\(\Rightarrow m\le12\)
NV
Nguyễn Việt Lâm
Giáo viên
12 tháng 7 2021
14.
\(\Leftrightarrow\left[{}\begin{matrix}cosx=1\\cosx=-\dfrac{5}{3}\left(loại\right)\end{matrix}\right.\)
\(\Leftrightarrow x=k2\pi\)
15.
\(\Leftrightarrow\left[{}\begin{matrix}tanx=-1\\tanx=3\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=arctan\left(3\right)+k\pi\end{matrix}\right.\)
Đáp án A
16.
\(\dfrac{\sqrt{3}}{2}sinx-\dfrac{1}{2}cosx=\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{6}\right)=\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-\dfrac{\pi}{6}=\dfrac{\pi}{6}+k2\pi\\x-\dfrac{\pi}{6}=\dfrac{5\pi}{6}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{3}+k2\pi\\x=\pi+k2\pi\end{matrix}\right.\)
\(\left[{}\begin{matrix}2\pi\le\dfrac{\pi}{3}+k2\pi\le2018\pi\\2\pi\le\pi+k2\pi\le2018\pi\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}1\le k\le1008\\1\le k\le1008\end{matrix}\right.\)
Có \(1008+1008=2016\) nghiệm
TT
21 tháng 3 2022
1. \(limu_n=-8\)
2. \(lim(-n+6)=\)\(-\infty\)
3. \(lim\left(u_n.v_n\right)=8.\dfrac{7}{2}=4.7=28\)
4. \(lim\dfrac{6n}{n+5}=lim\dfrac{6}{1+\dfrac{5}{n}}=6\)
5. \(lim\left(\dfrac{2}{9}\right)^n=\dfrac{2^n}{9^n}=\dfrac{\left(\dfrac{2}{9}\right)^n}{\left(\dfrac{9}{9}\right)^n}=0\)
TV
giúp mình với ạ Mình cảm ơn
3
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 10 2021
5.
ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(\left(1-\dfrac{sinx}{cosx}\right)\left(sin^2x+cos^2x+2sinx.cosx\right)=1+\dfrac{sinx}{cosx}\)
\(\Leftrightarrow\dfrac{\left(cosx-sinx\right)}{cosx}\left(sinx+cosx\right)^2=\dfrac{sinx+cosx}{cosx}\)
\(\Leftrightarrow\dfrac{\left(sinx+cosx\right)\left(cos^2x-sin^2x\right)}{cosx}=\dfrac{sinx+cosx}{cosx}\)
\(\Leftrightarrow\dfrac{cos2x\left(sinx+cosx\right)}{cosx}-\dfrac{sinx+cosx}{cosx}=0\)
\(\Leftrightarrow\dfrac{sinx+cosx}{cosx}\left(cos2x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=0\\cos2x=1\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+k\pi\\x=k\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 10 2021
6.
\(\Leftrightarrow\dfrac{1}{2}-\dfrac{1}{2}cos6x-\left(\dfrac{1}{2}+\dfrac{1}{2}cos8x\right)=\dfrac{1}{2}-\dfrac{1}{2}cos10x-\left(\dfrac{1}{2}+\dfrac{1}{2}cos12x\right)\)
\(\Leftrightarrow cos6x-cos10x+cos8x-cos12x=0\)
\(\Leftrightarrow2sin2x.sin8x+2sin2x.sin10x=0\)
\(\Leftrightarrow sin2x\left(sin8x+sin10x\right)=0\)
\(\Leftrightarrow2sin2x.sin9x.cosx=0\)
\(\Leftrightarrow\left[{}\begin{matrix}sin2x=0\\sin9x=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{k\pi}{2}\\x=\dfrac{k\pi}{9}\end{matrix}\right.\)
TV
help me giúp giúp mình với ạ
0
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
a.
\(0< x< \dfrac{\pi}{2}\Rightarrow cosx>0\Rightarrow cosx=\sqrt{1-sin^2x}=\dfrac{\sqrt{6}}{3}\)
\(cos\left(x+\dfrac{\pi}{3}\right)=cosx.cos\left(\dfrac{\pi}{3}\right)-sinx.sin\left(\dfrac{\pi}{3}\right)=\dfrac{\sqrt{6}-3}{6}\)
b.
\(\pi< x< \dfrac{3\pi}{2}\Rightarrow sinx< 0\)
\(\Rightarrow sinx=-\sqrt{1-cos^2x}=-\dfrac{5}{13}\)
\(B=sin\left(\dfrac{\pi}{3}-x\right)=sin\left(\dfrac{\pi}{3}\right).cosx-cos\left(\dfrac{\pi}{3}\right).sinx=...\) (bạn tự thay số bấm máy)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
c.
\(A=cos^2x+cos^2y+2cosx.cosy+sin^2x+sin^2y+2sinx.siny\)
\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)+2\left(cosx.cosy+sinx.siny\right)\)
\(=1+1+2cos\left(x-y\right)\)
\(=2+2cos\left(\dfrac{\pi}{3}\right)=...\)
d.
\(B=cos^2x+sin^2y+2cosx.siny+cos^2y+sin^2x-2sinx.cosy\)
\(=\left(cos^2x+sin^2x\right)+\left(cos^2y+sin^2y\right)-2\left(sinx.cosy-cosx.siny\right)\)
\(=2-2sin\left(x-y\right)=2-2sin\left(\dfrac{\pi}{3}\right)=...\)
PK
5
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2021
1.
\(\Leftrightarrow\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow sin\left(x-\dfrac{\pi}{4}\right)=0\)
\(\Leftrightarrow x-\dfrac{\pi}{4}=k\pi\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+k\pi\)
2.
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=1\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=\dfrac{\pi}{4}+k2\pi\\x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=k2\pi\\x=\dfrac{\pi}{2}+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 7 2021
3.
\(\Leftrightarrow\left(sin^2x+cos^2x\right)^2-2sin^2x.cos^2x=\dfrac{5}{8}\)
\(\Leftrightarrow1-\dfrac{1}{2}sin^22x=\dfrac{5}{8}\)
\(\Leftrightarrow1-\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{2}cos4x\right)=\dfrac{5}{8}\)
\(\Leftrightarrow\dfrac{3}{4}+\dfrac{1}{4}cos4x=\dfrac{5}{8}\)
\(\Leftrightarrow cos4x=-\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=\dfrac{2\pi}{3}+k2\pi\\4x=-\dfrac{2\pi}{3}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{6}+\dfrac{k\pi}{2}\\x=-\dfrac{\pi}{6}+\dfrac{k\pi}{2}\end{matrix}\right.\)
TA
6 tháng 8 2021
b) `sin^2 3x=1`
`<=> (1-cos6x)/2=1`
`<=> 1-cos6x=2`
`<=> cos6x=-1`
`<=> 6x=π +k2π`
`<=>x=π/6 +k π/3 ( k \in ZZ)`
c) `tan^2 2x=3`
`<=> (1-cos4x)/(1+cos4x)=3`
`<=> 1-cos4x=3+3cos4x`
`<=>cos4x = -1/2`
`<=>4x= \pm (2π)/3 +k2π`
`<=>x = \pm π/6 + k π/2 (k \in ZZ)`
GM
1
11.
\(\lim\limits_{x\rightarrow-\infty}\dfrac{5x+\sqrt{3x^2-2}}{\sqrt{9x^2+1}-\left|x\right|}=\lim\limits_{x\rightarrow-\infty}\dfrac{5x+\left|x\right|\sqrt{3-\dfrac{2}{x^2}}}{\left|x\right|\sqrt{9+\dfrac{1}{x^2}}-\left|x\right|}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{5x-x\sqrt{3-\dfrac{2}{x^2}}}{-x\sqrt{9+\dfrac{1}{x^2}}+x}=\lim\limits_{x\rightarrow-\infty}\dfrac{5-\sqrt{3-\dfrac{2}{x^2}}}{-\sqrt{9+\dfrac{1}{x^2}}+1}\)
\(=\dfrac{5-\sqrt{3}}{-\sqrt{9}+1}=\dfrac{\sqrt{3}-5}{2}\)
\(\Rightarrow bc=-5.2=-10\)
12.
\(\lim\limits_{x\rightarrow-\infty}\dfrac{ax^2-4x+5}{3x^2+3x+1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(a-\dfrac{4}{x}+\dfrac{5}{x^2}\right)}{x^2\left(3+\dfrac{3}{x}+\dfrac{1}{x^2}\right)}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{a-\dfrac{4}{x}+\dfrac{5}{x^2}}{3+\dfrac{3}{x}+\dfrac{1}{x^2}}=\dfrac{a}{3}\)
\(\Rightarrow\dfrac{a}{3}=-2\Rightarrow a=-6\)