Question

\(\frac{x+1}{19}\)+   \(\frac{x+2}{18}\)=   \(\frac{x+3}{17}\)+    \(\frac{x+4}{16}\)

tìm x

Answer 1

\(\frac{x+1}{19}+\frac{x+2}{18}=\frac{x+3}{17}+\frac{x+4}{16}\)

\(\Rightarrow\left(\frac{x+1}{19}+1\right)+\left(\frac{x+2}{18}+1\right)=\left(\frac{x+3}{17}+1\right)+\left(\frac{x+4}{16}+1\right)\)

\(\Rightarrow\frac{x+20}{19}+\frac{x+20}{18}=\frac{x+20}{17}+\frac{x+20}{16}\)

\(\Rightarrow\frac{x+20}{19}+\frac{x+20}{18}-\frac{x+20}{17}-\frac{x+20}{16}=0\)

\(\Rightarrow\left(x+20\right).\left(\frac{1}{19}+\frac{1}{18}-\frac{1}{17}-\frac{1}{16}\right)=0\)

Vì : \(\frac{1}{19}+\frac{1}{18}-\frac{1}{17}-\frac{1}{16}\ne0\)nên \(x+20=0\)

Suy ra : \(x=0-20=-20\)

Vậy \(x=-20\)

Answer 2

-20 nha 

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