=[(3x1-2-4)^2+(3x2-2-4)^2+(3x3-2-4)^2+(3x4-2-4)^2+(3x5-2-4)^2]/5=9[(x1-2)^2+(x2-2)^2+(x3-2)^2+(x4-2)^2+(x5-2)^2]/5这步怎么算的

[(3x1-2-4)^2+(3x2-2-4)^2+(3x3-2-4)^2+(3x4-2-4)^2+(3x5-2-4)^2]/5=[(3x1-6)^2+(3x2-6)^2+(3x3-6)^2+(3x4-6)^2+(3x5-6)^2]/5=[(3x1-3*2)^2+(3x2-3*2)^2+(3x3-3*2)^2+(3x4-3*2)^2+(3x5-3*2)^2]/5={[3(X1-2)]^2+[3（x2-2)]^2+[3(x3-2)]^2+[3(x4-2)]^2+[3(x5-2)]^2}/5=9[(x1-2)^2+(x2-2)^2+(x3-2)^2+(x4-2)^2+(x5-2)^2]/5
简单说就是把-2-4变成-6,然后变成-（2*3）,再把3提出来,平方得9,后面的就是（X-2）了.