cho abc=1 rút gọn :A= (a/ab+a+1)+(b/bc+b+1)+(ac+c+1)
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NN
1
UN
3
13 tháng 12 2017
\(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ac+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}\)
\(=1\)
12 tháng 1 2019
Ta có:
\(N=\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ac+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a}+\frac{c}{ac+c+abc}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{c}{c\left(a+1+ab\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{1+ab+a}+\frac{1}{a+1+ab}\)
\(=\frac{a+ab+1}{ab+a+1}=1\)
Vậy N = 1
NV
1
KN
11 tháng 11 2019
\(A=\frac{a}{ab+a+3}+\frac{b}{bc+b+1}+\frac{3c}{ac+3c+3}\)
\(=\frac{a}{ab+a+abc}+\frac{b}{bc+b+1}+\frac{3bc}{b\left(ac+3c+3\right)}\)
\(=\frac{a}{a\left(b+1+bc\right)}+\frac{b}{bc+b+1}+\frac{3bc}{abc+3bc+3b}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{3bc}{3+3bc+3b}\)
\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)
\(=\frac{bc+b+1}{bc+b+1}=1\)
NH
3 tháng 10 2018
\(\frac{a}{ab}+a+1+\frac{b}{bc}+b+1+\frac{c}{ca}+c+1\)
\(=\frac{1}{b}+a+1+\frac{1}{c}+b+1+\frac{1}{c}+c+1\)
\(=3+a+b+c+\frac{1}{a}+\frac{1}{c}+\frac{1}{b}\)
\(=3+\frac{a^2+1}{a}+\frac{b^2+1}{b}+\frac{c^2+1}{c}\)
\(...............................................................\)
CT
3
KT
23 tháng 12 2017
\(\frac{a}{ab+a+2}\)+ \(\frac{b}{bc+b+1}\)+ \(\frac{2c}{ac+2c+2}\)
= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{a\left(bc+b+1\right)}\)+ \(\frac{2abc}{ab\left(ac+2c+2\right)}\)
= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{abc+ab+a}\)+ \(\frac{2abc}{a^2bc+2abc+2ab}\)
= \(\frac{a}{ab+a+2}\)+ \(\frac{ab}{ab+a+2}\)+ \(\frac{2}{ab+a+2}\) (vì abc = 2 )
= \(\frac{ab+a+2}{ab+a+2}\)= 1
M
27 tháng 12 2020
M\(=\dfrac{a}{ab+a+2}+\dfrac{b}{bc+b+1}+\dfrac{2c}{ac+2c+2}\)
\(M=\dfrac{a}{ab+a+abc}+\dfrac{b}{bc+b+1}+\dfrac{2bc}{b\left(ac+2c+2\right)}\)
M = \(\dfrac{a}{a\left(b+1+bc\right)}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{abc+2bc+2b}\)
M=\(\dfrac{1}{b+1+bc}+\dfrac{b}{b+1+bc}+\dfrac{2bc}{2+2bc+2b}\)
M = \(\dfrac{1+b}{b+1+bc}+\dfrac{2bc}{2\left(1+bc+b\right)}\)
M = \(\dfrac{1+b}{b+1+bc}+\dfrac{bc}{b+1+bc}=\dfrac{1+b+bc}{b+1+bc}=1\)
