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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
\(A-B=1-1=0\)
C1) 17/25.13/19+17/25.6/19 C2) 17/25.13/19+17/25.6/19
=17/25.(13/19+6/19) =221/475+102/475
=17/25.1=17/25 =17/25
Học tốt nha
\(C1:\)
\(\frac{17}{25}x\frac{13}{19}+\frac{17}{25}x\frac{6}{19}\)
\(=\frac{17}{25}x\left(\frac{13}{19}+\frac{6}{19}\right)\)
\(=\frac{17}{25}x1\)
\(=\frac{17}{25}\)
\(C2:\)
\(\frac{17}{25}x\frac{13}{19}+\frac{17}{25}x\frac{6}{19}\)
\(=\frac{221}{475}+\frac{102}{475}\)
\(=\frac{323}{475}=\frac{17}{25}\)
\(\frac{5}{7}\)<\(\frac{15}{12}\)
\(\frac{17}{21}\)<\(\frac{17}{19}\)
a) $\frac{2}{3} - \frac{1}{3} = \frac{{2 - 1}}{3} = \frac{1}{3}$
b) $\frac{7}{{12}} - \frac{5}{{12}} = \frac{{7 - 5}}{{12}} = \frac{2}{{12}} = \frac{1}{6}$
c) $\frac{{17}}{{21}} - \frac{{10}}{{21}} = \frac{{17 - 10}}{{21}} = \frac{7}{{21}} = \frac{1}{3}$
1.
a) \(\frac{16}{24}-\frac{1}{3}=\frac{16}{24}-\frac{8}{24}=\)\(\frac{8}{24}=\frac{1}{3}\)
b) \(\frac{4}{5}-\frac{12}{60}=\frac{48}{60}-\frac{12}{60}=\frac{36}{60}=\frac{9}{15}\)
3.
a)\(\frac{17}{6}-\frac{2}{6}=\frac{17-2}{6}=\frac{15}{6}\)
b) \(\frac{16}{15}-\frac{11}{15}=\frac{16-11}{15}=\frac{5}{15}=\frac{1}{3}\)
c) \(\frac{19}{12}-\frac{13}{12}=\frac{19-13}{12}=\frac{6}{12}=\frac{1}{2}\)
25 phần 24 , 19 phần 18 , 21 phần 20 , 24 phần 23 , 18 phần 17
\(A=\frac{17}{19}x\frac{1}{6}+\frac{17}{19}x\frac{5}{6}\)
\(=\frac{17}{19}x\left(\frac{1}{6}+\frac{5}{6}\right)=\frac{17}{19}x1=\frac{17}{19}\)
A= \(\frac{1717}{1919}\text{ *}\frac{303}{1818}+\frac{17}{19}\text{ *}\frac{1515}{1818}\)
A= \(\frac{17}{19}\text{ *}\frac{3}{18}+\frac{17}{19}\text{ *}\frac{15}{18}\)
A= \(\frac{17}{19}\text{ *}\left(\frac{3}{18}+\frac{15}{18}\right)\)
A= \(\frac{17}{19}\text{ *}1\)
A= \(\frac{17}{19}\)
Vậy A= \(\frac{17}{19}\)
vVv Học tốt nhé bạn, mình sẽ rất vui nếu bạn k đúng cho mình, cảm ơn ban nhiều vVv
a,\(\frac{26}{7}\)
b,\(\frac{78}{19}\)
c,\(\frac{13}{11}\)
d,\(1\)
a, \(4-\frac{2}{7}\)=\(\frac{26}{7}\)
b,5-\(\frac{17}{19}=\frac{78}{19}\)
c,\(\frac{35}{11}-2=\frac{13}{11}\)
d,\(\frac{68}{17}-3=1\)
k mình nha
12/17+19/17=31/17
31/17 bạn nhé