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NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
a.
Đặt \(sinx+cosx=t\in\left[-\sqrt{2};\sqrt{2}\right]\)
\(\Rightarrow1+2sinx.cosx=t^2\Rightarrow2sinx.cosx=t^2-1\)
Phương trình trở thành:
\(3t=2\left(t^2-1\right)\)
\(\Leftrightarrow2t^2-3t-2=0\)
\(\Rightarrow\left[{}\begin{matrix}t=2>\sqrt{2}\left(loại\right)\\t=-\dfrac{1}{2}\end{matrix}\right.\)
\(\Rightarrow sinx+cosx=-\dfrac{1}{2}\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{1}{2}\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=-\dfrac{\sqrt{2}}{8}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{\pi}{4}=arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x+\dfrac{\pi}{4}=\pi-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{\pi}{4}+arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\\x=\dfrac{3\pi}{4}-arcsin\left(-\dfrac{\sqrt{2}}{8}\right)+k2\pi\end{matrix}\right.\)
NV
Nguyễn Việt Lâm
Giáo viên
23 tháng 10 2021
b.
ĐKXĐ: \(x\ne\dfrac{\pi}{2}+k\pi\)
\(1+\dfrac{sinx}{cosx}=2\sqrt{2}sinx\)
\(\Rightarrow sinx+cosx=2\sqrt{2}sinx.cosx\)
\(\Leftrightarrow\sqrt{2}sin\left(x+\dfrac{\pi}{4}\right)=\sqrt{2}sin2x\)
\(\Leftrightarrow sin\left(x+\dfrac{\pi}{4}\right)=sin2x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x=x+\dfrac{\pi}{4}+k2\pi\\2x=\dfrac{3\pi}{4}-x+k2\pi\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{\pi}{4}+k2\pi\\x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\end{matrix}\right.\)
\(\Leftrightarrow x=\dfrac{\pi}{4}+\dfrac{k2\pi}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
a.
\(90^0< a< 180^0\Rightarrow cosa< 0\)
\(\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{2\sqrt{2}}{3}\)
\(tana=\dfrac{sina}{cosa}=-\dfrac{\sqrt{2}}{4}\)
b.
\(0< a< 90^0\Rightarrow cosa>0\)
\(\Rightarrow cosa=\sqrt{1-sin^2a}=\dfrac{4}{5}\)
\(tana=\dfrac{sina}{cosa}=\dfrac{3}{4}\)
\(cota=\dfrac{1}{tana}=\dfrac{4}{3}\)
NV
Nguyễn Việt Lâm
Giáo viên
10 tháng 7 2021
c.
\(A=\dfrac{\dfrac{sina}{cosa}+\dfrac{3cosa}{sina}}{\dfrac{sina}{cosa}+\dfrac{cosa}{sina}}=\dfrac{sin^2a+3cos^2a}{sin^2a+cos^2a}=1+2cos^2a=\dfrac{17}{8}\)
d.
\(A=\dfrac{\dfrac{cosa}{sina}+\dfrac{3sina}{cosa}}{\dfrac{2cosa}{sina}+\dfrac{sina}{cosa}}=\dfrac{cos^2a+3sin^2a}{2cos^2a+sin^2a}=\dfrac{cos^2a+3\left(1-cos^2a\right)}{2cos^2a+\left(1-cos^2a\right)}\)
\(=\dfrac{3-2cos^2a}{1+cos^2a}=\dfrac{19}{13}\)
VA
2
PN
0
TL
2
16 tháng 11 2021
Câu 1.
Tờ vé số có dạng \(\overline{a_1a_2a_3a_4a_5a_6}\in A=\left\{0;1;2;3;4;5;6;7;8;9\right\}\)
\(;a_i\ne a_j\)
Chọn \(a_1\ne0\) nên \(a_1\) có 9 cách chọn.
5 số còn lại là chỉnh hợp chập 5 của 8 số còn lại \(\in A\backslash\left\{a_1\right\}\)
\(\Rightarrow\)Có \(A_8^5\) cách.
Vậy có tất cả \(A_8^5\cdot9=60480\) vé số.
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 3 2022
5.
\(\lim\dfrac{n+1}{n^2-2}=\lim\dfrac{n^2\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{n^2\left(1-\dfrac{2}{n^2}\right)}=\lim\dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{1-\dfrac{2}{n^2}}=\dfrac{0+0}{1-0}=0\)
\(\lim\dfrac{n\left(n+1\right)}{\left(n+4\right)^3}=\lim\dfrac{n^3\left(\dfrac{1}{n}+\dfrac{1}{n^2}\right)}{n^3\left(1+\dfrac{4}{n}\right)^3}=\lim\dfrac{\dfrac{1}{n}+\dfrac{1}{n^2}}{\left(1+\dfrac{4}{n}\right)^3}=\dfrac{0+0}{\left(1+0\right)^3}=0\)
\(\lim\dfrac{3n^3-2n+5}{2n^2+5n-3}=\lim\dfrac{n^3\left(3-\dfrac{2}{n^2}+\dfrac{5}{n^3}\right)}{n^3\left(\dfrac{2}{n}+\dfrac{5}{n^2}-\dfrac{3}{n^3}\right)}=\lim\dfrac{3-\dfrac{2}{n^2}+\dfrac{5}{n^3}}{\dfrac{2}{n}+\dfrac{5}{n^2}-\dfrac{3}{n^3}}=\dfrac{3}{0}=+\infty\)
\(\lim\dfrac{2n^3}{n^4+3n^2+1}=\lim\dfrac{n^4\left(\dfrac{2}{n}\right)}{n^4\left(1+\dfrac{3}{n^2}+\dfrac{1}{n^4}\right)}=\lim\dfrac{\dfrac{2}{n}}{1+\dfrac{3}{n^2}+\dfrac{1}{n^4}}=\dfrac{0}{1}=0\)
NV
Nguyễn Việt Lâm
Giáo viên
14 tháng 3 2022
6.
\(\lim\dfrac{3^n-4^n+5^n}{3^n+4^n-5^n}=\lim\dfrac{5^n\left[\left(\dfrac{3}{5}\right)^n-\left(\dfrac{4}{5}\right)^n+1\right]}{5^n\left[\left(\dfrac{3}{5}\right)^n+\left(\dfrac{4}{5}\right)^n-1\right]}=\lim\dfrac{\left(\dfrac{3}{5}\right)^n-\left(\dfrac{4}{5}\right)^n+1}{\left(\dfrac{3}{5}\right)^n+\left(\dfrac{4}{5}\right)^n-1}=\dfrac{0+0+1}{0+0-1}=-1\)
\(\lim\dfrac{1+3^n}{4+3^n}=\lim\dfrac{3^n\left[\left(\dfrac{1}{3}\right)^n+1\right]}{3^n\left[4.\left(\dfrac{1}{3}\right)^n+1\right]}=\lim\dfrac{\left(\dfrac{1}{3}\right)^n+1}{4.\left(\dfrac{1}{3}\right)^n+1}=\dfrac{0+1}{4.0+1}=1\)
\(\lim\dfrac{4.3^n+7^{n+1}}{2.5^n+7^n}=\lim\dfrac{7^n\left[4.\left(\dfrac{3}{7}\right)^n+7\right]}{7^n\left[\left(\dfrac{5}{7}\right)^n+1\right]}=\lim\dfrac{4.\left(\dfrac{3}{7}\right)^n+7}{\left(\dfrac{5}{7}\right)^n+1}=\dfrac{4.0+7}{0+1}=7\)
\(\lim\dfrac{4^{n+1}+6^{n+2}}{5^n+8^n}=\lim\dfrac{8^n\left[4.\left(\dfrac{4}{8}\right)^n+36.\left(\dfrac{6}{8}\right)^n\right]}{8^n\left[\left(\dfrac{5}{8}\right)^n+1\right]}=\lim\dfrac{4.\left(\dfrac{4}{8}\right)^n+36\left(\dfrac{6}{8}\right)^n}{\left(\dfrac{5}{8}\right)^n+1}=\dfrac{0}{1}=0\)
