Question

cho a,b,c thỏa mãn abc=2000 Tính

P=\(\frac{200a}{ab+2000a+2000}\)+\(\frac{b}{bc+b+2000}\)+\(\frac{c}{ac+c+1}\)

 

Answer 1

\(P=\frac{2000a}{ab+2000a+2000}+\frac{b}{bc+b+2000}+\frac{c}{ac+c+1}\)

\(=\frac{a\cdot abc}{ab+abc\cdot a+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(=\frac{a^2bc}{ab+a^2bc+abc}+\frac{b}{bc+b+abc}+\frac{c}{ac+c+1}\)

\(=\frac{a^2bc}{ab\left(ac+c+1\right)}+\frac{b}{b\left(ac+c+1\right)}+\frac{c}{ac+c+1}\)

\(=\frac{ac}{ac+c+1}+\frac{1}{ac+c+1}+\frac{c}{ac+c+1}\)

\(=\frac{ac+c+1}{ac+c+1}=1\)

Answer 2

Đặt bt là P ta có

P = 2000a/(ab + 2000a + 2000) + b/(bc + b + 2000) + c/(ac + c + 1) 
= 2000ac/(abc + 2000ac + 2000c) + b/(bc + b + abc) + c/(ac + c + 1) 
= 2000ac/(2000 + 2000ac + 2000c) + 1/(1 + c + ac) + c/(ac + c + 1) 
= ac/(1 + ac + c) + 1/(ac + c + 1) + c/(ac + c + 1) 
= (ac + c + 1)/(ac + c + 1) = 1