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Áp dụng t/c dtsbn ta có:
\(\dfrac{2b+c-a}{a}=\dfrac{2c-b+a}{b}=\dfrac{2a+b-c}{c}=\dfrac{2b+c-a+2c-b+a+2a+b-c}{a+b+c}=\dfrac{2b+2c+2a}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\dfrac{2b+c-a}{a}=2\Rightarrow2b+c-a=2a\Rightarrow2b=3a-c\)\(\dfrac{2c-b+a}{b}=2\Rightarrow2c-b+a=2b\Rightarrow2c=3b-a\)
\(\dfrac{2a+b-c}{c}=2\Rightarrow2a+b-c=2c\Rightarrow2a=3c-b\)
\(P=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{2a.2b.2c}=\dfrac{\left(2a-b\right)\left(2b-c\right)\left(2c-a\right)}{8abc}\)
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{a+c-b}{b}+2=\dfrac{b+c-a}{a}+2\)
\(\Rightarrow\dfrac{a+b-c}{c}+\dfrac{2c}{c}=\dfrac{a+c-b}{b}+\dfrac{2b}{b}+\dfrac{b+c-a}{a}+\dfrac{2a}{a}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+c+b}{b}=\dfrac{b+c+a}{a}\)
\(\Rightarrow a=b=c\) Thay \(M\) vào ta được:
\(M=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{2a.2a.2a}{a^3}=\dfrac{8a^3}{a^3}=8\)
Vậy .............
Chúc bạn học tốt!
Áp dụng t/c dtsbn ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)
\(\dfrac{a+b-2c}{c}=\dfrac{c+a-2b}{b}=\dfrac{b+c-2a}{a}=\dfrac{a+b-2c+c+a-2b+b+c-2a}{c+b+a}=0\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-2c}{c}=0\\\dfrac{c+a-2b}{b}=0\\\dfrac{b+c-2a}{a}=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b-2c=0\\a+c-2b=0\\b+c-2a=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc}=\dfrac{2c.2a.2b}{abc}=\dfrac{8abc}{abc}=8\)
Ta có:
\(a+b+c-abc=\left(a+b+c\right)\left(ab+bc+ca\right)-abc\)
\(=\left(a+b+c\right)\left(ab+c\left(a+b\right)\right)-abc\)
\(=\left(a+b\right)ab+\left(a+b\right)^2c+abc+c^2\left(a+b\right)-abc\)
\(=\left(a+b\right)\left(ab+c^2+c\left(a+b\right)\right)\)
\(=\left(a+b\right)\left(ab+ac+c^2+bc\right)\)
\(=\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Đồng thời:
\(a^2+1=a^2+ab+bc+ac=a\left(a+b\right)+c\left(a+b\right)=\left(a+b\right)\left(a+c\right)\)
Tương tự:
\(b^2+1=\left(a+b\right)\left(b+c\right)\)
\(c^2+1=\left(a+c\right)\left(b+c\right)\)
Từ đó:
\(P=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left(a+b\right)\left(a+c\right)\left(a+b\right)\left(b+c\right)\left(a+c\right)\left(b+c\right)}\)
\(=\dfrac{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}{\left[\left(a+b\right)\left(b+c\right)\left(a+c\right)\right]^2}=1\)
Áp dụng tính chất dãy tủ số bằng nhau, ta có:
\(\frac{a+b-c}{c}\) = \(\frac{a-b+c}{b}\) = \(\frac{-a+b+c}{a}\) = \(\frac{a+b+c}{a+b+c}\) = 1
=>\(\frac{a+b-c}{c}\) = 1
a+b-c = c
a+b =2c
=>\(\frac{a-b+c}{b}\) = 1
a-b+c = c
a+c =2b
=>\(\frac{-a+b+c}{a}\) = 1
-a+b+c = a
b+c =2a
Thay a+b =2c , a+c =2b , b+c =2a vào biểu thức:
M=\(\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\) = \(\frac{2c.2b.2a}{abc}\) = \(\frac{2^3abc}{abc}\) = 23 =8
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{a-b+c}{b}+2=\dfrac{-a+b+c}{a}+2\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{b}=\dfrac{a+b+c}{a}\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
\(\circledast\) Với \(a+b+c=0\) thì \(\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(m=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{-abc}{abc}=-1\)
\(\circledast\) Với \(a=b=c\) thì \(m=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{8a^3}{a^3}=8\)
Biết chế liền!