Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{25}+\frac{1}{41}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)\)
< \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{4}{20}+\frac{5}{20}+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)(đpcm)
ê cho hỏi tại sao lại ra < \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)
a)\(S=\left(\frac{1}{5}+\frac{1}{13}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)<\left(\frac{1}{5}+\frac{1}{5}\right)+\left(\frac{1}{60}+\frac{1}{60}+\frac{1}{60}\right)=\frac{2}{5}+\frac{1}{20}=\frac{9}{20}<\frac{10}{20}=\frac{1}{2}\)b) \(2.S=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\)
=> 2S - S = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{19}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{20}}\right)\)
=> S = \(1-\frac{1}{2^{20}}<1\) đpcm
Mấy câu trên dễ , bạn có thể tự làm được
Chứng minh \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< 1\)
Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2\cdot2}< \frac{1}{1\cdot2}\)
\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)
\(\frac{1}{4^2}=\frac{1}{4\cdot4}< \frac{1}{3\cdot4}\)
...
\(\frac{1}{10^2}=\frac{1}{10\cdot10}< \frac{1}{9\cdot10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{9\cdot10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{1}{1}-\frac{1}{10}\)
=> \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{10^2}< \frac{9}{10}\)
Lại có : \(\frac{9}{10}< 1\)
=> \(A< \frac{9}{10}< 1\)
=> \(A< 1\left(đpcm\right)\)
Mik lười quá bạn tham khảo câu 3 tại đây nhé:
Câu hỏi của nguyen linh nhi - Toán lớp 6 - Học toán với OnlineMath
\(S=\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{37\cdot38\cdot39}\)
\(2S=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{37\cdot38}-\frac{1}{38\cdot39}\)
\(2S=\frac{1}{2}-\frac{1}{38\cdot39}\)
\(S=\frac{1}{4}-\frac{1}{2\cdot38\cdot39}< \frac{1}{4}\)
1) \(\frac{2}{3}+x=-\frac{4}{5}\)
\(x=\left(-\frac{4}{5}\right)-\frac{2}{3}\)
\(x=-1\frac{7}{15}\)
Vậy \(x=-1\frac{7}{15}\)
2) \(\frac{2}{5}-x=-\frac{1}{3}\)
\(x=\frac{2}{5}-\left(-\frac{1}{3}\right)\)
\(x=\frac{11}{15}\)
Vậy \(x=\frac{11}{15}\)
3) \(1-\frac{x}{3}=1\frac{1}{2}\)
\(\frac{x}{3}=1-1\frac{1}{2}\)
\(\frac{x}{3}=-\frac{1}{2}\)
\(\Rightarrow x=\frac{\left(-1\right)\cdot3}{2}\)
\(x=-1\frac{1}{2}\)
4) \(1-\left(\frac{2x}{3}+2\right)=-1\)
\(\frac{2x}{3}+2=1-\left(-1\right)\)
\(\frac{2x}{3}+2=2\)
\(\frac{2x}{3}=2-2\)
\(\frac{2x}{3}=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
1. \(\frac{25}{100}x+x-\frac{1}{5}x=\frac{1}{5}\)
\(\Leftrightarrow\frac{1}{4}x+x-\frac{1}{5}x=\frac{1}{5}\)
\(\Leftrightarrow\left(\frac{1}{4}+1-\frac{1}{5}\right)x=\frac{1}{5}\)
\(\Leftrightarrow\frac{21}{20}x=\frac{1}{5}\)
\(\Leftrightarrow x=\frac{1}{5}:\frac{21}{20}\)
\(\Leftrightarrow x=\frac{4}{21}\)
Ta có:
\(\hept{\begin{cases}\frac{1}{5}=\frac{1}{5}\\\frac{1}{16}< \frac{1}{5}\\\frac{1}{113}< \frac{1}{5}\end{cases}}...\)\(\Rightarrow\frac{1}{5}+\frac{1}{16}+\frac{1}{25}+\frac{1}{41}+\frac{1}{60}+\frac{1}{85}+\frac{1}{113}< \frac{1}{5}.7=\frac{7}{5}< \frac{10}{5}=2\)(ĐPCM)
ai giúp mình cho