Question

Tìm x, biết:

Câu 1: (x+2)/18+(x+2)/19+(x+2)/20=(x+2)/21+(x+2)/22

Câu 2: 1+3+3^2+...+3^x=40

Answer 1

Câu 1 : \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}=\frac{x+2}{21}+\frac{x+2}{22}\)
      => \(\frac{x+2}{18}+\frac{x+2}{19}+\frac{x+2}{20}-\frac{x+2}{21}-\frac{x+2}{22}=0\)
      =>  x+2 . ( \(\frac{1}{18}+\frac{1}{19}+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\)) = 0
     Vì  \(\frac{1}{18}+\frac{1}{19}_{ }+\frac{1}{20}-\frac{1}{21}-\frac{1}{22}\ne0\)nên  x+2=0 
                                                                              => x= 0 - 2 = -2
                       Vậy x = -2