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\(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
\(\Rightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=2\)
\(\Rightarrow P=2+2+2=6\)
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:
$P=\frac{c+c}{c}+\frac{a+a}{a}+\frac{b+b}{b}=2+2+2=6$
Theo T/C dãy tỉ số bằng nhau
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\frac{a+b}{c}=2\Rightarrow a+b=2c\)
Tương tự ta có
\(b+c=2a\)
\(c+a=2b\)
Xét \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)\)
\(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)
Áp dụng t/c dtsbn ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)
Áp dụng t/c dãy tỉ số bằng nhau ta có:
\(\dfrac{a}{b+c}=\dfrac{b}{a+c}=\dfrac{c}{a+b}=\dfrac{a+b+c}{b+c+a+c+a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
Do \(\dfrac{a}{b+c}=\dfrac{1}{2}\Rightarrow b+c=2a\) (1)
\(\dfrac{b}{a+c}=\dfrac{1}{2}\Rightarrow a+c=2b\) (2)
\(\dfrac{c}{a+b}=\dfrac{1}{2}\Rightarrow a+b=2c\) (3)
Thay (1); (2) và (3) vào \(P\) ta có:
\(P=\dfrac{2a}{a}+\dfrac{2b}{b}+\dfrac{2c}{c}\)
\(\Rightarrow P=2+2+2=6\)
Vậy \(P=6.\)
mk còn cách khác