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\(\frac{252525}{373737}=\frac{252525:10101}{373737:10101}=\frac{25}{37}\)
\(\frac{272727}{353535}=\frac{272727:10101}{353535:10101}=\frac{27}{35}>\frac{27}{37}>\frac{25}{37}\)
vậy \(\frac{252525}{373737}
\(\frac{252525}{373737}=\frac{25.10101}{37.10101}=\frac{25}{37}=\frac{875}{1295}\)
\(\frac{272727}{353535}=\frac{27.10101}{35.10101}=\frac{27}{35}=\frac{999}{1295}\)
Vì \(\frac{875}{1295}
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\(A=\frac{595959}{454545}-\frac{14141414}{15151515}\)
\(A=\frac{595959:10101}{454545:10101}-\frac{14141414:1010101}{15151515:1010101}\)
\(A=\frac{59}{45}-\frac{14}{15}\)
\(A=\frac{59}{45}-\frac{42}{45}=\frac{17}{45}\)
\(B=\frac{65+891+135+909}{731-47+69-253}\)
\(B=\frac{\left(65+135\right)+\left(891+909\right)}{\left(731+69\right)-\left(47+253\right)}\)
\(B=\frac{200+1800}{800-300}=\frac{2000}{500}=4\)
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a) Ta có: \( - 2 = \frac{{ - 2}}{1} = \frac{{ - 40}}{{20}}\)
\(\frac{{ - 11}}{5} = \frac{{ - 44}}{{20}} < \frac{{ - 40}}{{20}}\) nên \(\frac{{ - 11}}{5} < -2\).
\(\frac{{ - 7}}{4} = \frac{{ - 7.5}}{{4.5}} = \frac{{ - 35}}{{20}} > \frac{{ - 40}}{{20}}\) nên \(\frac{{ - 7}}{4} > -2\)
Vậy \(\frac{{ - 11}}{5} < \frac{{ - 7}}{4}\).
b) Ta có: \(\frac{{2020}}{{ - 2021}} = \frac{{ - 2020}}{{2021}} > \frac{{ - 2022}}{{2021}}\)
Vậy \(\frac{{2020}}{{ - 2021}} > \frac{{ - 2022}}{{2021}}\)
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A = \(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{49.50}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{49}-\frac{1}{50}\)
=\(1-\frac{1}{50}\)
Vì \(1-\frac{1}{50}< 1\)nên A < 1
B = \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
=\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
=\(\frac{1}{2}-\frac{1}{100}\)
Vì \(\frac{1}{2}-\frac{1}{100}< \frac{1}{2}\)nên B < \(\frac{1}{2}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(A=1-\frac{1}{50}\)
\(\Rightarrow A< 1\)
\(B=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)
\(B=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)
\(B=\frac{1}{2}-\frac{1}{100}\)
\(\Rightarrow B< \frac{1}{2}\)
nhanh giup mk voi
Phân số A ta rút gọn tử số 353535 chia cho 7 thành 50505 ,mẫu số 707077 chia cho 7 thành 101011.
Phân số B ta rút gọn tử số chia 9 thành 50505 , mẫu số thành 101010,7777....(ta có thể để số thập phân so sánh)
=> Vì mẫu số B bé hơn nên B lớn hơn A.