Cho ( a+b+c)*(a*b+b*c+c*a)=2017
a*b*c=2017
Tính P=(b^2*c+2017)*(c^2*a+2017)*(a^2*b+2017)
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P=\(\frac{2017a}{ab+2017a+2017}\)+\(\frac{b}{bc+b+2017}\)+\(\frac{c}{ac+c+1}\)chứ bạn
Với abc=2017 ta có:
P=\(\frac{a^2bc}{ab+a^2bc+abc}\)+\(\frac{b}{bc +b+abc}\)+\(\frac{c}{ac+c+1}\)
P=\(\frac{ac}{ac+c+1}\)+\(\frac{1}{ac+c+1}\)+\(\frac{c}{ac+c+1}\)
P=1
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