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Ta có :
\(\frac{17}{16}>1\)
\(\frac{22}{25}< 1\)
Vậy \(\frac{17}{16}>\frac{22}{25}\)
a) \(\frac{4}{7}=\frac{16}{28}\)
\(\frac{9}{12}=\frac{3}{4}=\frac{21}{28}\)
b) \(\frac{13}{12}=\frac{39}{36}\)
\(\frac{19}{18}=\frac{38}{36}\)
c) \(\frac{1}{5}=\frac{2}{10}\)
d) \(\frac{1}{3}=\frac{21}{63}\)
\(\frac{2}{7}=\frac{18}{63}\)
\(\frac{4}{9}=\frac{28}{63}\)
a/ 4/7 = 1-3/7 và 9/12 = 1-3/12
vì 3/7>3/12 nên 1-3/7<1-3/12
Vậy 4/7<9/12
b/ 13/12 = 1+1/12 và 19/18 = 1+1/18
Vì 1/12>1/18 nên 13/12>19/18
b . 3/7 . 16/33 + 16/33 . 4/7
= 16/33 . ( 3/7 + 4/7 )
= 16/33 . 7/7
= 16/33 . 1
= 16/33
c2 3/7 . 16/33 + 16/33 . 4/7
= 16/77 + 64/231
= 48/231 + 64/231
= 112/231
= 16/33
chúc em học giỏi !!!
a . 5/24 . 5/12 . 24
= ( 5/24 . 24 ) .5/12
= 5 . 5/12
= 25/12
C2 : 5/24 . 5/12 . 24
= 5/24 . ( 5/12 . 24 )
= 5/24 . 10
= 50/24
= 25/12
ta có \(\frac{3}{7}=\frac{3\times3}{7\times3}=\frac{9}{21}\)(quy đồng tử)
So sánh \(\frac{9}{21}\)và \(\frac{9}{17}\)ta có:
\(21>17\Rightarrow\frac{9}{17}>\frac{9}{21}\Rightarrow\frac{9}{17}>\frac{3}{7}\)
VẬY: \(\frac{9}{17}>\frac{3}{7}\)
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
\(A-B=1-1=0\)
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) $\frac{{17}}{{21}} - \frac{{10}}{{21}} = \frac{{17 - 10}}{{21}} = \frac{7}{{21}} = \frac{1}{3}$
a: 4/7=48/84
5/12=35/84
b: 3/8=9/24
19/24=19/24
c: 21/22=21/22
7/11=14/22
d: 8/15=128/240
11/16=165/240
e: 4/25=16/100
72/100=72/100
f: 17/60=17/60
4/5=48/60
a ,Ta có : 14/25 < 15/25 = 3/5
Ta có : 1 - 3/5 = 2/5
1 - 5/7 = 2/7
Nên 2/5 > 2/7
Vậy 14/25 < 5/7
a, vì 5/7<1<15/12
nên 5/7<15/12
b,17/21<17/19
\(\frac{5}{7}\)<\(\frac{15}{12}\)
\(\frac{17}{21}\)<\(\frac{17}{19}\)