Question

Daca a+b=3radical 3 si b+c=-2radical din 3 calculati 2a+5b+3c

Answer 1

$\left \{ {{a+b=3\sqrt3 \ |\cdot2 } \atop {b+c=-2\sqrt3 \ |\cdot3}} \right. \Leftrightarrow \left \{ {{2a+2b=6\sqrt3} \atop {3b+3c=-6\sqrt3}} \right.$
Adunam cele doua ecuatii si avem:
$\left \{ {{2a+2b=6\sqrt3} \atop {3b+3c=-6\sqrt3}} \right.\\ ------\\ 2a+2b+3b+3c=6\sqrt3-6\sqrt3\\ 2a+5b+3c=0$

Answer 2

a + b = 3√3  si b + c = - 2√3

2 a + 5 b + 3 c = 0
= 2 a + 2 b + 3 b + 3 c = → l-am descompus pe 5b= 2 b + 3 b
= 2 × ( a + b ) + 3 × ( b + c) = → am dat factor comun pe 2, respectiv pe 3
= 2 × 3√3 + 3× ( - 2 √3) =
= 6√3 - 6√3=
= 0