Question

Cho biểu thức M=(x-a)(x-b)+(x-b)(x-c)+(x-c)(x-a)+x^2.
Tính M biết x=(1/2)a+(1/2)b+(1/2)c

Answer 1

\(M=\left(x-a\right)\left(x-b\right)+\left(x-b\right)\left(x-c\right)+\left(x-c\right)\left(x-a\right)+x^2\)

\(=x^2-bx-ax+ab+x^2-cx-bx+bc+x^2-ax-cx+ac+x^2\)

\(=4x^2-\left(bx+ax+cx+bx+ax+cx\right)+\left(ab+bc+ac\right)\)

\(=4x^2-2x\left(a+b+c\right)+\left(ab+bc+ac\right)\)

Thay \(x=\dfrac{1}{2}a+\dfrac{1}{2}b+\dfrac{1}{2}c\) vào M ta được:

\(M=4.\dfrac{1}{4}\left(a+b+c\right)^2-2.\dfrac{1}{2}\left(a+b+c\right)^2+ab+bc+ac=\left(a+b+c\right)^2-\left(a+b+c\right)^2+ab+bc+ac=ab+bc+ac\)