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Anusca130
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Răspuns :

(x1-x2)²+(x3-x2)²+(x1-x2)²=2(x1²+x2²+x3²)-2(x1x2+x2x3+x3x1)

dar
(x1²+x2²+x3²) = (x1+x2+x3)²-2(x1x2+x2x3+x3x1)
atunci
(x1-x2)²+(x3-x2)²+(x1-x2)²=
2((x1+x2+x3)²-2(x1x2+x2x3+x3x1))-2(x1x2+x2x3+x3x1)=
2((x1+x2+x3)²-6(x1x2+x2x3+x3x1)= 2*(-3/1)²-6*(-5/1)=2*9+6*5=18+30=48