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16 tháng 4 2022

\(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

\(=\left(\dfrac{1}{10}+\dfrac{9}{10}\right)+\left(\dfrac{2}{10}+\dfrac{8}{10}\right)+\left(\dfrac{3}{10}+\dfrac{7}{10}\right)+\left(\dfrac{4}{10}+\dfrac{6}{10}\right)+\dfrac{5}{10}\)

\(=1+1+1+1+\dfrac{5}{10}\)

\(=4+\dfrac{5}{10}\)

\(=\dfrac{45}{10}\)

\(13,25:0,5+13,25:0,25+13,25:0,125+13,25\times6\)

\(=13,25:\dfrac{1}{2}+13,25:\dfrac{1}{4}+13,25:\dfrac{1}{8}+13,25\times6\)

\(=13,25\times2+13,25\times4+13,25\times8+13,25\times6\)

\(=13,25\times\left(2+4+8+6\right)\)

\(=13,25\times20\)

\(=265\)

5 tháng 3 2023

`a)1/10 + 2/10 + 3/10 + 4/10 + 5/10 + 6/10 + 7/10 + 8/10 + 9/10`

`=(1/10+9/10)+(2/10+8/10)+(3/10+7/10)+(4/10+6/10)+5/10`

`=10/10 + 10/10+10/10+10/10+1/2`

`=1+1+1+1+1/2`

`=2+2+1/2`

`=4+1/2`

`=8/2+1/2`

`=9/2`

__

`13,25:0,5+13,25:0,25+13,25:0,125+13,25×6`

`=13,25×1/(0,5)+13,25×1/(0,25)+13,25×1/(0,125)+13,25×6`

`=13,25×(1/(0,5)+1/(0,25)+1/(0,125)+6)`

`=13,25×(2+4+8+6)`

`=13,25×20`

`=265`

`#QiN`

20 tháng 11 2021

a)=2

b)=1/3

29 tháng 3 2022

´  là sao v bn?

AH
Akai Haruma
Giáo viên
11 tháng 2

Lời giải:

$\frac{1}{4}:0,25-\frac{1}{8}:0,125:0,5-\frac{1}{10}$

$=\frac{1}{4}\times 4-\frac{1}{8}\times 8\times 2-0,1$

$=1-2-0,1=-1-0,1=-1,1$

27 tháng 12 2022

1)\(y\times7:5+4\times8=134\)

  \(\Leftrightarrow y\times7:5+32=134\)

  \(\Leftrightarrow y\times7:5=102\)

  \(\Leftrightarrow y\times7=510\)

  \(\Leftrightarrow y=72,86\)

2) \(\dfrac{1}{4}:0,25-\dfrac{1}{8}:0,125+\dfrac{1}{2}:0,5-\dfrac{1}{10}\)

\(=0,25:0,25-0,125:0,125+0,5:0,5-\dfrac{1}{10}\)

\(=1-1+1-\dfrac{1}{10}\)

\(=\dfrac{9}{10}\)

\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8}:0,125-\frac{1}{10}:0,1\)

\(=\frac{1}{2}\times2-\frac{1}{4}\times4+\frac{1}{8}\times8-\frac{1}{10}\times10\)

\(=1-1+1-1\)

\(=0\)

15 tháng 5 2022

1 - (32/5 + x - 53/10) = 0

32/5 + x - 53/10 = 1

32/5 + x = 63/10

x = 63/10 - 32/5

x = -1/10

Giải:

\(\left(1-\dfrac{3}{4}\right).\left(1-\dfrac{3}{7}\right).\left(1-\dfrac{3}{10}\right).\left(1-\dfrac{3}{13}\right).....\left(1-\dfrac{3}{97}\right).\left(1-\dfrac{3}{100}\right)\) 

\(=\dfrac{1}{4}.\dfrac{4}{7}.\dfrac{7}{10}.\dfrac{10}{13}.....\dfrac{94}{97}.\dfrac{97}{100}\) 

\(=\dfrac{1.4.7.10.....94.97}{4.7.10.13.....97.100}\) 

\(=\dfrac{1}{100}\)