`x xx6/9=13/18+8/6`

`x xx6/9=13/18+24/18`

`x xx6/9=37/18`

`x=37/18:6/9`

`x=37/18xx9/6`

`x=37/12`

a: =y x15/5=3y

b: =(4/6+14/6)+(7/13+19/13)+(17/9-8/9)

=3+2+1

=6

= 6

mk chỉnh lại đề nha

     \(\frac{x-6}{2012}+\frac{x-8}{2010}=\frac{x-2000}{18}+\frac{x-2005}{13}\)

\(\Leftrightarrow\)\(\frac{x-6}{2012}-1+\frac{x-8}{2010}-1=\frac{x-2000}{18}-1+\frac{x-2005}{13}-1\)

\(\Leftrightarrow\)\(\frac{x-2018}{2012}+\frac{x-2018}{2010}=\frac{x-2018}{18}+\frac{x-2018}{13}\)

\(\Leftrightarrow\)\(\left(x-2018\right)\left(\frac{1}{2012}+\frac{1}{2010}-\frac{1}{18}-\frac{1}{13}\right)=0\)

\(\Leftrightarrow\)\(x-2018=0\)   (1/2012 + 1/2010 - 1/18 - 1/13 # 0)

\(\Leftrightarrow\)\(x=2018\)

Vậy...

\(a.\)

\(-18-x=-8+\left(13\right)\)

\(\Rightarrow-18-x=5\)

\(\Rightarrow x=-18-5=-23\)

\(b.\)

\(x-43=\left(57-x\right)-50\)

\(\Rightarrow x-43=57-x-50\)

\(\Rightarrow x+x=43+57-50\)

\(\Rightarrow2x=50\)

\(\Rightarrow x=25\)

\(c.\)

\(-4\left(2x+9\right)-\left(-8x+3\right)-\left(x+13\right)=0\)

\(\Rightarrow-8x-36+8x-3-x-13=0\)

\(\Rightarrow-x-52=0\)

\(\Rightarrow-x=52\)

\(\Rightarrow x=-52\)

\(\dfrac{1}{7}-\left(-\dfrac{3}{5}\right)+\dfrac{6}{7}-\left|-\dfrac{1}{5}\right|=1+\dfrac{3}{5}+\dfrac{1}{5}=\dfrac{9}{5}\\ \dfrac{8}{13}\cdot\dfrac{11}{18}+\dfrac{7}{18}\cdot\dfrac{8}{13}-1\dfrac{8}{13}=\dfrac{8}{13}\left(\dfrac{11}{18}+\dfrac{7}{18}+1\right)=\dfrac{8}{13}\cdot2=\dfrac{16}{13}\)

\(\dfrac{1}{7}-\left(-\dfrac{3}{5}\right)+\dfrac{6}{7}-\left|-\dfrac{1}{5}\right|\)

\(=\dfrac{1}{7}+\dfrac{3}{5}+\dfrac{6}{7}-\dfrac{1}{5}\)

\(=1+\dfrac{2}{5}=\dfrac{7}{5}\)

8 - 6 + 2 = 4

12 - 5 + 8 = 15

Suy ra :13 - 10 + x = 18

13 - 10 = 18 - x

        3 = 18 - x

        x = 18 - 3

        x = 15