Tinh
A).11/14-5/7
5/8:8/12
2/5+1/6x2/7
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A. C1: (6/11 + 5/11) x 3/7 = 1 x 3/7 = 3/7
C2: (6/11 + 5/11) x 3/7 = 6/11 x 3/7 + 5/11 x 3/7 = 18/77 + 15/77 = 3/7
a)Cách 1:
(6/11+5/11)x3/7
=11/11x3/7
=1x3/7
=3/7
Cách 2:
(6/11+5/11)x3/7
=6/11x3/7+5/11x3/7
=18/77 + 15/77
=33/77
=3/7 (mink đã rút gọn)
k nhé
(Dấu . là dấu nhân)
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+...+\dfrac{3}{119.122}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{119}-\dfrac{1}{122}\)
\(=\dfrac{1}{5}-\dfrac{1}{122}\)
\(=\dfrac{122}{610}-\dfrac{5}{610}\)
\(=\dfrac{117}{610}\)
#Sahara |
\(\dfrac{3}{5.8}+\dfrac{3}{8.11}+\dfrac{3}{11.14}+.....+\dfrac{3}{119.122}\)
\(=\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+......+\dfrac{1}{119}-\dfrac{1}{122}\)
\(=\dfrac{1}{5}-\dfrac{1}{122}=\dfrac{122-5}{610}=\dfrac{117}{610}\)
Doi 12 km = 120000cm
Quang duong do dai :
120000 : 100000 = 12 (cm)
Dap so : 12 cm
độ dài thật của quãng đường từ HCM - Quy Nhơn là :
27 x 2500000= 67500000 cm=67,5 km
đọ dài thật của quãng đường là :
2500000 nhân 27 =67500000 (cm)
= 675 km
a.(x+2)+(x+5)+(x+8)+(x+11)+(x+14)=75
=>(x+x+x+x+x)+(2+5+8+11+14)=75
=>Xx5+40=75
=>Xx5=75-40
=>Xx5=35
=>x=35:5
=>x=7
a) \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+........+\frac{1}{99.100}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+.........+\frac{1}{99}-\frac{1}{100}\)
\(=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
b) \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+..........+\frac{2}{73.75}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.......+\frac{1}{73}-\frac{1}{75}\)
\(=\frac{1}{3}-\frac{1}{75}=\frac{8}{25}\)
c) \(\frac{4}{4.6}+\frac{4}{6.8}+\frac{4}{8.10}+..........+\frac{4}{64.66}\)
\(=2.\left(\frac{2}{4.6}+\frac{2}{6.8}+\frac{2}{8.10}+..........+\frac{2}{64.66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+\frac{1}{8}-\frac{1}{10}+.....+\frac{1}{64}-\frac{1}{66}\right)\)
\(=2.\left(\frac{1}{4}-\frac{1}{66}\right)=2.\frac{31}{132}=\frac{31}{66}\)
d) \(\frac{9}{5.8}+\frac{9}{8.11}+\frac{9}{11.14}+........+\frac{9}{497.500}\)
\(=3.\left(\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+..........+\frac{3}{497.500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{14}+......+\frac{1}{497}-\frac{1}{500}\right)\)
\(=3.\left(\frac{1}{5}-\frac{1}{500}\right)=3.\frac{99}{500}=\frac{297}{500}\)
e) \(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+......+\frac{1}{93.95}\)
\(=\frac{1}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+........+\frac{2}{93.95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+........+\frac{1}{93}-\frac{1}{95}\right)\)
\(=\frac{1}{2}.\left(\frac{1}{5}-\frac{1}{95}\right)=\frac{1}{2}.\frac{18}{95}=\frac{9}{95}\)
g) \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+..........+\frac{1}{200.203}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+........+\frac{3}{200.203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{200}-\frac{1}{203}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{203}\right)=\frac{1}{3}.\frac{201}{406}=\frac{67}{406}\)
\(a;\frac{11}{14}-\frac{5}{7}=\frac{11}{14}-\frac{10}{14}=\frac{1}{14}\)
\(b;\frac{5}{8}:\frac{8}{12}=\frac{5}{8}:\frac{2}{3}=\frac{5}{8}\times\frac{3}{2}=\frac{15}{16}\)
\(c;\frac{2}{5}+\frac{1}{6}\times\frac{2}{7}=\frac{2}{5}+\frac{2}{42}=\frac{2}{5}+\frac{1}{21}=\frac{42}{105}+\frac{5}{105}=\frac{47}{105}\)
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