Question

Fie a,b,c trei numere naturale direct proportionale cu 3,5 respectiv 7.
a) demonstrați ca a+c=2b
b) aflați numerele , știind ca a^2-b^2+c^2=297

Answer 1

[tex](a,b,c)d.p(3,5,7)=\ \textgreater \ \frac{a}{3} = \frac{b}{5} = \frac{c}{7} =k \\ =\ \textgreater \ a=3k \\ =\ \textgreater \ b=5k \\ =\ \textgreater \ c=7k \\ a+c=2b=\ \textgreater \ 3k+7k=2*5k=\ \textgreater \ 10k=10k=\ \textgreater \ a+c=2b [/tex] [tex]a^2-b^2+c^2=297=\ \textgreater \ 3k^2-5k2+7k^2=297 \\ 9kk -25kk+49kk=297=\ \textgreater \ 33kk=297 \\ =\ \textgreater \ kk=9=\ \textgreater \ k=3 \\ a=3k=3*3=9 =\ \textgreater \ a=9 \\ b=5k=5*3=\ \textgreater \ b=15 \\ c=7k=7*3=\ \textgreater \ c=21 [/tex]