Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
2013/2014=1-1/2014
2003/2004=1-1/2004
vì 1/2014<1/2004
=) 1-1/2014>1-1/2004
hay 2013/2014>2003/2004
Ta có:
\(A=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1.2}+\frac{1}{3.4}+...+\frac{1}{3999.4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{3999}-\frac{1}{4000}}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{3}+...+\frac{1}{3999}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-2.\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{4000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{3999}+\frac{1}{4000}\right)-\left(1+\frac{1}{2}+...+\frac{1}{2000}\right)}\)
\(=\frac{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}{\frac{1}{2001}+\frac{1}{2002}+...+\frac{1}{4000}}=1\)
Ta lại có:
\(B=\frac{\left(17+1\right)\left(\frac{17}{2}+1\right)...\left(\frac{17}{19}+1\right)}{\left(1+\frac{19}{17}\right)\left(1+\frac{19}{16}\right)...\left(1+19\right)}\)
\(=\frac{\frac{18}{1}.\frac{19}{2}.\frac{20}{3}...\frac{36}{19}}{\frac{36}{17}.\frac{35}{16}.\frac{34}{15}...\frac{20}{1}}\)
\(=\frac{1.2.3...36}{1.2.3...36}=1\)
Từ đây ta suy ra được
\(A-B=1-1=0\)
\(\frac{19}{20}\)và \(\frac{4}{3}\)
Ta có:
\(\frac{19}{20}\)< 1 ; \(\frac{4}{3}\)> 1
Vậy phân số \(\frac{4}{3}\)lớn hơn phân số \(\frac{19}{20}\)
\(\frac{19}{20}\)và \(\frac{15}{29}\)
\(\Rightarrow\frac{19}{20}>\frac{19}{29}>\frac{15}{29}\)
Vậy : \(\frac{19}{20}>\frac{15}{29}\)
\(\frac{5}{7}\)<\(\frac{15}{12}\)
\(\frac{17}{21}\)<\(\frac{17}{19}\)
a) Ta có :\(\frac{12}{18}< \frac{12}{17}\)
Mà : \(\frac{12}{17}< \frac{13}{17}\)
Từ đó : \(\frac{12}{18}< \frac{13}{17}\)
Bài 1 :
12/19 + 4 = 88/19
1234 + 9800 - 9700 = 1334
Bài 2:
10/32 < 9/8
21/14 > 20/12
1)
a)\(\frac{12}{19}\)+4=\(\frac{12}{19}\)+ \(\frac{76}{19}\)=\(\frac{88}{19}\) 1234+9800-9700
=1234+100=1334
2)\(\frac{10}{32}\)<\(\frac{9}{8}\) \(\frac{21}{14}\)<\(\frac{20}{12}\)
a)\(\frac{19}{20}+\frac{1}{20}=1\)
\(\frac{20}{21}+\frac{1}{21}=1\)
vi \(\frac{1}{20}>\frac{1}{21}\) nen \(\frac{19}{20}<\frac{20}{21}\)
b) \(\frac{89}{88}-\frac{1}{88}=1\)
\(\frac{90}{89}-\frac{1}{89}=1\)
vi \(\frac{1}{88}>\frac{1}{89}nen\frac{89}{88}>\frac{90}{89}\)
c)\(\frac{2005}{2003}-\frac{2}{2003}=1\)
\(\frac{2003}{2001}-\frac{2}{2001}=1\)
vi \(\frac{2}{2003}<\frac{2}{2001}nen\frac{2005}{2003}<\frac{2003}{2001}\)
1 <
2 <
3 >