K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
HT
Làm giúp mình 2 bài này với, có giải chi tiết. Mình cảm ơn nhiều
0
NV
Nguyễn Việt Lâm
Giáo viên
8 tháng 3 2022
\(IM=\dfrac{1}{4}IB\Rightarrow IM=\dfrac{1}{5}BM\Rightarrow\overrightarrow{MI}=\dfrac{1}{5}\overrightarrow{MB}=-\dfrac{1}{10}\left(\overrightarrow{BC}+\overrightarrow{BD}\right)\)
\(\Rightarrow\overrightarrow{DI}=\overrightarrow{DM}+\overrightarrow{MI}=\dfrac{1}{2}\overrightarrow{DC}-\dfrac{1}{10}\left(\overrightarrow{BC}+\overrightarrow{BD}\right)=\dfrac{1}{2}\overrightarrow{DB}+\dfrac{1}{2}\overrightarrow{BC}-\dfrac{1}{10}\overrightarrow{BC}-\dfrac{1}{10}\overrightarrow{BD}\)
\(\Rightarrow\overrightarrow{DI}=\dfrac{2}{5}\overrightarrow{BC}-\dfrac{3}{5}\overrightarrow{BD}\)
\(\overrightarrow{DJ}=\overrightarrow{DC}+\overrightarrow{CJ}=\overrightarrow{DB}+\overrightarrow{BC}+x.\overrightarrow{CB}=\left(1-x\right)\overrightarrow{BC}-\overrightarrow{BD}\)
D; I; J thẳng hàng \(\Rightarrow\dfrac{1-x}{\dfrac{2}{5}}=\dfrac{1}{\dfrac{3}{5}}\Rightarrow x=\dfrac{1}{3}\)
\(\Rightarrow CJ=\dfrac{1}{3}CB\Rightarrow BJ=\dfrac{2}{3}BC\Rightarrow\dfrac{BJ}{BC}=\dfrac{2}{3}\)
Gọi N là trung điểm AD \(\Rightarrow\dfrac{BG}{BN}=\dfrac{2}{3}\) (theo t/c trọng tâm)
\(\Rightarrow\dfrac{BJ}{BC}=\dfrac{BG}{BN}\Rightarrow JG||CN\)
\(\Rightarrow\widehat{\left(JG;CD\right)}=\widehat{\left(CN;CD\right)}=\widehat{NCD}=30^0\) (do tam giác ACD đều)
TV
giúp
1
HP
17 tháng 10 2021
2.
\(cosx+cos3x=1+\sqrt{2}sin\left(2x+\dfrac{\pi}{4}\right)\)
\(\Leftrightarrow2cos2x.cosx=1+cos2x+sin2x\)
\(\Leftrightarrow2cos2x.cosx=2cos^2x+2sinx.cosx\)
\(\Leftrightarrow cosx\left(cos2x-cosx-sinx\right)=0\)
\(\Leftrightarrow cosx\left(cos^2x-sin^2x-cosx-sinx\right)=0\)
\(\Leftrightarrow cosx\left(cosx+sinx\right)\left(cosx-sinx-1\right)=0\)
\(\Leftrightarrow cosx.\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right).\left[\sqrt{2}cos\left(x+\dfrac{\pi}{4}\right)-1\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}cosx=0\\sin\left(x+\dfrac{\pi}{4}\right)=0\\cos\left(x+\dfrac{\pi}{4}\right)=\dfrac{1}{\sqrt{2}}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x+\dfrac{\pi}{4}=\pm\dfrac{\pi}{4}+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{2}+k\pi\\x=-\dfrac{\pi}{4}+k\pi\\x=k2\pi\end{matrix}\right.\)
25 tháng 4 2023
c: \(\left(x^2+x-3\right)'=2x+1\)
(2x-1)'=2
\(y'=\dfrac{\left(2x+1\right)\cdot\left(2x-1\right)-\left(x^2+x-3\right)\cdot2}{\left(2x-1\right)^2}\)
\(=\dfrac{4x^2-1-2x^2-2x+6}{\left(2x-1\right)^2}=\dfrac{2x^2-2x+5}{\left(2x-1\right)^2}\)
20 tháng 6 2019
\(\orbr{\begin{cases}2x+\frac{\pi}{6}=x+k2\pi\\2x+\frac{\pi}{6}=\pi-x+k2\pi\end{cases}}\) \(\Rightarrow\orbr{\begin{cases}x=-\frac{\pi}{6}+k2\pi\\x=\frac{5\pi}{18}+\frac{k2\pi}{3}\end{cases}}\)
NV
Nguyễn Việt Lâm
Giáo viên
22 tháng 4 2023
a.
\(y'=4x^3-8x\)
b.
\(y'=\dfrac{2\left(5x+1\right)-5\left(2x-3\right)}{\left(5x+1\right)^2}=\dfrac{17}{\left(5x+1\right)^2}\)
c.
\(y'=\dfrac{\left(2x+1\right)\left(2x-1\right)-2\left(x^2+x-3\right)}{\left(2x-1\right)^2}=\dfrac{2x^2-2x+5}{\left(2x-1\right)^2}\)
d.
\(y'=\dfrac{\left(x^2-2x\right)'}{2\sqrt{x^2-2x+5}}=\dfrac{x-1}{\sqrt{x^2-2x+5}}\)
NV
Nguyễn Việt Lâm
Giáo viên
11 tháng 3 2023
\(\lim\limits_{x\rightarrow4}\dfrac{2x-8}{4x^2-18x+8}=\lim\limits_{x\rightarrow4}\dfrac{2\left(x-4\right)}{2\left(x-4\right)\left(2x-1\right)}=\lim\limits_{x\rightarrow4}\dfrac{1}{2x-1}=-\dfrac{1}{7}\)
b.
\(\lim\limits_{x\rightarrow-\infty}\left(\sqrt{x^2+3x-1}+x\right)=\lim\limits_{x\rightarrow-\infty}\dfrac{\left(\sqrt{x^2+3x-1}+x\right)\left(\sqrt{x^2+3x-1}-x\right)}{\sqrt{x^2+3x-1}-x}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{3x-1}{\sqrt{x^2+3x-1}-x}=\lim\limits_{x\rightarrow-\infty}\dfrac{3-\dfrac{1}{x}}{-\sqrt{1+\dfrac{3}{x}-\dfrac{1}{x^2}}-1}=\dfrac{3-0}{-\sqrt{1+0-0}-1}=-\dfrac{3}{2}\)
c.
\(\lim\limits_{x\rightarrow1^+}\dfrac{2x^2-3x+1}{x^2-1}=\lim\limits_{x\rightarrow1^+}\dfrac{\left(x-1\right)\left(2x-1\right)}{\left(x-1\right)\left(x+1\right)}=\lim\limits_{x\rightarrow1^+}\dfrac{2x-1}{x+1}=\dfrac{1}{2}\)
d.
\(\lim\limits_{x\rightarrow-\infty}\dfrac{2x^2+3x}{\sqrt{4x^4+2x^2}+3x^2-1}=\lim\limits_{x\rightarrow-\infty}\dfrac{x^2\left(2+\dfrac{3}{x}\right)}{x^2\left(\sqrt{4+\dfrac{2}{x^2}}+3-\dfrac{1}{x^2}\right)}\)
\(=\lim\limits_{x\rightarrow-\infty}\dfrac{2+\dfrac{3}{x}}{\sqrt{4+\dfrac{2}{x^2}}+3-\dfrac{1}{x^2}}=\dfrac{2+0}{\sqrt{4+0}+3-0}=\dfrac{2}{5}\)
NH
1
NM
Nguyễn Minh Quang
Giáo viên
2 tháng 9 2021
ta có \(x\in\left[-\frac{\pi}{4};0\right]\Rightarrow2x\in\left[-\frac{\pi}{2},0\right]\Rightarrow sin2x\in\left[-1,0\right]\)
Vậy \(\hept{\begin{cases}GTNN=-1\\GTLN=0\end{cases}}\)