a) So sánh \(\frac{33}{131}\) và \(\frac{53}{217}\)
b) Cho B = \(\frac{15}{26}\) + \(\frac{10}{17}\) + \(\frac{8}{21}\). CMR: B < 2
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so sanh\(\frac{33}{131}va\frac{53}{217}\)
so sánh :a, \(63^7và16^{12}\)
b\(\frac{21}{52}và\frac{213}{523}\)
c,\(\frac{13}{19}và\frac{1}{1000000}\)
d -33/131 và 53/-217
1/ So sánh các số hữu tỉ sau
a/ \(\frac{13}{17}và\frac{46}{50}\)
b/ \(\frac{33}{131}và\frac{53}{217}\)
c/ \(\frac{41}{91}và\frac{411}{911}\)
d/ \(\frac{2001}{2002}và\frac{2005}{2003}\)
e/ \(\frac{-2005}{2010}và\frac{2001}{2002}\)
a.\(\frac{13}{17}\)=1-\(\frac{4}{17}\); \(\frac{46}{50}\)=1-\(\frac{4}{50}\)
Vì \(\frac{4}{17}\)>\(\frac{4}{50}\)=> 1-\(\frac{4}{17}\)<1-\(\frac{4}{50}\)
Vậy\(\frac{13}{17}\)<\(\frac{46}{50}\)
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c.\(\frac{41}{91}\)=1-\(\frac{50}{91}\)=1-\(\frac{500}{910}\); \(\frac{411}{911}\)=1-\(\frac{500}{911}\)
Vì \(\frac{500}{910}\)>\(\frac{500}{911}\)=>1-\(\frac{500}{910}\)<1-\(\frac{500}{911}\)=>\(\frac{41}{91}\)<\(\frac{411}{911}\)
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d. \(\frac{2001}{2002}< \frac{2002}{2002}=1;\frac{2005}{2003}>\frac{2003}{2003}=1\text{ hay }\frac{2001}{2002}< 1< \frac{2005}{2003}\)
Vậy \(\frac{2001}{2002}< \frac{2005}{2003}\).
e. \(-\frac{2005}{2010}< 0;\frac{2001}{2002}>0\text{ hay }-\frac{2005}{2010}< 0< \frac{2001}{2002}\)
Vậy \(-\frac{2005}{2010}< \frac{2001}{2002}\).
b. \(\frac{33}{131}>\frac{33}{132}=\frac{1}{4};\frac{53}{217}< \frac{53}{212}=\frac{1}{4}\text{ hay }\frac{53}{217}< \frac{1}{4}< \frac{33}{131}\)
Vậy \(\frac{53}{217}< \frac{33}{131}\).
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Hông quy đồng mẫu số, hãy so sánh A và B, biết
A= \(\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}\)
B= \(\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)
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Ta có: \(A=\frac{-9}{10^{2010}}+\frac{-19}{10^{2011}}=\frac{-9}{10^{2010}}-\frac{9}{10^{2011}}-\frac{10}{10^{2011}}\)
\(=\frac{-9}{10^{2010}}-\frac{9}{10^{1011}}-\frac{1}{10^{2010}}=\frac{-9}{10^{2011}}+\frac{-10}{10^{2010}}\)
Ta thấy : \(\frac{10}{10^{2010}}< \frac{19}{10^{2010}}\Rightarrow\frac{-10}{10^{2010}}>\frac{-19}{10^{2010}}\)
\(\Rightarrow\frac{-9}{10^{2011}}+\frac{-10}{10^{2010}}>\frac{-9}{10^{2011}}+\frac{-19}{10^{2010}}\)
Hay \(A>B\)
Vậy ...
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So sanh bằng cách hợp lí:
a.\(\frac{33}{131}\) và \(\frac{53}{217}\)
b.\(\frac{41}{91}\) và \(\frac{411}{911}\)
a. \(\frac{33}{131}>\frac{33}{132}=\frac{1}{4}\)
\(\frac{53}{217}< \frac{53}{212}=\frac{1}{4}\)
Suy ra \(\frac{33}{131}>\frac{53}{217}\)
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Cho \(A=\frac{3^3}{1}-\frac{5^3}{3}+\frac{7^3}{6}-\frac{9^3}{10}+\frac{11^3}{15}-\frac{13^3}{21}+\frac{15^3}{28}-\frac{17^3}{36}+...+\frac{199^3}{4950}\)
So sánh A với 814
Ta có:
\(\frac{A}{2}=\frac{3^3}{2}-\frac{5^3}{6}+\frac{7^3}{12}-\frac{9^3}{20}+\frac{11^3}{30}-\frac{13^3}{42}+\frac{15^3}{56}-\frac{17^3}{72}+...+\frac{199^3}{9900}\)
\(=3^2.\left(1+\frac{1}{2}\right)-5^2.\left(\frac{1}{2}+\frac{1}{3}\right)+7^2.\left(\frac{1}{3}+\frac{1}{4}\right)-9^2.\left(\frac{1}{4}+\frac{1}{5}\right)+...+199^2.\left(\frac{1}{99}+\frac{1}{100}\right)\)
\(=3^2+\left(\frac{3^2}{2}-\frac{5^2}{2}\right)-\left(\frac{5^2}{3}-\frac{7^2}{3}\right)+\left(\frac{7^2}{4}-\frac{9^2}{4}\right)-\left(\frac{9^2}{5}-\frac{11^2}{5}\right)+...+\left(\frac{197^2}{99}-\frac{199^2}{99}\right)+\frac{199^2}{100}\)
\(=3^2-8+8-8+...+8+\frac{199^2}{100}=3^2+\frac{199^2}{100}< 3^2+\frac{199.200}{100}=9+398=407\)
\(\Rightarrow A< 407.2=814\)
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bài 1: So sánh
a,\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{10^2} và 1 \)
b,\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+\frac{1}{12^2} và 1\)
Bài 5 : Chững minh rẳng :
a) S frac{3}{10}+frac{3}{11}+frac{3}{12}+frac{3}{13}+frac{3}{14} CMR :1 S 2
b) frac{1}{4^2}+frac{1}{6^2}+frac{1}{8^2}+frac{1}{10^2}+...+frac{1}{left(2nright)^2} frac{1}{4}
c) frac{3}{4}+frac{8}{9}+frac{15}{16}+...+frac{2499}{2500}48
d) frac{1}{5}+frac{1}{13}+frac{1}{14}+frac{1}{15}+frac{1}{61}+frac{1}{62}+frac{1}{63} frac{1}{2}
Bài :So sánh phân số sau:
a)frac{1985.1987-1}{1980+1985.1986}và1
b) A frac{13^{15}+1}{13^{16}+1}và...
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Bài 5 : Chững minh rẳng :
a) S= \(\frac{3}{10}+\frac{3}{11}+\frac{3}{12}+\frac{3}{13}+\frac{3}{14}\) CMR :1< S <2
b) \(\frac{1}{4^2}+\frac{1}{6^2}+\frac{1}{8^2}+\frac{1}{10^2}+...+\frac{1}{\left(2n\right)^2}< \frac{1}{4}\)
c) \(\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{2499}{2500}>48\)
d) \(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)
Bài :So sánh phân số sau:
a)\(\frac{1985.1987-1}{1980+1985.1986}và1\)
b) A= \(\frac{13^{15}+1}{13^{16}+1}\)và B = \(\frac{13^{16}+1}{13^{17}+1}\)
c)\(\frac{18}{53}và\frac{26}{79}\)
d)\(\frac{5}{8}và\frac{14}{17}\)
e)\(\frac{1}{5^{199}}và\frac{1}{3^{300}}\)
g)\(\frac{1}{3^{17}}và\frac{1}{5^{10}}\)
h) \(\frac{18}{109}và\frac{5}{30}\)
So sánh A và B biết:
A=\(\frac{10^{17}+1}{10^{18}+1}\), B=\(\frac{10^{18}+1}{10^{19}+1}\)
Ta có: \(A=\frac{10^{18}+1}{10^{19}+1}>\frac{10.\left(10^{17}+1\right)}{10.\left(10^{18}+1\right)}=\frac{10^{17}+1}{10^{18}+1}\)
Vậy A < B
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