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a) 4/25 < 4/23 ; b) 14/1 > 1 c) 20/19 > 20/27 ; 1 < 15/14 chắc chắn 100% đúng
\(\frac{5}{6}+\frac{3}{4}< \frac{19}{11}\)
\(\frac{17}{12}-\frac{9}{8}< \frac{1}{3}\)
\(\frac{4}{9}\times\frac{6}{5}=\frac{8}{15}\)
\(\frac{15}{4}:\frac{3}{8}>8\)
a) 19 x 64 + 76 x 34
= 19 x 64 + 19 x 4 x 34
= 19 x ( 64 + 34 x 4 )
= 19 x 200
= 3800
b ) 135 x 12 + 165 x 13
= 135 x 12 + 165 x ( 12 + 1 )
= 135 x 12 + 165 x 12 + 165
= 12 x ( 135 + 165 ) + 165
= 12 x 300 + 165
= 3600 + 165
= 3765
c ) 136 x 68 + 16 x 272
= 136 x 68 + 32 x 136
= 136 x ( 68 + 32 )
= 136 x 100
= 13600
d ) 1 + 2 + 3 + 4 + ...... + 19 + 20
= ( 1 + 20 ) x 20 : 2
= 21 x 10
= 210
e ) 1 + 3 + 5 + ......+ 17 + 19
= 10 : 2 x ( 1 + 19 )
= 5 x 20
= 100
f ) 2 + 4 + 6 + ..... + 18 + 20
= ( 20 + 2 ) x 10 : 2
= 22 x 5
= 110
a, 21/11-9/19-3/13-12/38-24/22 =21/11-(9/19-3/13)-12/38-24/22 =207/143-12/38-24/22 =3075/2717-24/22 =111/2717 b,2/3+3/4+4/5+12/9+28/16+48/15 =17/12+4/5+12/9+28/16+48/15 =133/60+12/9+28/16+48/15 =71/20+28/16+48/15 =53/10+48/15 =17/2 c, 4/5-1/2-1/4-1/8-1/16-1/36 =3/10-1/4-1/8-1/36 =1/20-1/8-1/16-1/36 =-3/40-1/16-1/36 =-11/60-1/36=-119/720
19/36=95/180
28/45=112/180
vì 95/180<112/180
nên 19/36<28/45
\(\frac{5}{7}\)<\(\frac{7}{9}\)
\(\frac{7}{8}\)<\(\frac{5}{6}\)
\(\frac{10}{15}\)=\(\frac{16}{24}\)
\(\frac{19}{43}\)<\(\frac{19}{34}\)
\(\frac{5}{7}< \frac{7}{9};\)\(\frac{7}{8}>\frac{5}{6};\)\(\frac{10}{15}=\frac{16}{24};\)\(\frac{19}{43}< \frac{19}{34}\)
mk nhé
chúc bạn học tốt
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
\(A-B=1-1=0\)
C.=1 nhé
C. = 1 nhé