-18-x=-8+(-13)
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`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
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
-18-x=-21
-x=-18-21
-x=-39
x=39
-18-x=-21
x=-18-(-21)
x=3