\(\frac{x-39}{19}+\frac{x-49}{29}=\frac{x-59}{39}+\frac{x-69}{49}\)
Công mỗi vế cho 2 ta được:
\(\frac{x-39}{19}+\frac{x-49}{29}+2=\frac{x-59}{39}+\frac{x-69}{49}+2\)
\(\frac{x-39}{19}+1+\frac{x-49}{29}+1=\frac{x-59}{39}+1+\frac{x-69}{49}+1\)
\(\frac{x-39}{19}+\frac{19}{19}+\frac{x-49}{29}+\frac{29}{29}=\frac{x-59}{39}+\frac{39}{39}+\frac{x-69}{49}+\frac{49}{49}\)
\(\frac{x-20}{19}+\frac{x-20}{29}=\frac{x-20}{39}+\frac{x-20}{49}\)
\(\frac{x-20}{19}+\frac{x-20}{29}-\frac{x-20}{39}-\frac{x-20}{49}=0\)
\(\left(x-20\right).\frac{1}{19}+\left(x-20\right).\frac{1}{29}-\left(x-20\right).\frac{1}{39}-\left(x-20\right).\frac{1}{49}=0\)
Đặt thừa số chung (x-20) ra ngoài
\(\left(x-20\right).\left(\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\right)=0\)
\(\text{Vì }\frac{1}{19}+\frac{1}{29}-\frac{1}{39}-\frac{1}{49}\ne0\text{ nên }x-20=0\Rightarrow x=20\)
