Question

să se determine x, y si z din relatiile.​

Answer 1

Răspuns:

Explicație pas cu pas:

$\frac{x}{6}=\frac{y}{10}=\frac{z}{18}=k,~coeficient~de~proportionalitate,~deci~x=6k,~y= 10k,~z=18k.~Atunci\\x*y+y*z+x*z=2784,~inlocuim~6k*10k+10k*18k+6k*18k=2784,~60k^{2}+180k^{2}+108k^{2}=2784,~348k^{2}=2784,~k^{2}=8, deci k=2\sqrt{2} \\atunci~x=12\sqrt{2},~y=20\sqrt{2},~z=36\sqrt{2} .\\b)~\frac{x}{3}=\frac{y}{5}=\frac{z}{9}=k, ~coeficient~de~proportionalitate,~deci~x=3k,~y= 5k,~z=9k.~Atunci,\\7x^{2}+9y^{2}-2z^{2}=4032,~7*(3k)^{2}+9*(5k)^{2}-2*(9k)^{2}=4032,~\\63k^{2}+225k^{2}-162k^{2}=4032,$

$=126k^{2}=4032,~k^{2}=4032:126=32.~Deci~k=\sqrt{32}=\sqrt{16*2}=4\sqrt{2}.\\Deci~x=3*4\sqrt{2}=12\sqrt{2},~y=5*4\sqrt{2}=20\sqrt{2},~z=9*4\sqrt{2}=36\sqrt{2}$