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12 tháng 8 2019

a)85793-36841+3826=62778

b)84/100-29/100+30/100=85/100=17/50

c)325,97+86,54+103,46=515,97

d)1/2+1/3+1/4+1/5=30+20+14+12/60=76/60=19/14

1/2+1/3+1/4+1/5

mơn trc nha!!!

2 tháng 8 2023

bài 2 tìm x

a,106- ( x+ 7) =9

x+7 = 106 - 9

x+7 = 107

x= 107 - 7

x=100

b, 2 x ( x+ 4) + 5 =65

2 x (x+4) = 65 - 5

2 x (x+4) = 60

x+4 = 60:2

x+4= 30

x= 30 - 4

x=26

c, (16x x -32) x 45=0

16 x X - 32 = 0: 45

16 x X - 32 =0

16 x X = 0 + 32

16 x X = 32

X= 32:16

X=2

d, x+4 x x = 100 : 5

X + 4 x X = 20

(1+4) x X = 20

5 x X  = 20

X= 20:5

X=4

2 tháng 8 2023

ủa bài 1 đâu =]]

Bài 1: 

a) \(\dfrac{65}{91}+\dfrac{-33}{55}=\dfrac{5}{7}+\dfrac{-3}{5}=\dfrac{25}{35}+\dfrac{-21}{35}=\dfrac{4}{35}\)

b) \(\dfrac{36}{-84}+\dfrac{100}{450}=\dfrac{-3}{7}+\dfrac{2}{9}=\dfrac{-27}{63}+\dfrac{14}{63}=\dfrac{-13}{63}\)

 

4 tháng 4 2021

14 tháng 2 2019

c)

\(\left(1-\frac{1}{2}\right)+\left(1-\frac{1}{6}\right)+....+\left(1-\frac{1}{42}\right)+\left(1-\frac{1}{56}\right)\)

\(\left(1+1+1+....+1+1\right)+\left(\frac{1}{1\times2}+\frac{1}{2\times3}+...+\frac{1}{6\times7}+\frac{1}{7\times8}\right)\)(Có  7 số 1)

\(7+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)

\(7+1-\frac{1}{8}=\frac{63}{8}\)

Gợi ý 1 bài c) còn d) e) cũng làm như vậy nhé

Chúc bạn học tốt !!!

1 tháng 10 2021

\(a,=75+60-13=122\\ b,=250-25+40=265\\ c,=4+100-9=95\\ d,=45+80-27=98\\ e,=5^3-100=125-100=25\\ g,=5+5=10\)

1 tháng 10 2021

h h i nữa

20 tháng 12 2021

a, = 3.25 + 15.4 - 16 : 2 = 75 + 60 - 8 = 135 - 8 = 127

20 tháng 12 2021

a: =75+60-8=127

Giải:

a) \(75\%+1,2-2+\dfrac{1}{5}+2018^0\) 

=\(\dfrac{3}{4}+\dfrac{6}{5}-2+\dfrac{1}{5}+1\) 

=\(\left(\dfrac{6}{5}+\dfrac{1}{5}\right)+\left(\dfrac{3}{4}-2+1\right)\) 

=\(\dfrac{7}{5}+\dfrac{-1}{4}\) 

=\(\dfrac{23}{20}\) 

b) \(\left(\dfrac{-4}{3}+0,75\right):\dfrac{2017}{2018}+\left(1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left(\dfrac{-4}{3}+0,75+1+\dfrac{1}{3}-75\%\right):\dfrac{2017}{2018}\) 

=\(\left[\left(\dfrac{-4}{3}+1+\dfrac{1}{3}\right)+\left(0,75-75\%\right)\right]:\dfrac{2017}{2018}\) 

=\(\left[0+0\right]:\dfrac{2017}{2018}\) 

=0\(:\dfrac{2017}{2018}\) 

=0

c)\(\left(2018-\dfrac{1}{3}-\dfrac{2}{4}-\dfrac{3}{5}-\dfrac{4}{6}-...-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\)

=\(\left(1-\dfrac{1}{3}-1-\dfrac{2}{4}-1-\dfrac{3}{5}-1-\dfrac{4}{6}-...-1-\dfrac{2018}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) 

=\(\left(\dfrac{2}{3}-\dfrac{2}{4}-\dfrac{2}{5}-\dfrac{2}{6}-...-\dfrac{2}{2020}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left[2.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[\dfrac{5}{5}.\left(\dfrac{1}{3}-\dfrac{1}{4}-\dfrac{1}{5}-\dfrac{1}{6}-...-\dfrac{1}{2020}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(\left\{2.\left[5.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right)\right]\right\}:\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =\(10.\left(\dfrac{1}{15}-\dfrac{1}{20}-\dfrac{1}{25}-\dfrac{1}{30}-...-\dfrac{1}{10100}\right):\left(\dfrac{1}{15}+\dfrac{1}{20}+\dfrac{1}{25}+\dfrac{1}{30}+...+\dfrac{1}{10100}\right)\) =-10