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\(A=\dfrac{a}{a+b+c-c}+\dfrac{b}{a+b+c-a}+\dfrac{c}{a+b+c-b}\\ A=\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{c+a}\\ \Rightarrow A>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=1\left(1\right)\\ A< \dfrac{a+c}{a+b+c}+\dfrac{b+a}{a+b+c}+\dfrac{c+b}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\left(2\right)\\ \left(1\right)\left(2\right)\Rightarrow1< A< B\\ \Rightarrow A\notin Z\)
Lời giải:
$\frac{2022a+b+c}{a}=\frac{a+2022b+c}{b}=\frac{a+b+2022c}{c}$
$=2021+\frac{a+b+c}{a}=2021+\frac{a+b+c}{b}=2021+\frac{a+b+c}{c}$
$\Rightarrow \frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$
$\Rightarrow a+b+c=0$ hoặc $\frac{1}{a}=\frac{1}{b}=\frac{1}{c}$
$\Rightarrow a+b+c=0$ hoặc $a=b=c$
Nếu $a+b+c=0$ thì:
$P=\frac{a+b}{c}+\frac{b+c}{a}+\frac{a+c}{b}=\frac{(-c)}{c}+\frac{(-b)}{b}+\frac{(-a)}{a}=-1+(-1)+(-1)=-3$
Nếu $a=b=c$ thì:
$P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=2+2+2=6$
Lời giải:
\(\frac{ab}{a+b}=\frac{bc}{b+c}=\frac{ca}{c+a}\Rightarrow \frac{abc}{c(a+b)}=\frac{abc}{a(b+c)}=\frac{bca}{b(c+a)}\)
\(\Leftrightarrow c(a+b)=a(b+c)=b(c+a)\)
\(\Leftrightarrow ac+bc=ab+ac=bc+ab\Leftrightarrow ab=bc=ac\)
\(\Rightarrow a=b=c\) (do $a,b,c>0$)
$\Rightarrow M=\frac{a^2+a^2+a^2}{a^2+a^2+a^2}=1$
\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=2\)
\(\Rightarrow P=2+2+2=6\)
\(\dfrac{ab}{a+b}=\dfrac{bc}{b+c}=\dfrac{ca}{c+a}\Rightarrow\dfrac{a+b}{ab}=\dfrac{b+c}{bc}=\dfrac{c+a}{ca}\)
\(\Rightarrow\dfrac{1}{a}+\dfrac{1}{b}=\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{c}+\dfrac{1}{a}\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{a}=\dfrac{1}{c}\\\dfrac{1}{a}=\dfrac{1}{b}\end{matrix}\right.\) \(\Rightarrow a=b=c\)
\(\Rightarrow M=\dfrac{a^2+a^2+a^2}{a^2+a^2+a^2}=1\)
a+b−cc=b+c−aa=c+a−bb
⇒a+b−cc+1=b+c−aa+1=c+a−bb+1
⇒a+bc=b+ca=c+ab
+)Nếu a+b+c=0⇒a+b=−c;b+c=−a;c+a=−b
⇒B=a+ba.c+ac.b+cb=−ca.−bc.−ab=−(abc)abc=−1
Nếu a+b+c≠0
Áp dụng tính chất dãy tỉ số bằng nhau ta có
a+bc=b+ca=c+ab=2(a+b+c)a+b+c=2
⇒a+b=2c
b+c=2a
c+a=2b
⇒B=2ca.2bc.2ab=2.2.2=8
Ta có:
\(\dfrac{a}{a+b}+\dfrac{b}{b+c}+\dfrac{c}{a+c}>\dfrac{a}{a+b+c}+\dfrac{b}{a+b+c}+\dfrac{c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)\(\Rightarrow\)\(M>1\left(1\right)\)
M=\(\dfrac{a+b-b}{a+b}+\dfrac{b+c-c}{b+c}+\dfrac{c+a-a}{c+a}\)
= \(3-\left(\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{c+a}\right)< 2\) \(\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{c+a}>1\)
(Vì \(\dfrac{b}{a+b}+\dfrac{c}{b+c}+\dfrac{a}{c+a}>1\)
\(\Rightarrow1< M< 2\)
Vậy M không có giá trị nguyên(đpcm)