CMR \(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+...+\)\(\frac{1}{2020^2+2021^2}\)< \(\frac{1}{2}\)
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11 tháng 12 2019
a) \(\frac{15}{12}+\frac{5}{13}-\frac{3}{12}-\frac{18}{13}=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{11}{24}-\frac{5}{41}+\frac{13}{24}+0,5-\frac{36}{41}=\left(\frac{11}{24}+\frac{13}{24}\right)+\left(-\frac{5}{41}-\frac{36}{41}\right)+0,5\)
\(=1+\left(-1\right)+0,5\)
\(=0,5\)
_Học tốt nha_
11 tháng 12 2019
a, \(\frac{15}{12}\)+ \(\frac{5}{13}\)- \(\frac{3}{12}\)-\(\frac{18}{13}\)
= \(\frac{5}{4}\)+ \(\frac{5}{13}\) - \(\frac{1}{4}\) - \(\frac{18}{13}\)
= \(\left(\frac{5}{4}-\frac{1}{4}\right)\)+ \(\left(\frac{5}{13}-\frac{18}{13}\right)\)
= 1 - 1 = 0
b, \(\frac{11}{24}\)- \(\frac{5}{41}\)+ \(\frac{13}{24}\)+ 0,5 - \(\frac{36}{41}\)
= \(\left(\frac{11}{24}+\frac{13}{24}\right)\)- \(\left(\frac{5}{41}+\frac{36}{41}\right)\)+ 0,5
= 1 - 1 + 0,5 = 0,5
c, \(\left(-\frac{3}{4}+\frac{2}{3}\right):\frac{5}{11}+\left(-\frac{1}{4}+\frac{1}{3}\right):\frac{5}{11}\)
=\(\left(-\frac{3}{4}+\frac{2}{3}\right).\frac{11}{5}+\left(-\frac{1}{4}+\frac{1}{3}\right).\frac{5}{11}\)
= \(\frac{11}{5}.\left(-\frac{3}{4}+\frac{2}{3}-\frac{1}{4}+\frac{1}{3}\right)\)
= \(\frac{11}{5}.\left[\left(-\frac{3}{4}-\frac{1}{4}\right)+\left(\frac{2}{3}+\frac{1}{3}\right)\right]\)
= \(\frac{11}{5}.\left[\left(-1\right)+1\right]\)
= 0
d, \(\left(-3\right)^2.\left(\frac{3}{4}-0,25\right)-\left(3\frac{1}{2}-1\frac{1}{2}\right)\)
= \(9.\left(0,75-0,25\right)-2\)
= 9. 0,5 - 2 = 2,5
e, \(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
= -1 + 1 - \(\frac{1}{2}\)
= \(-\frac{1}{2}\)
13 tháng 5 2018
\(\frac{13}{25}+\frac{6}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
\(=\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
\(=-1+1-\frac{1}{2}=0-\frac{1}{2}\)
\(=\frac{-1}{2}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
\(=\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
\(=1+1+0,5=2,5\)
13 tháng 5 2018
\(\frac{13}{25}+\frac{4}{41}-\frac{38}{25}+\frac{35}{41}-\frac{1}{2}\)
= \(\left(\frac{13}{25}-\frac{38}{25}\right)+\left(\frac{6}{41}+\frac{35}{41}\right)-\frac{1}{2}\)
= \(-1+1-\frac{1}{2}=-\frac{1}{2}\)
\(1\frac{4}{23}+\frac{5}{21}-\frac{4}{23}+0,5+\frac{16}{21}\)
=\(\left(1\frac{4}{23}-\frac{4}{23}\right)+\left(\frac{5}{21}+\frac{16}{21}\right)+0,5\)
= \(1+1+0,5=2,5\)
29 tháng 3 2020
Đặt \(A=\frac{\frac{1}{2020}+\frac{2}{2019}+\frac{3}{2018}+...+\frac{2019}{2}+\frac{2020}{1}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{1+\left(\frac{1}{2020}+1\right)+\left(\frac{2}{2019}+1\right)+\left(\frac{3}{2018}+1\right)+...+\left(\frac{2019}{2}+1\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{\frac{2021}{2021}+\frac{2021}{2020}+\frac{2021}{2019}+...+\frac{2021}{2}}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}\)
\(A=\frac{2021\left(\frac{1}{2021}+\frac{1}{2020}+\frac{1}{2019}+...+\frac{1}{2}\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2021}}=2021\)
19 tháng 12 2018
\(a,=\left(\frac{15}{12}-\frac{3}{12}\right)+\left(\frac{5}{13}-\frac{18}{13}\right)\)
\(=1+-1\)
\(=0\)
D
1 tháng 5 2015
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{25}+\frac{1}{41}\right)+\left(\frac{1}{61}+\frac{1}{85}+\frac{1}{113}\right)\)
< \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3=\frac{1}{5}+\frac{1}{4}+\frac{1}{20}=\frac{4}{20}+\frac{5}{20}+\frac{1}{20}=\frac{10}{20}=\frac{1}{2}\)(đpcm)
LT
23 tháng 3 2017
ê cho hỏi tại sao lại ra < \(\frac{1}{5}+\frac{1}{12}.3+\frac{1}{60}.3\)
UI
19 tháng 8 2020
a)
\(P=a\sqrt{1+\frac{1}{a^2}+\frac{1}{\left(a+1\right)^2}}+\frac{a}{b}=a\sqrt{\frac{a^2\left(a+1\right)^2+\left(a+1\right)^2+a^2}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}\)
=\(a\sqrt{\frac{a^2\left(a+1\right)^2+2a\left(a+1\right)+1}{a^2\left(a+1\right)^2}}+\frac{a}{a+1}=a\sqrt{\frac{\left[a\left(a+1\right)+1\right]^2}{\left[a\left(a+1\right)\right]^2}}+\frac{a}{a+1}\)
\(=a.\frac{a\left(a+1\right)+1}{a\left(a+1\right)}+\frac{a}{a+1}=a+\frac{1}{a+1}+\frac{a}{a+1}=a+1\)
Vay P=a+1
phan b,c ap dung phan a la ra
8 tháng 10 2020
CM bài toán phụ: \(x+y+z=0\)
CM: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\) với x,y,z dương
Ta có: \(I=\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)
\(=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2-2\cdot\frac{x+y+z}{xyz}}=\sqrt{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}\)
\(=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Áp dụng vào ta được: \(Q=1+1-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)
\(Q=2021-\frac{1}{2021}=...\)
18 tháng 8 2020
c) Áp dụng công thức \(\sqrt{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{\left(a+b\right)^2}}=\frac{1}{a}+\frac{1}{b}-\frac{1}{a+b}\),ta được:
\(Q=1+\frac{1}{1}-\frac{1}{2}+1+\frac{1}{2}-\frac{1}{3}+...+1+\frac{1}{2020}-\frac{1}{2021}\)
\(=1+1+1+...+1-\frac{1}{2021}\)
\(=2021-\frac{1}{2021}=\frac{4084440}{2021}\)
Trả lời:
CMR A=\(\frac{1}{1^2+2^2}\)+\(\frac{1}{2^2+3^2}\)+....+\(\frac{1}{n^2+\left(n+1\right)^2}\)<\(\frac{1}{2}\)
Ta có bất đẵng thức:
\(\frac{1}{n^2+\left(n+1\right)^2}\)<\(\frac{1}{2n\left(n+1\right)}\)
Thay A, ta có:
A=\(\frac{1}{1^2+2^2}\)+ ......+\(\frac{1}{n^2+\left(n+1\right)^2}\)<\(\frac{1}{2.1.2}\)+\(\frac{1}{2.2.3}\)+\(\frac{1}{2.3.4}\)+....+\(\frac{1}{2n.\left(n+1\right)}\)= \(\frac{1}{2}\)(\(\frac{1}{1.2}\)+\(\frac{1}{2.3}\)+....+\(\frac{1}{n.\left(n+1\right)}\)=\(\frac{1}{2}\)(1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-.....+\(\frac{1}{n}\)-\(\frac{1}{n+1}\))=\(\frac{1}{2}\)(1-\(\frac{1}{n+1}\))<\(\frac{1}{2}\) (ĐPCM)