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

`=(1/3+2/3)+(1/4+3/4)+(2/5+3/5)`

`=1+1+1`

`=3`

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đáp án:

9/27: -2/9 +1/2-3/2= -3/2+ -2/2=-5/2

 K HỘ MK NHA

y x 2/3 + 1/3 = 8/9 : 2

Y x 2/3 +1/3 = 4/9

Y x 1/3 = 4/9 - 1/3

y x 1/3 = 1/9

Y = 1/9 : 1/3

y = 1/3

y x 2/3 + 1/3 = 8/9 : 2

y x 2/3 + 1/3 = 4/9

                y   = (4/9 - 1/3) : 2/3

                y   = 1/6

( dùng thứ lại biểu thức cộng trừ trước nhân chia sau)

y x 2/3 + 1/3 = 8/9 : 2

Y x 2/3 +1/3 = 4/9

Y x 1/3 = 4/9 - 1/3

y x 1/3 = 1/9

Y = 1/9 : 1/3

y = 1/3

1/2* x+2/3=9/2

1/2 * x = 9/2 - 2/3 

1/2 * x= 23/6

x= 23/6 : 1/2

x= 23/6 x 2= 23/3

___

1/2*x-1/3=2/3

1/2*x = 2/3 + 1/3

1/2 * x= 1

x= 1: 1/2 

x= 2

____

1/4+3/4:x=3

3/4 : x = 3 - 1/4

3/4 : x= 11/4

x= 11/4 : 3/4

x= 11/3

\(\dfrac{1}{2}\)\(\times\)\(x\) + \(\dfrac{2}{3}\) = \(\dfrac{9}{2}\)

\(\dfrac{1}{2}\)\(\times\)\(x\)        = \(\dfrac{9}{2}\) - \(\dfrac{2}{3}\)

\(\dfrac{1}{2}\)\(\times\)\(x\)       = \(\dfrac{23}{6}\)

      \(x\)       = \(\dfrac{23}{6}\):\(\dfrac{1}{2}\)

      \(x\)      = \(\dfrac{23}{3}\) 

\(\dfrac{1}{2}\)\(\times\)\(x\) - \(\dfrac{1}{3}\) = \(\dfrac{2}{3}\) 

\(\dfrac{1}{2}\)\(\times\)\(x\)       = \(\dfrac{2}{3}\) + \(\dfrac{1}{3}\)

\(\dfrac{1}{2}\times\)\(x\)      =  1

     \(x\)       = 1 : \(\dfrac{1}{2}\)

   \(x\)         = 2

\(\dfrac{1}{4}\) + \(\dfrac{3}{4}\): \(x\) = 3

          \(\dfrac{3}{4}\): \(x\) = 3 - \(\dfrac{1}{4}\) 

          \(\dfrac{3}{4}\):\(x\) = \(\dfrac{11}{4}\)

              \(x\) = \(\dfrac{3}{4}\): \(\dfrac{11}{4}\)

             \(x\) = \(\dfrac{3}{11}\)

` @ L I N H `

1/2* x+2/3=9/2

1/2 * x = 9/2 - 2/3 

1/2 * x= 23/6

x= 23/6 : 1/2

x= 23/6 x 2= 23/3

___

1/2*x-1/3=2/3

1/2*x = 2/3 + 1/3

1/2 * x= 1

x= 1: 1/2 

x= 2

____

1/4+3/4:x=3

3/4 : x = 3 - 1/4

3/4 : x= 11/4

x= 11/4 : 3/4

x= 11/3

\(\dfrac{1}{2}\) \(\times\) ( \(x\) - \(\dfrac{2}{3}\)) - \(\dfrac{1}{3}\) \(\times\) ( 2\(x\)  - 3) = \(x\)

\(\dfrac{1}{2}\)  \(\times\) \(\dfrac{3x-2}{3}\) -   \(\dfrac{2x-3}{3}\) = \(x\)

\(\dfrac{3x-2}{6}\) - \(\dfrac{4x-6}{6}\) = \(\dfrac{6x}{6}\)

3\(x-2-4x\) + 6 = 6\(x\) 

-\(x\) + 4 - 6\(x\) = 0

7\(x\)  = 4

    \(x\) =  \(\dfrac{4}{7}\) 

`#3107.101107`

\(\dfrac{1}{5}+\dfrac{2}{3}\times x=\dfrac{1}{3}\)

\(\dfrac{2}{3}\times x=\dfrac{1}{3}-\dfrac{1}{5}\)

\(\dfrac{2}{3}\times x=\dfrac{2}{15}\)

\(x=\dfrac{2}{15}\div\dfrac{2}{3}\)

\(x=\dfrac{1}{5}\)

Vậy, \(x=\dfrac{1}{5}.\)

1/5 + 2/3 . x = 1/3

2/3 . x = 1/3 - 1/5

2/3 . x = 2/15

x = 2/15 : 2/3

x = 1/5

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}=1-\dfrac{1}{101}=\dfrac{100}{101}\)

\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\\ =1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\\ =1-\dfrac{1}{101}=\dfrac{100}{101}\)

a: Ta có: \(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{99\cdot101}\)

\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)

\(=1-\dfrac{1}{101}=\dfrac{100}{101}\)

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