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a; (15 - \(\dfrac{121}{18}\)) : \(\dfrac{297}{27}\) - \(\dfrac{17}{8}\) : \(\dfrac{51}{40}\)
(\(\dfrac{270}{18}\) - \(\dfrac{121}{18}\)) : \(\dfrac{297}{27}\) - \(\dfrac{17}{8}\) x \(\dfrac{40}{51}\)
= \(\dfrac{149}{18}\) : \(\dfrac{297}{27}\) - \(\dfrac{5}{3}\)
= \(\dfrac{149}{18}\) x \(\dfrac{27}{297}\) - \(\dfrac{5}{3}\)
= \(\dfrac{149}{198}\) - \(\dfrac{5}{3}\)
= \(\dfrac{149}{198}\) - \(\dfrac{330}{198}\)
= \(\dfrac{-181}{198}\)
b; (- 3,2) x (- \(\dfrac{15}{64}\)) + (0,8 - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)
= (\(\dfrac{-16}{5}\)) x ( \(\dfrac{-15}{64}\)) + (\(\dfrac{4}{5}\) - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)
= \(\dfrac{3}{4}\) + (\(\dfrac{4}{5}\) - \(\dfrac{34}{15}\)): \(\dfrac{11}{3}\)
= \(\dfrac{3}{4}\) + (\(\dfrac{12}{15}\) - \(\dfrac{34}{15}\)) : \(\dfrac{11}{3}\)
= \(\dfrac{3}{4}\) + \(\dfrac{-22}{15}\) : \(\dfrac{11}{3}\)
= \(\dfrac{3}{4}\) - \(\dfrac{22}{15}\) x \(\dfrac{3}{11}\)
= \(\dfrac{3}{4}\) - \(\dfrac{2}{5}\)
= \(\dfrac{15}{20}\) - \(\dfrac{8}{20}\)
= \(\dfrac{7}{20}\)
a)15/8+3/4-5/12
=45+18-10/24
=53/24
b)11/24.12/33+5/6
=11.12/12.2.11.3+5/6
=1/6+5/6
=6/6=1
c)15/8+7/24:5/8
=15/8+7/24.8/5
=15/8+7.8/3.8.5
=15/8+7/15
=đề sai, nếu đúng thì như này
=8/15+7/15
=15/15=1
\(c)\) \(C=\frac{\frac{2}{7}+\frac{2}{5}+\frac{2}{17}-\frac{2}{293}}{\frac{3}{7}+\frac{3}{5}+\frac{3}{17}-\frac{3}{293}}\)
\(C=\frac{2\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{293}\right)}{3\left(\frac{1}{7}+\frac{1}{5}+\frac{1}{17}-\frac{1}{193}\right)}\)
\(C=\frac{2}{3}\)
Bạn Cô nàng Thiên Bình làm đúng hết òi =.=
a=7.[1/8+1/27-1/49]
------------------------
11.[1/8+1/27-1/49]
=7/11
cau b,c tuong tu nha h mk
A = - 2 / 5 + ( - 2 / 5 + 2 )
A = - 2 / 5 + 8 / 5
A = 6 / 5
B = ( - 5 / 24 + 0, 75 + 7 / 12 ) : - 17/8
B = 9 / 8 : - 17/8
B = -9 / 17
C = 3 / 7 + ( - 1 / 5 + - 3 / 7 )
C = 3 / 7 - 1 / 5 - 3 / 7
C = - 1 / 5
D = ( 6 - 14 / 5 ) . 25 / 8 - 8 / 5 : 1 / 4
D = 16 / 5 . 25 / 8 - 8 / 5 . 4
D = 10 - 6 , 4
D = 3,6
\(A=-\frac{2}{5}+\left[\left(-\frac{2}{5}\right)+2\right]=\left[\left(-\frac{2}{5}\right)+\left(-\frac{2}{5}\right)\right]+2=\left(-\frac{4}{5}\right)+2=\left(-\frac{4}{5}\right)+\frac{10}{5}=\frac{6}{5}\)\(\frac{10}{5}=\frac{6}{5}\)
\(B=\left[\left(-\frac{5}{24}\right)+0,75+\frac{7}{12}\right]:\left(\frac{-17}{8}\right)\)\(=\left[\left(-\frac{5}{24}\right)+\frac{3}{4}+\frac{7}{12}\right].\frac{-8}{17}\)
\(=\left[\left(-\frac{5}{24}\right)+\frac{18}{24}+\frac{14}{24}\right].\frac{-8}{17}\)\(=\frac{27}{24}.\frac{-8}{17}=\frac{9}{8}.\frac{-8}{17}=\frac{9.\left(-8\right)}{8.17}=\frac{-9}{17}\)
\(C=\frac{3}{7}+\left[\left(-\frac{1}{5}\right)+\left(-\frac{3}{7}\right)\right]=\left[\frac{3}{7}+\left(-\frac{3}{7}\right)\right]+\left(-\frac{1}{5}\right)=-\frac{1}{5}\)
\(D=\left(6-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}:\frac{1}{4}=\left(\frac{30}{5}-\frac{14}{5}\right).\frac{25}{8}-\frac{8}{5}.4\)
\(=\frac{16}{5}.\frac{25}{8}-\frac{32}{5}=\frac{16.25}{5.8}-\frac{32}{5}=10-\frac{32}{5}\)\(=\frac{50}{5}-\frac{32}{5}=\frac{18}{5}\)
a) \(\dfrac{-5}{9}+\dfrac{8}{15}+\dfrac{-2}{11}+\dfrac{4}{-9}+\dfrac{7}{15}\)
=\(\left(\dfrac{-5}{9}+\dfrac{-4}{9}\right)+\left(\dfrac{8}{15}+\dfrac{7}{15}\right)+\dfrac{-2}{11}\)
=\(\left(-1\right)+1+\dfrac{-2}{11}\)
=\(\dfrac{-2}{11}\)
b) \(\left(\dfrac{-5}{12}+\dfrac{6}{11}\right)+\left(\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\right)\)
=\(\dfrac{-5}{12}+\dfrac{6}{11}+\dfrac{7}{17}+\dfrac{5}{11}+\dfrac{5}{12}\)
=\(\left(\dfrac{-5}{12}+\dfrac{5}{12}\right)+\left(\dfrac{6}{11}+\dfrac{5}{11}\right)+\dfrac{7}{17}\)
=\(0+0+\dfrac{7}{17}\)
=\(\dfrac{7}{17}\)
c) A= \(49\dfrac{8}{23}-\left(5\dfrac{7}{32}+14\dfrac{8}{23}\right)\)
A=\(49\dfrac{8}{23}-5\dfrac{7}{32}-14\dfrac{8}{23}\)
A=\(\left(49\dfrac{8}{23}-14\dfrac{8}{23}\right)-5\dfrac{7}{32}\)
A=\(35-5\dfrac{7}{32}\)
A=\(35-\dfrac{167}{32}=\dfrac{953}{32}\)
d) C=\(\dfrac{-3}{7}.\dfrac{5}{9}+\dfrac{4}{9}.\dfrac{-3}{7}+2\dfrac{3}{7}\)
C=\(\dfrac{-3}{7}.\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}.1+\dfrac{17}{7}\)
C=\(\dfrac{-3}{7}+\dfrac{17}{7}=2\)
a, `(-5)/9+8/15+(-2)/11+4/(-9)+7/15`
`=-5/9+8/15-2/11-4/9+7/15`
`=(-5/9-4/9)+(8/15+7/15)-2/11`
`=-9/9+15/15-2/11`
`=-1+1-2/11`
`=-2/11`
b, `((-5)/12+6/11)+(7/17+5/11+5/12)`
`=-5/12+6/11+7/17+5/11+5/12`
`=(-5/12+5/12)+(6/11+5/11)+7/17`
`=0+11/11+7/17`
`=1+7/17`
`=17/17+7/17`
`=24/17`
c, `A=49 8/23 - (5 7/32 + 14 8/23)`
`A=49 8/23 - 5 7/32 - 14 8/23`
`A=(49 8/23 - 14 8/23)-5 7/32`
`A=35 - 167/32`
`A=953/32`
d, `C=(-3)/7.5/9+4/9.(-3)/7+2 3/7`
`C=-3/7 . 5/9-4/9 . 3/7+17/7`
`C=-3/7.(5/9+4/9)+17/7`
`C=-3/7 . 1+17/7`
`C=2`
a) \(\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{-11}{2}\right)\)
\(=\dfrac{17}{8}.\dfrac{-8}{17}\)
\(=-1\)
b) \(7,63.21,15+7,63.\left(-121,15\right)\)
\(=7,63.\left(21,15-121,15\right)\)
\(=7,63.\left(-100\right)\)
\(=-763\)
c) \(\left(-\dfrac{5}{24}-0,75\right):\left(-2\dfrac{7}{8}\right)\)
\(=\left(-\dfrac{5}{24}-\dfrac{3}{4}\right):\left(-\dfrac{7}{4}\right)\)
\(=\dfrac{-23}{24}.\dfrac{-4}{7}\)
\(=\dfrac{23}{42}\).
\(a.\dfrac{17}{8}:\left(\dfrac{27}{8}+\dfrac{-11}{2}\right)\text{=}\dfrac{17}{8}:\dfrac{-17}{8}\text{=}-1\)
\(b.7,63.21,15+7,63.-121,15\text{=}7,63.\left(21,15-121,15\right)\text{=}7,63.-100\text{=}-763\)
\(c.\left(\dfrac{-5}{24}-0,75\right):\dfrac{-23}{8}\text{=}\dfrac{-23}{24}.\dfrac{-8}{23}\text{=}\dfrac{1}{3}\)