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có:
(1994-1)+1=1994
Tổng là:
1994x(1994+1):2=1989015
Đáp số:1989015
a) 1/7 . 5/6 + 1/7 . 1/6 + -8/7
= 1/7 . ( 5/6 + 1/6 ) + -8/7
= 1/7 . 1 + -8/7
= 1/7 + -8/7 = -1
b) 3/5 . -4/9 . 5/3 . 18/7
= ( 3/5 . 5/3 ) . ( -4/9 . 18/7 )
= 1 . -8/7 = -8/7
c) 1/2 + -3/4 . 16/9
= 1/2 + -4/3 = -5/6
d) ( 1/7 + 5/14 ) . -28/3
= 1/2 . -28/3 = -14/3
\(\frac{3}{5}-\frac{-7}{10}-\frac{13}{-20}=\frac{3}{5}+\frac{7}{10}+\frac{13}{20}=\frac{12}{20}+\frac{14}{20}+\frac{13}{20}=\frac{39}{20}\)
\(\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}-\frac{-1}{6}=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{6+(-4)+3+2}{12}=\frac{7}{12}\)
\(\frac{9}{4}.\frac{8}{27}.\frac{5}{7}=\frac{9.8.5}{4.27.7}=\frac{1.2.5}{1.3.7}=\frac{10}{21}\)
\(\frac{2}{5}.(\frac{2}{3}-\frac{1}{4})+\frac{1}{2}=\frac{2}{5}.(\frac{8}{12}-\frac{3}{12})+\frac{1}{2}=\frac{2}{5}.\frac{5}{12}+\frac{1}{2}=\frac{1}{6}+\frac{1}{2}=\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\)
\((\frac{1}{3}-\frac{1}{6}):(\frac{1}{3}+\frac{1}{6})=(\frac{2}{6}-\frac{1}{6}):(\frac{2}{6}+\frac{1}{6})=\frac{1}{6}:\frac{3}{6}=\frac{1}{6}.\frac{6}{3}=\frac{1.6}{6.3}=\frac{1.1}{1.3}=\frac{1}{3}\)
Hok tốt
Đặt A = 1 - 2 + 3 - 4 + 5 - 6 + ......+ 97 - 98 + 99 - 100
<=> A = ( 1 - 2 ) + ( 3 - 4 ) + ( 5 - 6 ) + ...... + ( 97 - 98 ) + ( 99 - 100 )
<=> A = ( - 1 ) + ( - 1 ) + ( - 1 ) + ...... + ( - 1 ) + ( - 1 ) ( Có 50 số )
=> A = ( - 1 ) . 50 = - 50
Vậy A = - 50
S1=1+(-2)+(-3)+4+5+(-6)+(-7)+8+...+1997+(-1998)+(-1999)+2000
S1=(1+4-2-3)+(5+8-6-7)+...+(1997+2000-1998-1999)
S1=0+0+...+0
S1=0
câu 2
S2=1+3+4+5+...+99-(2+4+6+...+100)
S2=51.50-(50.51)
S2=0
tich nha
Ta có:
1 = \(\frac{1}{10}+\frac{1}{10}+\frac{1}{10}+............+\frac{1}{10}\)(10 phân số \(\frac{1}{10}\))
Mà \(\frac{1}{2}>\frac{1}{10};\frac{2}{3}>\frac{1}{10};............;\frac{9}{10}>10\)
\(\Rightarrow M>1\)
Vậy M > 1
Ta có:\(\frac{1}{2}>\frac{1}{8};\frac{1}{3}>\frac{1}{8};...;\frac{1}{6}>\frac{1}{8};\frac{1}{7}+\frac{1}{8}+\frac{1}{9}>\frac{3}{8}\)
\(\Rightarrow\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{3}{8}\)
\(=\frac{8}{8}=1\)
Vậy\(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{9}>1\)