Question

Fie numerele a,b,c,astfel încât a+b+c=12 si a^2 +b^2+c^2 =50. Sa se culculeze (-a+b+c)^2 + ( a-b+c)^2 + (a+b-c)^2

Answer 1

(a+b+c)²=a²+b²+c²+2(ab+ac+bc)⇒2(ab+ac+bc)=(a+b+c)²-(a²+b²+c²)⇒
2(ab+ac+bc)=12²-50=144-50=[tex](-a+b+c)^2 =\mathbf{a^{2} - 2 \; a \; b - 2 \; a \; c + b^{2} + 2 \; b \; c + c^{2}}\\ ( a-b+c)^2=\mathbf{a^{2} - 2 \; a \; b + 2 \; a \; c + b^{2} - 2 \; b \; c + c^{2}}\\ (a+b-c)^2=\mathbf{a^{2} + 2 \; a \; b - 2 \; a \; c + b^{2} - 2 \; b \; c + c^{2}}\\ (-a+b+c)^2 + ( a-b+c)^2 + (a+b-c)^2=\\ =\mathbf{3 \; a^{2} - 2 \; a \; b - 2 \; a \; c + 3 \; b^{2} - 2 \; b \; c + 3 \; c^{2}}=\\ =3(a^2+b^2+c^2)-2(ab+ac+bc)=\\ =3\cdot 50-94=\\ =150-94=\\ =56\\ [/tex]