K
Khách

Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.

4 tháng 9 2015

( 16/5 : 16/5 + 9/2 . 76/45 ) : (-43/2) = (1+38/5).(-2/43)=43/5.(-2/43)=-2/5

Tóm lại là = -2/5

15 tháng 8 2017

Bài 1 :

\(\dfrac{1}{2}-\left(\dfrac{1}{3}+\dfrac{1}{4}\right)< ..................< \dfrac{1}{48}-\left(\dfrac{1}{16}-\dfrac{1}{6}\right)\)

\(\Leftrightarrow\dfrac{1}{2}-\dfrac{7}{12}< .....................< \dfrac{1}{48}-\left(\dfrac{-5}{48}\right)\)

\(\Leftrightarrow\dfrac{5}{12}< .............< \dfrac{1}{8}\)

\(\Leftrightarrow0,41\left(6\right)< ...........< 0,125\)

\(\Leftrightarrow\) ko tìm dc số nguyên thích hợp vào chỗ chấm

29 tháng 5 2017

bài này dài lắm

\(A=\frac{\frac{1}{1.101}+\frac{1}{2.102}+\frac{1}{3.103}+...+\frac{1}{25.125}}{\frac{1}{1.26}+\frac{1}{2.27}+\frac{1}{3.28}+...+\frac{1}{100.125}}\)

\(A=\frac{\frac{1}{100}.\left(1-\frac{1}{101}+\frac{1}{2}-\frac{1}{102}+\frac{1}{3}-\frac{1}{103}+...+\frac{1}{25}-\frac{1}{125}\right)}{\frac{1}{25}.\left(1-\frac{1}{26}+\frac{1}{2}-\frac{1}{27}+\frac{1}{3}-\frac{1}{28}+...+\frac{1}{100}-\frac{1}{125}\right)}\)

\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-\frac{1}{28}-...-\frac{1}{125}\right)}\)

\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+...+\frac{1}{25}+\frac{1}{26}+\frac{1}{27}+...+\frac{1}{100}-\frac{1}{26}-\frac{1}{27}-...-\frac{1}{100}-\frac{1}{101}-...-\frac{1}{125}\right)}\)

\(A=\frac{\frac{1}{100}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}{\frac{1}{25}.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{25}-\frac{1}{101}-\frac{1}{102}-\frac{1}{103}-...-\frac{1}{125}\right)}\)

\(A=\frac{\left(\frac{1}{100}\right)}{\left(\frac{1}{25}\right)}=\frac{1}{4}\)

\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{125}{42}}{\frac{2000}{43}-\frac{250}{252}-\frac{2000}{257}}\)

\(B=\frac{\frac{16}{9}-\frac{16}{127}+\frac{16}{2017}}{\frac{5}{2017}+\frac{5}{9}-\frac{5}{127}}-\frac{\frac{6000}{43}-\frac{6000}{257}-\frac{6000}{2016}}{\frac{2000}{43}-\frac{2000}{2016}-\frac{2000}{257}}\)

\(B=\frac{16.\left(\frac{1}{9}-\frac{1}{127}+\frac{1}{2017}\right)}{5.\left(\frac{1}{2017}+\frac{1}{9}-\frac{1}{127}\right)}-\frac{6000.\left(\frac{1}{43}-\frac{1}{257}-\frac{1}{2016}\right)}{2000.\left(\frac{1}{43}-\frac{1}{2016}-\frac{1}{257}\right)}\)

\(B=\frac{16}{5}-3=\frac{1}{5}\)

Đặt \(C=\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}\)

\(C=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{2005^2}+\frac{1}{2006^2}+\frac{1}{2007^2}\)

\(C< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{2004.2005}+\frac{1}{2005.2006}+\frac{1}{2006.2007}\)

\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{2005}-\frac{1}{2006}+\frac{1}{2006}-\frac{1}{2007}\)

\(=\frac{1}{4}-\frac{1}{2017}\left(đpcm\right)\)

\(C>\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{2005.2006}+\frac{1}{2006.2007}+\frac{1}{2007.2008}\)

\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{2006}-\frac{1}{2007}+\frac{1}{2007}-\frac{1}{2008}\)

\(=\frac{1}{5}-\frac{1}{2008}\left(đpcm\right)\)

Vậy \(A>\frac{1}{2007^2}+\frac{1}{2006^2}+\frac{1}{2005^2}+...+\frac{1}{7^2}+\frac{1}{6^2}+\frac{1}{5^2}>B\)

13 tháng 9 2021

\(a,=\dfrac{3^6\cdot5^4\cdot9^4-5^{13}\cdot3^{13}\cdot5^{-9}}{3^{12}\cdot5^6+9^6\cdot5^6}=\dfrac{3^{14}\cdot5^4-5^4\cdot3^{13}}{3^{12}\cdot5^6+3^{12}\cdot5^6}\\ =\dfrac{3^{13}\cdot5^4\cdot2}{2\cdot3^{12}\cdot5^6}=\dfrac{3}{5^2}=\dfrac{3}{25}\)

13 tháng 9 2021

\(b,=\dfrac{\left(\dfrac{2}{5}\cdot5\right)^7+\left(\dfrac{9}{4}\cdot\dfrac{16}{3}\right)^3}{2^7\cdot5^2+2^9}=\dfrac{2^7+12^3}{2^7\left(5^2+2^2\right)}=\dfrac{2^7+4^3\cdot3^3}{2^7\cdot29}=\dfrac{2^6\left(2+3^3\right)}{2^7\cdot29}=\dfrac{1}{2}\)

17 tháng 7 2016

đăng từng bài rồi mình giải cho nha

17 tháng 7 2016

Câu 3,57-56+55=55.52-55.5+55=55.(52-5+1)=55.21 chia hết cho 21

Câu:4:76+75-74=74.72+74.7-74=74.(72+7-1)=74.55=74.11.5=73.7.11.5=73.77.5 chia hết cho 77

Các câu khác tương tự

3: \(=5^5\left(5^2-5+1\right)=5^2\cdot21⋮21\)

4: \(=7^4\left(7^2+7-1\right)=7^4\cdot55=7^3\cdot5\cdot77⋮77\)

5: \(=\left(2^{26}+2^{25}-2^{24}\right)=2^{24}\left(2^2+2-1\right)=2^{24}\cdot5⋮5\)