Question

Radical din (a^2+6radical din 3a+28)+radical din( b^2-12b+40)+radical din(c^2-6c+25)=7 a?b?c?

Answer 1

[tex] \sqrt{a^2+6\sqrt{3}a+28}+\sqrt{b^2+12b+40}+\sqrt{c^2-6c+25}=7\\ \sqrt{a^2+6\sqrt{3}a+27+1}+\sqrt{b^2+12b+36+4}+\sqrt{c^2-6c+9+16}=7\\ \sqrt{(a+3\sqrt{3})^2+1}+\sqrt{(b+6)^2+4}+\sqrt{(c-3)^2+16}=7\\ \sqrt{(a+3\sqrt{3})^2+1} \geq1\\ \sqrt{(b+6)^2+4} \geq2\\ \sqrt{(c-3)^2+16} \geq 4[/tex]
Le adunam si vom obtine:
[tex]\sqrt{(a+3\sqrt{3})^2+1}+\sqrt{(b+6)^2+4}+\sqrt{(c-3)^2+16} \geq 7 [/tex]
Dar:
$\sqrt{(a+3\sqrt{3})^2+1}+\sqrt{(b+6)^2+4}+\sqrt{(c-3)^2+16}=7$
Din ambele rezulta:
a+3√3=0⇒a=-3√3
b+6=0⇒b=-6
c-3=0⇒c=3

Answer 2

√(a² + 6√3 ·a + 28) + √(b² - 12b + 40) + √(c² - 6c +25) = 7
√[(a + 3√3)² +1²] + √[(b - 6)² + 2²] + √[(c - 3)² + 4²] = 7
1+ 2 + 4 = 7 ⇒
⇒ a + 3√3 = 0    a = - 3√3
b - 6 = 0   b = 6
c - 3 = 0   c = 3