\(\dfrac{6}{7}km^2+\dfrac{9}{9}km^2+\dfrac{12}{14}km^2\)
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Sao lại hỏi quãng đường em nhỉ? Vì đây là con số đi kèm đơn vị vận tốc mà???
a: =4/5+1/5+2/3+1/3=1+1=2
b: =17/12+7/12+29/7-8/7=3+2=5
c: =3/5+2/5+16/7-1/7-1/7
=1+2=3
d: =2/5+3/5+2/3+1/3+7/4+1/4
=1+1+2
=4
a: =>x-5/14=6/14-1/14=5/14
=>x=10/14=5/7
b; =>2/9:x=3/6+4/6=7/6
=>x=2/9:7/6=2/9*6/7=4/21
c: =>32:x=8
=>x=4
`@` `\text {Answer}`
`\downarrow`
`a,`
`x - 5/14 = 3/7 - 1/14`
`x - 5/14 = 5/14`
`=> x = 5/14 + 5/14`
`=> x = 5/7`
Vậy, `x = 5/7`
`b,`
`2/9 \div x = 1/2 + 2/3`
`2/9 \div x = 7/6`
`x = 2/9 \div 7/6`
`x = 4/21`
Vậy, `x = 4/21`
`c,`
\(\dfrac{6}{32\div x}=\dfrac{12}{16}\)
`6/(32 \div x) = 3/4`
`32 \div x = 6 \div 3/4`
`32 \div x = 8`
` x = 32 \div 8`
`x = 4`
Vậy, `x = 4`
`3/5` giờ `=0,6` giờ
`5/4m=1,25m`
`5/8` giờ `=0,625` giờ
`9/5km=1,8km`
`3/5` phút `=0,6` giờ
`7/8kg=0,875kg`
b: =12+5/14-3-5/7-5-5/14
=4-5/7
=28/7-5/7=23/7
c: =(-2/5-11/10)+(7/11-7/11)
=-4/10-11/10=-15/10=-3/2
\(a,\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{5}{13}\cdot\dfrac{2}{9}\)
\(=\dfrac{5}{9}\cdot\dfrac{7}{13}+\dfrac{5}{9}\cdot\dfrac{8}{13}-\dfrac{2}{13}\cdot\dfrac{5}{9}\)
\(=\dfrac{5}{9}\cdot\left(\dfrac{7}{13}+\dfrac{8}{13}-\dfrac{2}{13}\right)\)
\(=\dfrac{5}{9}\cdot\dfrac{14}{13}\)
\(=\dfrac{70}{117}\)
\(d,\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}+\dfrac{-2}{5}\)
\(=\left(\dfrac{1}{2}+\dfrac{-2}{3}+\dfrac{1}{6}\right)+\dfrac{-2}{5}\)
\(=0+\dfrac{-2}{5}\)
\(=\dfrac{-2}{5}\)
a: \(\dfrac{5}{6}-\dfrac{4}{6}=\dfrac{5-4}{6}=\dfrac{1}{6}\)
b: \(\dfrac{7}{12}-\dfrac{6}{12}=\dfrac{7-6}{12}=\dfrac{1}{12}\)
c: \(\dfrac{7}{9}-\dfrac{2}{9}=\dfrac{7-2}{9}=\dfrac{5}{9}\)
d: \(\dfrac{16}{5}-\dfrac{9}{5}=\dfrac{16-9}{5}=\dfrac{7}{5}\)
a)\(\dfrac{2}{3}giờ< \dfrac{3}{4}giờ\)
b)\(\dfrac{7}{10}m< \dfrac{3}{4}m\)
c)\(\dfrac{7}{8}kg< \dfrac{9}{10}kg\)
d)\(\dfrac{5}{6}\)km/h >\(\dfrac{7}{9}\)km/h
\(=\left(\dfrac{6}{7}+\dfrac{9}{9}+\dfrac{12}{14}\right)km^2\\ =\left(\dfrac{6}{7}+1+\dfrac{6}{7}\right)km^2\\ =\left(\dfrac{12}{7}+1\right)km^2\\ =\left(\dfrac{12+7}{7}\right)km^2\\ =\dfrac{19}{7}km^2\)