giúp tớ với
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
LH
1 tháng 8 2016
\(B=1.3+3.5+5.7+....+27.29+29.31\)
\(6B=1.3.6+3.5.6+5.7.6+....+27.29.6+29.31.6\)
\(6B=1.3.\left(5+1\right)+3.5.\left(7-1\right)+5.7.\left(9-3\right)+...+27.29.\left(31-25\right)+29.31.\left(33-27\right)\)
\(6B=1.3.5+3+3.5.7-1.3.5+5.7.9-3.5.7+...+27.29.31-25.27.29+29.31.33-27.29.31\)
\(6B=3+29.31.33\)
\(B=\frac{3+29.21.33}{6}=4945\)
TN
4
27 tháng 11 2021
Gọi 3 số tự nhiên đó lần lượt là a,b,c(\(a,b,c\ge0\))
Áp dụng t/c dtsbn ta có:
\(\dfrac{a}{\dfrac{1}{3}}=\dfrac{b}{\dfrac{1}{4}}=\dfrac{c}{\dfrac{1}{5}}=\dfrac{a+b+c}{\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}}=\dfrac{94}{\dfrac{47}{60}}=120\)
\(\dfrac{a}{\dfrac{1}{3}}=120\Rightarrow a=40\\ \dfrac{b}{\dfrac{1}{4}}=120\Rightarrow b=30\\ \dfrac{c}{\dfrac{1}{5}}=120\Rightarrow c=24\)
ND
5
\(a,cos\alpha=\dfrac{5}{13}\)
\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\dfrac{5}{13}\right)^2}=\dfrac{12}{13}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{5}{13}\right)^2}\Leftrightarrow tan^2\alpha=\dfrac{144}{25}\Leftrightarrow tan\alpha=\dfrac{12}{5}\)
\(cot\alpha=\dfrac{1}{tan\alpha}=1:\dfrac{12}{5}=\dfrac{5}{12}\)
\(b,sin\alpha=\dfrac{7}{12}\)
\(cos\alpha=\sqrt{1-sin^2\alpha}=\sqrt{1-\left(\dfrac{7}{12}\right)^2}=\dfrac{\sqrt{95}}{12}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+tan^2\alpha=\dfrac{1}{\left(\dfrac{\sqrt{95}}{12}\right)^2}\Leftrightarrow tan\alpha=\dfrac{49}{95}\)
\(cot\alpha=1:\dfrac{49}{95}=\dfrac{95}{49}\)
\(c,tan\alpha=\dfrac{15}{4}\)
\(cot\alpha=1:\dfrac{15}{4}=\dfrac{4}{15}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(\dfrac{15}{4}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\sqrt{\dfrac{16}{241}}\)
\(sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\sqrt{\dfrac{16}{241}}\right)^2}\approx0,97\)
\(d,cot\alpha=-\dfrac{1}{\sqrt{3}}\\ tan\alpha=1:\left(-\dfrac{1}{\sqrt{3}}\right)=-\sqrt{3}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\Leftrightarrow1+\left(-\sqrt{3}\right)^2=\dfrac{1}{cos^2\alpha}\Leftrightarrow cos\alpha=\dfrac{1}{2}\)
\(sin\alpha=\sqrt{1-\left(\dfrac{1}{2}\right)^2}=\dfrac{\sqrt{3}}{2}\)