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mình nhầm câu b:
Áp dụng....
A=10^11-1/10^12-1<10^11-1+11/10^12-1+11=10^11+10/10^12+10=10.(10^10+1)/10.(10^11+1)
=10^10+1/10^11+1=B
Vậy A<B(câu này mới đúng còn câu b mình làm chung với câu a là sai)
a) Với a<b=>a+n/b+n >a/b
Với a>b=>a+n/b+n<a/b
Với a=b=>a+n/b+n=a/b
b) Áp dụng t/c a/b<1=>a/b<a+m/b+m(a,b,m thuộc z,b khác 0)ta có:
A=(10^11)-1/(10^12)-1=(10^11)-1+11/(10^12)-1+11=(10^11)+10/(10^12)+10=10.[(10^10)+1]/10.[(10^11)+1]
=(10^10)+1/(10^11)+1=B
Vậy A=B
A = \(\frac{20^{10}+1}{20^{10}-1}=1\) B = \(\frac{20^{10}-1}{20^{10}-3}=1\)
Nên A = B
\(A=\frac{20^{10}+1}{20^{10}-1}=\frac{20^{10}-1+2}{20^{10}-1}=1+\frac{2}{20^{10}-1}\)
\(B=\frac{20^{10}-1}{20^{10}-3}=\frac{20^{10}-3+2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì \(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\Rightarrow1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\Rightarrow A< B\)
Ta thấy : \(\frac{1}{11}>\frac{1}{100},\frac{1}{12}>\frac{1}{100},...,\frac{1}{100}=\frac{1}{100}\)
\(\Rightarrow\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{1}{100}+\frac{1}{100}+\frac{1}{100}+...+\frac{1}{100}=\frac{90}{100}=\frac{9}{10}\)
\(\Rightarrow\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>\frac{9}{10}+\frac{1}{10}=1\)
Do đó : \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}+\frac{1}{13}+...+\frac{1}{100}>1\)
\(A=\frac{\left(23\frac{11}{15}-26\frac{13}{20}\right)}{12^2+5^2}\cdot\frac{1-\frac{1}{30}-\frac{1}{42}-\frac{1}{56}}{3^2.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+\frac{11}{15}-26+\frac{13}{20}\right)}{144+25}\cdot\frac{1-\frac{1}{5.6}-\frac{1}{6.7}-\frac{1}{7.8}}{9.13.2-13.5}-\frac{19}{37}\)
\(A=\frac{\left(23+26+\frac{11}{15}-\frac{13}{20}\right)}{169}\cdot\frac{1-\left(\frac{1}{5}-\frac{1}{6}\right)-\left(\frac{1}{6}-\frac{1}{7}\right)-\left(\frac{1}{7}-\frac{1}{8}\right)}{13.\left(9.2-5\right)}-\frac{19}{37}\)
\(A=\frac{49+\frac{44}{60}-\frac{39}{60}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{6}-\frac{1}{6}+\frac{1}{7}-\frac{1}{7}+\frac{1}{8}}{13.13}-\frac{19}{37}\)
\(A=\frac{49+\frac{1}{20}}{169}\cdot\frac{1-\frac{1}{5}+\frac{1}{8}}{169}-\frac{19}{37}\)
\(A=\frac{49\frac{1}{20}}{169}\cdot\frac{\frac{4}{5}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{32}{40}+\frac{5}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981}{169}\cdot\frac{\frac{37}{40}}{169}-\frac{19}{37}\)
\(A=\frac{981.\frac{37}{40}}{169^2}-\frac{19}{37}\)
\(A=\frac{\frac{36297}{40}}{28561}-\frac{19}{37}\)
\(A=\frac{907,425}{28561}-\frac{19}{37}\)
\(A=\frac{33574,725}{1056757}-\frac{542659}{1056757}\)
\(A=\frac{-509084,275}{1056757}=-0,04604282...\)
Mik chỉ làm đc thế này thôi, ôn thi học kì II tốt nha bạn!
thì bạn lấy { (1.10) + [17. (2.10)] + [10. (3.10)] } : 10 = 65 (điểm 10) (tích nha!)
Ta có: A=1/11+1/12+1/13+...+1/30
=(1/11+1/12+1/13+..+1/20)+(1/21+1/22+1/23+...+1/30)
\(\Rightarrow\)A<(1/10+1/10+1/10+...+1/10)+(1/20+1/20+1/20+...1/20)
\(\Rightarrow\)A<(1/10)*10+(1/20)*10
\(\Rightarrow\)A<1+1/2
\(\Rightarrow\)A<3/2<11/6
A=20 mủ 10 - 1 +12/(20 mủ 10 -1)=1+12/20 MỦ 10 -1
B=20 mủ 10 - 3 + 2 /(20 mủ 10 - 3)=1+2/20 mủ 10 - 3
Vì ... bạn tự làm nha.nhớ k đấy
A=\(\frac{20^{10}+1}{20^{10}-1}\)=\(\frac{\left(20^{10}-1\right)+2}{20^{10}-1}\)=\(\frac{20^{10}-1}{20^{10}-1}+\frac{2}{20^{10}-1}\)=\(1+\frac{2}{20^{10}-1}\)
B= \(\frac{20^{10}-1}{20^{10}-3}=\frac{\left(20^{10}-3\right)+2}{20^{10}-3}\)=\(\frac{20^{10}-3}{20^{10}-3}+\frac{2}{20^{10}-3}=1+\frac{2}{20^{10}-3}\)
Vì 2010-1 > 2010-3
=>\(\frac{2}{20^{10}-1}< \frac{2}{20^{10}-3}\)
=> \(1+\frac{2}{20^{10}-1}< 1+\frac{2}{20^{10}-3}\)
=> A < B
Vậy A < B