Question

x+1/99+x+2/98+x+3/97+x+4/96=-4

Answer 1

(x+1)/99 + (x+2)/98 + (x+3)/97 + (x+4)/96 = -4`

$\frac{x+1}{99}+\frac{x+2}{98}+\frac{x+3}{97}+\frac{x+4}{96}+4=0$

$\frac{x+1}{99}+1+\frac{x+2}{98}+1+\frac{x+3}{97}+1+\frac{x+4}{96}+1=0$

$\frac{x+100}{99}+\frac{x+100}{98}+\frac{x+100}{97}+\frac{x+100}{96}=0$

$(x+100)(\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96})=0$

Vì $\frac{1}{99}+\frac{1}{98}+\frac{1}{97}+\frac{1}{96}\ne 0$

$\Rightarrow x+100=0$

$\Rightarrow x=0-100$

$\Rightarrow x=-100$

Vậy $x=-100$