Question

Numerele rationale poz9itive a si b sunt direct proportionale cu nr. 2,3,4 stiind ca 6a²+b²-2c²=9 determinati nr.a b si c

Answer 1

a/2=b/3=c/4=k
⇒a/2=k⇒a=2k;
⇒b/3=k⇒b=3k;
⇒c/4=k⇒c=4k;
⇒6a²+b²-2c²=9
⇒6·2k²+3k²-2·4k²=9
⇒6·4k+9k-2·16k=9
⇒24k+9k-32k=9
⇒33k-32k=9
⇒k=9;
⇒a=2·9⇒a=18;
⇒b=3·9⇒b=27;
⇒c=4·9⇒c=36;

Answer 2

$a,b,c (d.p.) cu 2,3,4=\ \textgreater \ \frac{a}{2} = \frac{b}{3} = \frac{c}{4} =k \\ \frac{a}{2} =k=\ \textgreater \ a=2k \\ \frac{b}{3} =k=\ \textgreater \ b=3k \\ \frac{c}{4} =k=\ \textgreater \ c=4k \\ 6 a^{2} + b^{2} -2 c^{2} =9 \\ 6 (2k)^{2} + (3k)^{2} -2 (4k)^{2} =9 \\ 6*4 k^{2} +9 k^{2} -2*16 k^{2} =9 \\ k^{2} (24+9-32)=9 \\ k^{2} (33-32)=9 \\ k^{2} *1=9 \\ k^{2} =9 \\ 3^{2} =9=\ \textgreater \ k=3 \\ a=2*3=6 \\ b=3*3=9 \\ c=4*3=12$