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Bài 2:1-2+3-4+...+2011-2012
=1+2+3+4+...+2011+2012-2(2+4+6+...+2012)
=2025078-2(1012036)
=2025078-2024072
=1006
Học giỏi!
a) \(\frac{2^{10}+1}{2^{10}-1}\)và \(\frac{2^{10}-1}{2^{10}-3}\)
Ta có chính chất phân số trung gian là \(\frac{2^{10}+1}{2^{10}-3}\)
\(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}\) ; \(\frac{2^{10}-1}{2^{10}-3}< \frac{2^{10}+1}{2^{10}-3}\)
Vì \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}+1}{2^{10}-3}>\frac{2^{10}-1}{2^{10}-3}\)
Nên \(\frac{2^{10}+1}{2^{10}-1}>\frac{2^{10}-1}{2^{10}-3}\)
b) \(A=\frac{2011}{2012}+\frac{2012}{2013}\)và \(B=\frac{2011+2012}{2012+2013}\)
Ta có : \(A=\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2013}+\frac{2012}{2013}=\frac{2011+2012}{2013}>\frac{2011+2012}{2012+2013}=B\)
Vậy A > B
Có gì sai cho sorry
a,
\(\frac{2^{10}+1}{2^{10}-1}=1+\frac{2}{2^{10}-1}< 1+\frac{2}{2^{10}-3}=\frac{2^{10}-1}{2^{10}-3}\)
b,
\(\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
b,Ta có
\(\frac{2010}{2011}>\frac{2010}{2011+2012+2013}\)
\(\frac{2011}{2012}>\frac{2011}{2011+2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2011+2012+2013}\)
\(\Rightarrow P>Q\)
\(A=\frac{-10}{20}+\frac{-10}{30}+\frac{-10}{42}+\frac{-10}{56}+\frac{-10}{72}+\frac{-10}{90}+\frac{-10}{110}\)
\(=-10\left(\frac{1}{20}+\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\right)\)
\(=-10\left(\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=-10\left(\frac{1}{4}-\frac{1}{11}\right)\)
\(=\frac{-35}{22}\)
Mình biết làm nhưng bạn nên viết rời ra.Viết liền làm người khác không muốn làm đó.
Làm thì dài quá nên mình gợi ý thôi nhé
a)quy đồng
b)Sử dụng phần bù
c)(1/80)^7>(1/81)^7=(1/3^4)^7=1/3^28
(1/243)^6=(1/3^5)^6=1/3^30
Vì 1/3^28>1/3^30 nên ......
d)Tương tự câu d
Mấy câu còn lại thì nhắn tin với mình,mình sẽ trả lời cho,mình đang mệt lắm rồi nha!!!
a, Ta có: \(\frac{2001}{2002}=\frac{2002-1}{2002}=\frac{2002}{2002}-\frac{1}{2002}=1-\frac{1}{2002}\)
\(\frac{2000}{2001}=\frac{2001-1}{2001}=\frac{2001}{2001}-\frac{1}{2001}=1-\frac{1}{2001}\)
Vì \(\frac{1}{2002}< \frac{1}{2001}\Rightarrow1-\frac{1}{2002}>1-\frac{1}{2001}\Rightarrow\frac{2001}{2002}>\frac{2000}{2001}\)
b, Ta có: \(\left(\frac{1}{80}\right)^7>\left(\frac{1}{81}\right)^7=\left(\frac{1}{3^4}\right)^7=\left(\frac{1}{3}\right)^{28}=\frac{1}{3^{28}}\)
\(\left(\frac{1}{243}\right)^6=\left(\frac{1}{3^5}\right)^6=\left(\frac{1}{3^5}\right)^6=\frac{1}{3^{30}}\)
Vì \(\frac{1}{3^{28}}>\frac{1}{3^{30}}\Rightarrow\left(\frac{1}{81}\right)^7>\left(\frac{1}{243}\right)^6\Rightarrow\left(\frac{1}{80}\right)^7>\left(\frac{1}{243}\right)^6\)
c, Ta có: \(\left(\frac{3}{8}\right)^5=\frac{3^5}{\left(2^3\right)^5}=\frac{243}{2^{15}}>\frac{243}{3^{15}}>\frac{125}{3^{15}}=\frac{5^3}{\left(3^5\right)^3}=\frac{5^3}{243^3}=\left(\frac{5}{243}\right)^3\)
Vậy \(\left(\frac{3}{8}\right)^5>\left(\frac{5}{243}\right)^3\)
d, Ta có: \(\frac{2011}{2012}>\frac{2011}{2012+2013}\)
\(\frac{2012}{2013}>\frac{2012}{2012+2013}\)
\(\Rightarrow\frac{2011}{2012}+\frac{2012}{2013}>\frac{2011}{2012+2013}+\frac{2012}{2012+2013}=\frac{2011+2012}{2012+2013}\)
e, \(C=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=\frac{20^{10}-1}{20^{10}-1}+\frac{2}{2^{10}-1}=1+\frac{2}{2^{10}-1}\)
\(D=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=\frac{20^{10}-3}{20^{10}-3}+\frac{2}{2^{10}-3}=1+\frac{2}{2^{10}-3}\)
Vì \(\frac{2}{10^{10}-1}< \frac{2}{10^{10}-3}\Rightarrow1+\frac{2}{10^{10}-1}< 1+\frac{2}{10^{10}-3}\Rightarrow C< D\)
g, \(G=\frac{10^{100}+2}{10^{100}-1}=\frac{10^{100}-1+3}{10^{100}-1}=\frac{10^{100}-1}{10^{100}-1}+\frac{3}{10^{100}-1}=1+\frac{3}{10^{100}-1}\)
\(H=\frac{10^8}{10^8-3}=\frac{10^8-3+3}{10^8-3}=\frac{10^8-3}{10^8-3}+\frac{3}{10^8-3}=1+\frac{3}{10^8-3}\)
Vì \(\frac{3}{10^{100}-1}< \frac{3}{10^8-3}\Rightarrow1+\frac{3}{10^{100}-1}< 1+\frac{3}{10^8-3}\Rightarrow G< H\)
h, Vì E < 1 nên:
\(E=\frac{98^{99}+1}{98^{89}+1}< \frac{98^{99}+1+97}{98^{89}+1+97}=\frac{98^{99}+98}{98^{89}+98}=\frac{98\left(98^{98}+1\right)}{98\left(98^{88}+1\right)}=\frac{98^{98}+1}{98^{88}+1}=F\)
Vậy E = F
So sánh 2 phân số sau $\frac{10^{2011}+10}{10^{2012}+10}v\text{à}\frac{10^{2012}-10}{10^{2013}-10}$102011+10102012+10 và102012−10102013−10
kick dzô chữ xanh là được!! OK
Ta có :
10. A = \(\frac{10.\left(10^{2011}+1\right)}{10^{2012}+1}\)
= \(\frac{10^{2012}+10}{10^{2012}+1}\)
= \(\frac{10^{2012}+1+9}{10^{2012}+1}\)
= \(\frac{10^{2012}+1}{10^{2012}+1}-\frac{9}{10^{2012}+1}\)
= 1 - \(\frac{9}{10^{2012}+1}\)
10 . B = \(\frac{10.\left(10^{2012}+1\right)}{10^{2013}+1}\)
= \(\frac{10^{2013}+10}{10^{2013}+1}\)
= \(\frac{10^{2013}+1+9}{10^{2013}+1}\)
= 1 - \(\frac{9}{10^{2013}+1}\)
Vì \(\frac{9}{10^{2012}+1}\) >\(\frac{9}{10^{2013}+1}\) nên 10.A > 10.B
=> A >B
Vậy ...........
a)
\(10A=\frac{10^{2002}+10}{10^{2002}+1}=1+\frac{9}{10^{2002}+1}\)
\(10B=\frac{10^{2003}+10}{10^{2003}+1}=1+\frac{9}{10^{2003}+1}\)
=> 10A > 10B => A > B