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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)[(a - b)2 + (b - c)2 + (c - a)2] = 0
=> a + b + c = 0
Hoặc (a - b)2 + (b - c)2 + (c - a)2 = 0
Mặt khác : (a - b)2 \(\ge\)0
(b - c)2 \(\ge\)0
(c - a)2 \(\ge\)0
=> (a - b)2 = 0 => a - b = 0 => a = b
(b - c)2 = 0 b - c = 0 b = c
(c - a)2 = 0 c - a = 0 c = a
=> a = b = c
Ta có :
\(B=\left(1+\frac{a}{b}\right).\left(1+\frac{b}{c}\right).\left(1+\frac{c}{a}\right)\)
\(B=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\) (quy đồng cho các hạng tử cùng mẫu rồi cộng)
\(B=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{bca}\)
Mà a = b = c
Thay vào , ta lại có :
\(B=\frac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a^3}=\frac{2a.2a.2a}{a^3}=\frac{8.a^3}{a^3}=8\)
=> B = 8
\(\frac{1}{c}=\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}\right)\)
\(\frac{1}{c}=\frac{1}{2}\left(\frac{a+b}{ab}\right)\)
\(2ab=c\left(a+b\right)\)
\(ab+ab=ca+bc\)
\(ab-cb=ac-ab\)
\(b\left(a-c\right)=a\left(c-b\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{c-b}\)
Ta có; \(\frac{a+b+c}{c}=\frac{a+b}{c}+1;\frac{b+c-a}{a}=\frac{b+c}{a}-1;\frac{c+a-b}{b}=\frac{c+a}{b}-1\)\(\Rightarrow\frac{a+b}{c}+1=\frac{b+c}{a}-1=\frac{c+a}{b}-1\)
\(\Rightarrow\frac{a+b-2c}{c}=\frac{b+c}{a}=\frac{c+a}{b}\)
\(\Rightarrow\frac{a}{c}+\frac{b}{c}-2=\frac{c}{b}+\frac{a}{b}=\frac{b}{a}+\frac{c}{a}\)
Ta có; a+b+cc =a+bc +1;b+c−aa =b+ca −1;c+a−bb =c+ab −1⇒a+bc +1=b+ca −1=c+ab −1
⇒a+b−2cc =b+ca =c+ab
⇒ac +bc −2=cb +ab =ba +ca
Từ \(gt\Leftrightarrow\frac{1}{c}=\frac{1}{2}.\frac{a+b}{ab}\)
\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\Leftrightarrow c\left(a+b\right)=2ab\Leftrightarrow ac+bc=ab+ab\)
\(\Leftrightarrow ac-ab=ab-bc\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{b-c}\) (đpcm)
Từ \(gt\Leftrightarrow\frac{1}{c}=\frac{1}{2}.\frac{a+b}{ab}\)
\(\Leftrightarrow\frac{1}{c}=\frac{a+b}{2ab}\Leftrightarrow c\left(a+b\right)=2ab\Leftrightarrow ac+bc=ab+ab\)
\(\Leftrightarrow ac-ab=ab-bc\Leftrightarrow a\left(c-b\right)=b\left(a-c\right)\)
\(\Rightarrow\frac{a}{b}=\frac{a-c}{b-c}\)
\(\Rightarrowđpcm\)
từ a - b- c = 0 .Suy ra:A=\(\frac{b}{a}.\frac{-c}{b}.\frac{a}{c}=1\)