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Có : a/ab+a+1 = a/ab+a+abc = 1/b+1+bc = 1/bc+b+1
c/ca+c+1 = bc/abc+bc+b = b/1+bc+b = b/bc+b+1
=> A = 1+bc+b/bc+b+1 = 1
Tk mk nha
BÀI 1:
\(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{a\left(bc+b+1\right)}+\frac{abc}{ab\left(ca+c+1\right)}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{abc+ab+a} +\frac{abc}{a^2bc+abc+ab}\)
\(=\frac{a}{ab+a+1}+\frac{ab}{ab+a+1}+\frac{1}{ab+a+1}\) (thay abc = 1)
\(=\frac{a+ab+1}{a+ab+1}=1\)
a, b, c khác 0
Áp dụng tính chất dãy tỉ số bằng nhau. Ta có:
\(\frac{a+3b-c}{c}\)=\(\frac{-a+b+3c}{a}\) =\(\frac{c-b+3a}{b}\)=\(\frac{a+3b-c-a+b+3c+c-b+3a}{a+b+c}=\frac{3a+3b+3c}{a+b+c}=3\)
=> \(\frac{a+3b-c}{c}=3\Rightarrow\frac{a+3b}{c}-\frac{c}{c}=3\Rightarrow\frac{a+3b}{c}=4\)
\(\frac{-a+b+3c}{a}=3\Rightarrow-1+\frac{b+3c}{a}=3\Rightarrow\frac{b+3c}{a}=4\)
\(\frac{c-b+3a}{b}=3\Rightarrow\frac{c+3a}{b}-\frac{b}{b}=3\Rightarrow\frac{c+3a}{b}=4\)
=> P =\(\left(3+\frac{a}{b}\right).\left(3+\frac{b}{c}\right).\left(3+\frac{c}{a}\right)=\frac{3b+a}{b}.\frac{3c+b}{c}.\frac{3a+c}{a}\)
= \(\frac{a+3b}{c}.\frac{b+3c}{a}.\frac{c+3a}{b}=4.4.4=64\)
Đặt \(\frac{a}{2016}=\frac{b}{2017}=\frac{c}{2018}=k\Rightarrow a=2016k;b=2017k;c=2018k\)
\(\frac{a}{24}+\frac{b}{4}=\frac{c}{2018}\)
\(\Rightarrow\frac{2016k}{24}+\frac{2017k}{4}=\frac{2018k}{2018}\)
\(\Rightarrow84k+504,25k=k\)
\(\Rightarrow k=0\)
\(\Rightarrow a,b,c=0\)
Do \(\frac{a}{b+c}=\frac{b}{c+a}=\frac{c}{a+b}\\ \)
=> \(\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b}{c}=\frac{a+b+c+a+b+c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
=> \(\frac{b+c}{a}+\frac{c+a}{b}+\frac{a+b}{c}=2+2+2=6\)
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow\frac{a+b}{ab}=\frac{b+c}{bc}=\frac{c+a}{ca}=\frac{1}{a}+\frac{1}{b}=\frac{1}{b}+\frac{1}{c}=\frac{1}{c}+\frac{1}{a}\Rightarrow a=b=c\Rightarrow M=1\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có
\(\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\Leftrightarrow a^3=c^3=b^3\)
Ta có : \(a^3=b^3=c^3=abc\)
\(\frac{a^3}{abc}=\frac{abc}{abc}=1\Leftrightarrow\frac{a^3+b^3+c^3}{3abc}=\frac{3abc}{3abc}=1\)
Vậy \(P=1\)