Đặt A = \(\frac{a}{ab+a+1}\)\(+\)\(\frac{b}{bc+b+1}\)\(+\)\(\frac{c}{ac+c+1}\)
= \(\frac{a}{ab+a+1}\)\(+\)\(\frac{ab}{a\left(bc+b+1\right)}\)\(+\)\(\frac{abc}{ab\left(ac+c+1\right)}\)
= \(\frac{a}{ab+a+1}\)\(+\)\(\frac{ab}{abc+ab+a}\)\(+\)\(\frac{abc}{abc.a+abc+ab}\)
Vì abc = 1 nên:
A = \(\frac{a}{ab+a+1}\)\(+\)\(\frac{ab}{ab+a+1}\)\(+\)\(\frac{1}{ab+a+1}\)
= \(\frac{a+ab+1}{ab+a+1}\)= 1
Cho a.b.c = 1 cmr: a/(ab + a + 1) + b/(bc + b + 1) + c/(ac + c + 1)?
Cho a.b.c=1
CMR \(\frac{1}{1+ab}+a+\frac{1}{1+bc}+b+\frac{1}{1+ac}+c=1\)
cho a.b.c=1
CM a/ab+a+1 + b/bc+b+1 + c/ac+c+1 = 1
Cho a.b.c=1
CMR \(\frac{1}{1+ab}\)+ a + \(\frac{1}{1+bc}\)+ b + \(\frac{1}{1+ac}\)+c =1
Cho A=\(\frac{4bc-a^2}{bc+2a^2}\),B=\(\frac{4ca-b^2}{ac+2b^2}\),C=\(\frac{4ab-c^2}{ab+2c^2}\).CMR nếu a+b+c=0 thì A.B.C=1
Cho a, b, c thuộc R sao cho a.b.c = 2008.
CMR : \(\frac{2008a}{ab+2008a+2008}+\frac{b}{bc+b+2008}+\frac{c}{ca+c+1}=1\)
CMR
nếu \(a^2+b^2+c^2=ab+ac+bc\)thì a=b=c
nếu\(a^2+b^2+c^{^2}+3=2\left(a.b.c\right)\)thì a=b=c=1
Cho \(a.b.c\ne1;-1\)và \(\frac{ab+1}{b}=\frac{bc+1}{c}=\frac{ca+1}{a}\).
Cmr a=b=c
Cho a,b,c là độ dài 3 cạnh của 1 tam giác có chu vi bằng 1. C/m rằng: ab+ac+bc>a.b.c
