Question

Se da numarul n=$1989^{2}$-1989-1988.Aflati x din proportia:$\frac{1988}{4}$=$\frac{497}{n}$.

Answer 1

[tex]n=1989^2-1989-1988\\ n=1989(1989-1)-1988\\ n=1989\cdot1988-1988\\ n=1988(1989-1)\\ n=1988^2\\ \frac{x}{4}=\frac{497}{n}\\ \frac{x}{4}=\frac{497}{1988^2}\\ 1988^2\cdot x=1988\\ x=\frac{1}{1988}\\[/tex]