Question

folosind sirul de rapoarte egale :a,n=b/m=c/p=d/q=7; m,n,p,q diferit de zero
calculati:7^-a+b+c+d/n+m+p+q si (a+b/n+m)+(b+c/m+p)+(c+d/p+q)+(d+a/q+n)-17
la prima =354 si a doua 39
nu sunt sigur de rezolvare ,va rog frumos

Answer 1

a/n=b/m=c/p=d/q=7; m,n,p,q ≠0

=> a=7n; b=7m; c=7p; d=7q

=> a+b+c+d=7n+7m+7p+7q=7(n+m+p+q)

a+b=7n+7m=7(m+n); b+c=7(m+p); c+d=7(p+q); d+a=7(q+n)

7³-(a+b+c+d)/(n+m+p+q)=7³- 7(n+m+p+q)/(n+m+p+q)=343 -7=336

(a+b)/(n+m)+(b+c)/(m+p)+(c+d)/(p+q)+(d+a)/(q+n)-17

=7(n+m)/(n+m)+7(m+p)/(m+p)+7(p+q)/(p+q)+7(q+n)/(q+n)-17

=7+7+7+7-17

=28-17

=11