\(\frac{5}{6}\cdot x+\frac{2}{3}\cdot\left(\frac{3}{5}\cdot x-\frac{6}{5}\right)=\frac{2}{9}-\frac{1}{2}\)

\(\Rightarrow\frac{5}{6}\cdot x+\frac{2}{3}\cdot\frac{3}{5}\cdot x-\frac{2}{3}\cdot\frac{6}{5}=-\frac{5}{18}\)

\(\Rightarrow\frac{5}{6}\cdot x+\frac{2}{5}\cdot x-\frac{4}{5}=-\frac{5}{18}\)

\(\Rightarrow x\cdot\left(\frac{5}{6}+\frac{2}{5}\right)=-\frac{5}{18}+\frac{4}{5}\)

\(\Rightarrow x\cdot\frac{37}{30}=\frac{47}{90}\)

\(\Rightarrow x=\frac{47}{90}\div\frac{37}{30}\)

\(\Rightarrow x=\frac{47}{111}\)

Vậy \(x=\frac{47}{111}\)

D= 1-2 + 3-4 + 5-6 +... - 2016 + 2017

D= (1-2) + (3-4) + (5-6) +...+ (2015 - 2016) + 2017

D= (-1) x [(2016-1+1):2] + 2017

D= -1 x 1008 + 2017

D= -1008 + 2017

D= 1009

Đáp án:-2017

Giải thích các bước giải:

A=-1+2+3-4-5+6+7-8-...+2015-2016-2017

A=(-1+2+3-4)+(-5+6+7-8)-...+(-2013+2014+2015-2016)-2017

A=       0         +      0       +....       +            0                           -2017

A=-2017

3) \(\left(x-3\right)\left(x+3\right)\left(x^2+9\right)-\left(x^2-2\right)\left(x^2+2\right)\)

\(=\left(x^2-9\right)\left(x^2+9\right)-\left(x^4-4\right)\)

\(=\left(x^4-81\right)-\left(x^4-4\right)\)

\(=x^4-81-x^4+4\)

=-77 =>đpcm

4)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)

\(=\left[\left(3x+1\right)-\left(3x+5\right)\right]^2\)

\(=\left(3x+1-3x-5\right)^2\)

=(-4)²

=16 => đpcm

1)\(\left(x-2\right)^2-\left(x-3\right)\left(x-1\right)=\left(x^2-4x+4\right)-\left(x^2-4x+3\right)=1\)

=>đpcm

2)\(\left(x-1\right)^3-\left(x+1\right)^3+6\left(x+1\right)\left(x-1\right)\)

\(=\left(x-1-x-1\right)\left[\left(x-1\right)^2+\left(x-1\right)\left(x+1\right)+\left(x+1\right)^2\right]+6\left(x^2-1\right)\)

\(=\left(-2\right)\left(x^2-2x+1+x^2-1+x^2+2x+1\right)+6x^2-6\)

\(=\left(-2\right)\left(3x^2+1\right)+6x^2-6=-6x^2-2+6x^2-6=-8\) => đpcm

a; \(\dfrac{2}{3}\)\(x\) - \(\dfrac{3}{2}\)\(x\) = \(\dfrac{5}{12}\)

    (\(\dfrac{2}{3}\) - \(\dfrac{3}{2}\))\(x\) = \(\dfrac{5}{12}\)

     - \(\dfrac{5}{6}\)\(x\) = \(\dfrac{5}{12}\)

     \(x\)   = \(\dfrac{5}{12}\) : (- \(\dfrac{5}{6}\))

     \(x=\) - \(\dfrac{1}{2}\)

Vậy \(x=-\dfrac{1}{2}\) 

b; \(\dfrac{2}{5}\) + \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\)

           \(\dfrac{3}{5}\).(3\(x\) - 3,7) = \(\dfrac{-53}{10}\) - \(\dfrac{2}{5}\)

          \(\dfrac{3}{5}\).(3\(x\) - 3,7) = - \(\dfrac{57}{10}\)

         3\(x\) - 3,7 = - \(\dfrac{57}{10}\) : \(\dfrac{3}{5}\)

         3\(x\)   - 3,7 = - \(\dfrac{19}{2}\)

         3\(x\)         = - \(\dfrac{19}{2}\) + 3,7

          3\(x\)        = - \(\dfrac{29}{5}\)

           \(x\)         = - \(\dfrac{29}{5}\) : 3

           \(x\)        = - \(\dfrac{29}{15}\)

Vậy \(x\) \(\in\) - \(\dfrac{29}{15}\) 

c; \(\dfrac{7}{9}\) : (2 + \(\dfrac{3}{4}\)\(x\)) + \(\dfrac{5}{9}\) = \(\dfrac{23}{27}\)

    \(\dfrac{7}{9}\): (2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{23}{27}\) - \(\dfrac{5}{9}\)

     \(\dfrac{7}{9}\):(2 + \(\dfrac{3}{4}\)\(x\)) = \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) = \(\dfrac{7}{9}\) : \(\dfrac{8}{27}\)

      2 + \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{21}{8}\)

             \(\dfrac{3}{4}x\) = \(\dfrac{21}{8}\) - 2

             \(\dfrac{3}{4}\)\(x\) =  \(\dfrac{5}{8}\)

               \(x\) = \(\dfrac{5}{8}\) : \(\dfrac{3}{4}\)

              \(x\) =  \(\dfrac{5}{6}\)

Vậy \(x=\dfrac{5}{6}\)

= 51/20

----------------------------------

= 3/14

------------------------------------

= 4/7 - 2/7 

= 2/7

--------------------------------------

= 17/45

---------------------------------------

= 2/3 

--------------------------------------

= 2 x 1/4 

= 1/2

`a, 2/3 +3/4 = (8+9)/12=17/12.`

`1 1/3+4/5 = 4/3 + 4/5 = (20+12)/15=32/15`.

`=> x=2.`

`b, 5/6-1/4=(20-6)/24=7/12`.

`2 1/3-2/5= 7/3-2/5 = (35-6)/15=29/15`.

`=> x=1`.

a) \(\dfrac{2}{3}+\dfrac{3}{4}=\dfrac{8+9}{12}=\dfrac{17}{12}\)

-> 1 1/3 + 4/5 = 4/3 + 4/5 =  20+12/15 = 32/15

vậy x có thể = 14/14 = 1 (x thuộc N)