Question

Cho a + b + c = 4m. Chứng minh rằng:

\(\left(\dfrac{a+b-c}{2}\right)^2+\left(\dfrac{a-b+c}{2}\right)^2+\left(\dfrac{-a+b+c}{2}\right)^2=a^2+b^2+c^2-4m^2\)

Answer 1

\(\left(\dfrac{a+b-c}{2}\right)^2+\left(\dfrac{a-b+c}{2}\right)^2+\left(\dfrac{-a+b+c}{2}\right)^2\)

\(=\left(\dfrac{4m-2c}{2}\right)^2+\left(\dfrac{4m-2b}{2}\right)^2+\left(\dfrac{4m-2a}{2}\right)^2\)

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

\(=4m^2-4mc+c^2+4m^2-4mb+b^2+4m^2-4ma+a^2\)

\(=a^2+b^2+c^2+12m^2-4m\left(a+b+c\right)\)

\(=a^2+b^2+c^2+12m^2-4m\cdot4m\)

\(=a^2+b^2+c^2+12m^2-16m^2\)

\(=a^2+b^2+c^2-4m^2\)