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Bài giải
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b, \(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
Ta có :
a, \(\left(\frac{1}{6}+\frac{1}{10}+\frac{1}{15}\right)\text{ : }\left(\frac{1}{6}+\frac{1}{10}-\frac{1}{15}\right)=\left(\frac{5}{30}+\frac{3}{30}+\frac{2}{30}\right)\text{ : }\left(\frac{5}{30}+\frac{3}{30}-\frac{2}{30}\right)=\frac{1}{3}-\frac{1}{5}=\frac{2}{15}\)
b,
\(\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)\text{ : }\left(\frac{1}{4}-\frac{1}{5}\right)=\left(\frac{60}{120}-\frac{40}{120}+\frac{30}{120}-\frac{24}{120}\right)\text{ : }\left(\frac{5}{20}-\frac{4}{20}\right)=\frac{13}{60}\text{ : }\frac{1}{20}=\frac{13}{3}\)
`4/5 - (6/15 + 1/5)`
`=4/5 - (6/15 + 3/15)`
`= 4/5 - 9/15`
`= 4/5 -3/5`
`=1/5`
1/10 + 3/5 : 2/3
=1/10 + 9/10
=10/10
=1
\((\)6/8 -2/6\()\) x 1/2
=18/48 X 1/2
=3/7 x 1/2
=2/14
\(\dfrac{1}{10}\) + \(\dfrac{3}{5}\):\(\dfrac{2}{3}\)
= \(\dfrac{1}{10}\) + \(\dfrac{3}{5}\) \(\times\) \(\dfrac{3}{2}\)
= \(\dfrac{1}{10}\) + \(\dfrac{9}{10}\)
= \(\dfrac{10}{10}\)
= 1
( \(\dfrac{6}{8}\) - \(\dfrac{2}{6}\)) \(\times\) \(\dfrac{1}{2}\)
=( \(\dfrac{18}{24}\) - \(\dfrac{8}{24}\)) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{10}{24}\) \(\times\) \(\dfrac{1}{2}\)
= \(\dfrac{5}{24}\)
Lời giải:
a.
$\frac{5}{15}-\frac{1}{6}\times \frac{2}{5}=\frac{5}{15}-\frac{1}{15}=\frac{4}{15}$
b.
$\frac{8}{24}+\frac{3}{4}:\frac{1}{8}=\frac{1}{3}+6=\frac{19}{3}$
c.
$\frac{1}{7}: \frac{2}{8}-\frac{1}{7}=\frac{1}{7}\times 4-\frac{1}{7}$
$=\frac{1}{7}\times (4-1)=\frac{1}{7}\times 3=\frac{3}{7}$
`1/3-2/15+14/15`
`=5/15-2/15+14/15`
`=17/15`
`3/5+4-6/7`
`=21/35+140/35-30/35`
`=131/35`
a: Thay a=9 và b=15 vào P, ta được:
\(P=\left(9+1\right)\cdot2+\left(15+1\right)\cdot3\)
\(=10\cdot2+16\cdot3=20+48=68\)
b: \(m=2\cdot a+3\cdot b+5=2\cdot9+3\cdot15+5=68\)
mà P=68
nên P=m
A)\(\dfrac{7}{20}-\left(\dfrac{5}{8}-\dfrac{2}{5}\right)\)
\(=\dfrac{7}{20}-\left(\dfrac{25}{40}-\dfrac{16}{40}\right)\)
\(=\dfrac{7}{20}-\dfrac{9}{40}\)
\(=\dfrac{14}{40}-\dfrac{9}{40}=\dfrac{5}{40}=\dfrac{1}{8}\)
B) \(\dfrac{5}{6}+\left(\dfrac{5}{9}-\dfrac{1}{4}\right)\)
\(=\dfrac{5}{6}+\left(\dfrac{20}{36}-\dfrac{9}{36}\right)\)
\(=\dfrac{5}{6}+\dfrac{11}{36}\).
\(=\dfrac{30}{36}+\dfrac{11}{36}=\dfrac{41}{36}\)
C) \(\dfrac{9}{10}-\left(\dfrac{2}{5}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)
\(=\dfrac{9}{10}-\left(\dfrac{4}{10}-\dfrac{3}{10}\right)+\dfrac{7}{20}\)
\(=\dfrac{9}{10}-\dfrac{1}{10}+\dfrac{7}{20}\)
\(=\dfrac{18}{20}-\dfrac{2}{20}+\dfrac{7}{20}=\dfrac{23}{20}\)
a: =7/20-5/8+2/5
=14/40-25/40+16/40
=5/40=1/8
b: =5/6+5/9-1/4
=30/36+20/36-9/36
=41/36
c: =9/10-2/5+3/10+7/20
=12/10-2/5+7/20
=7/20+6/5-2/5
=7/20+4/5
=7/20+16/20
=23/20
kết quả là 1/3
ta có
6=2×3
10=2×5
15=3×5
➡2^2×3^2×5^2=900
=150/900+90/900+60/900
=150+90+60/900=300/900=1/3
mk chắc chắn đúng 100% lun. cho mk xin 1 k nha bạn.