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Ta phân tích: 1/18=1/3x6;1/54=1/6x9;1/108=1/9x12;.........1/990=1/30x33, ta có Fx3=3/3x6+3/6x9+3/9x12+........+3/30x33=1/3-1/6+1/6-1/9+1/9-1/12+.............+1/30-1/33=1/3-1/33=10/33, suy ra F là: 10/33/3=10/99
F=1/18+1/54+1/108+...+1/990 F=1/3.6 + 1/6.9 + 1/9.12 +...+ 1/30.33 suy ra : 3F= 3/3.6 + 3/6.9 + 3/9.12 +...+3/30.33 3F= 3/3 - 3/6 + 3/6 - 3/9 + 3/9 - 3/12 +...+3/30 - 3/33 3F=1 - 3/33 = 33/33 - 3/33 = 30/33 F= 30/33 : 3 = 30/33 . 1/3 =10/99
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\left(\frac{1}{3}-\frac{1}{33}\right)\)
\(F=\frac{1}{3}.\frac{10}{33}\)
\(F=\frac{10}{99}\)
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\right)\)
\(=\frac{1}{3}\left(\frac{1}{3}-\frac{1}{33}\right)=\frac{1}{3}\cdot\frac{10}{33}=\frac{10}{99}\)
a) 38 + 41 + 117 + 159 + 62
= ( 38 + 62 ) + ( 41 + 159 ) + 117
= 100 + 200 + 117
= 300 + 117
= 417
b) 42 x 53 + 47 x 156 - 47 x 144
= 47 x ( 156 - 144 ) + 42 x 53
= 47 x 12 + 42 x 53
= 564 + 2226
= 2790
c) 341 x 67 + 341 x 16 + 659 x 83
= 341 x ( 67 + 16 ) + 659 x 83
= 341 x 83 + 659 x 83
= 83 x ( 341 + 659 )
= 83 x 1000
= 83000
hok tốt
1.a)341.67+341.16+659.82
=341.(67+16)+659.82
=341.83+659.82
=341+341.82+659.82
=341+82.(341+659)
=341+82.1000
=82341
b)42.53+47.156-47.114
=42.53+47.(156-114)
=42.53+47.42
=42.(53+47)
=42.100
=4200
1/6*3+1/6*9+1/9*12+........+1/30*33
=(1/3-1/6)+(1/6-1/9)+(1/9-1/12)+........+(1/30-1/33)
=1/3-1/6+1/6-1/9+1/9-1/12+........+1/30-1/33
=1/3-1/33
=10/33
nho k cho mink nha
CHUC BAN HOC GIOI !
Gợi ý: 18 = 3.6
54 = 6.9
108 = 9.12
.............
990 = 30.33
Gấp 3 lần R rồi dùng sai phân hữu hạn.
Tự làm tiếp nhé!!!
\(\frac{1}{24\cdot25}+\frac{1}{25\cdot26}+...+\frac{1}{29\cdot30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120.}\)
\(\frac{1}{24.25}+\frac{1}{25.26}+...+\frac{1}{29.30}\)
\(=\frac{1}{24}-\frac{1}{25}+\frac{1}{25}-\frac{1}{26}+...+\frac{1}{29}-\frac{1}{30}\)
\(=\frac{1}{24}-\frac{1}{30}\)
\(=\frac{1}{120}\)
Ủng hộ mk nha ^_-
1/18+1/54+1/108+...+1/990
=1/3 x 6+1/6 x 9+1/9 x 12+...+1/30 x 33
=(1/3-1/6)+(1/6-1/9)+...+1(/30-1/33)
=1/3-1/6+1/6-1/9+...+1/30-1/33
=1/3-1/33
=10/33
giải
1/18 + 1/54 +1/108 + ......+ 1/990
ta tách mẫu số ra thành 1 tích của 2 số :
1/3x6 + 1/6x9 + 1/9x12 +........ + 1/30x33
theo quy tắc ta có : nếu tử nhân với 3 thì mẩu cũng sẽ nhân với 3 :
1x3/3x6x3 +1x3/6x9x3 + 1x3/9x11x3 + .........+ 1x3/30x33x3
= 1/3 x ( 3/3x6 + 3/6x9 + 3/9x11 +.....+3/30x33
= 1/3 x ( 1/3 - 1/33 )
= 1/3 x 10/33
=10/99
Mình giải hơi khó hiểu 1 chút nha
1/18+1/54+1/108+...+1/990
=1/3*6+1/6*9+1/9*12+...+1/30*33\
=(1/3-1/6)+(1/6-1/9)+...+1(/30-1/33)
=1/3-1/6+1/6-1/9+...+1/30-1/33
=1/3-1/33
=10/33
\(F=\frac{1}{18}+\frac{1}{54}+\frac{1}{108}+...+\frac{1}{990}\)
\(F=\frac{1}{3.6}+\frac{1}{6.9}+\frac{1}{9.12}+...+\frac{1}{30.33}\)
\(F=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+\frac{1}{9}-\frac{1}{12}+...+\frac{1}{30}-\frac{1}{33}\)
\(F=\frac{1}{3}-\frac{1}{33}\)
\(F=\frac{11}{33}-\frac{1}{33}\)
\(F=\frac{10}{33}\)
Chúc bạn học tốt !!!!
f =1/18+1/54+1/108+...+1/990
F=1/3.6+1/6.9+1/9.12+...+1/30.33
F=3/3.6+3/6.9+3/9.12+...+3/30.33
F=1/3-1/6+1/6-1/9+...+1/30-1/33
F=1/3-1/33
F=11/33-1/33
F=10/33
Vậy F=10/33.
dấu chấm là dấu nhân nha bạn
chúc bạn học tốt ^^. cho mình nha