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Áp dụng t/c dtsbn:
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow P=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a\cdot a\cdot a}=\dfrac{8a^3}{a^3}=8\)
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)
\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)
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 \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)\)
\(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)
b)\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}\) và \(\dfrac{b+c}{a}=\dfrac{c+a}{b}\)
\(\Rightarrow1+\dfrac{a+b}{c}=1+\dfrac{b+c}{a}\)và \(1+\dfrac{b+c}{a}=1 +\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{c}{c}+\dfrac{a+b}{c}=\dfrac{a}{a}+\dfrac{b+c}{a}\)và \(\dfrac{a}{a}+\dfrac{b+c}{a}=\dfrac{b}{b}+\dfrac{c+a}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}\)và \(\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\dfrac{a+b+c}{c}-\dfrac{a+b+c}{a}=0\) \(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{c}-\dfrac{1}{a}\right)=0\)
và \(\dfrac{a+b+c}{a}-\dfrac{a+b+c}{b}=0\)
\(\Rightarrow\left(a+b+c\right)\cdot\left(\dfrac{1}{a}-\dfrac{1}{b}\right)=0\)
+) Vì a,b,c đôi một khác 0
\(\Rightarrow a+b+c=0\)
\(\rightarrow a+b=\left(-c\right)\)
\(\rightarrow a+c=\left(-b\right)\)
\(\rightarrow b+c=\left(-a\right)\)
+) Ta có:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
Lời giải:
Ta có:
\(\frac{b-c}{(a-b)(a-c)}+\frac{c-a}{(b-a)(b-c)}+\frac{a-b}{(c-a)(c-b)}=2013\)
\(\Leftrightarrow \frac{-(b-c)^2}{(a-b)(b-c)(c-a)}+\frac{-(c-a)^2}{(a-b)(b-c)(c-a)}+\frac{-(a-b)^2}{(a-b)(b-c)(c-a)}=2013\)
\(\Leftrightarrow \frac{-[(a-b)^2+(b-c)^2+(c-a)^2]}{(a-b)(b-c)(c-a)}=2013\)
\(\Rightarrow \frac{2(a^2+b^2+c^2-ab-bc-ac)}{(a-b)(b-c)(c-a)}=-2013(*)\)
Lại có:
\(P=\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\)
\(=\frac{(b-c)(c-a)+(c-a)(a-b)+(a-b)(b-c)}{(a-b)(b-c)(c-a)}\)
\(=\frac{bc-ba-c^2+ca+ca-bc-a^2+ab+ab-ac-b^2+bc}{(a-b)(b-c)(c-a)}\)
\(=\frac{ab+bc+ac-(a^2+b^2+c^2)}{(a-b)(b-c)(c-a)}=-\frac{1}{2}.\frac{2(a^2+b^2+c^2-ab-bc-ac)}{(a-b)(b-c)(c-a)}\)
\(=\frac{-1}{2}.-2013=\frac{2013}{2}\) (theo $(*)$)
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{a+c-b}{b}+2=\dfrac{b+c-a}{a}+2\)
\(\Rightarrow\dfrac{a+b-c}{c}+\dfrac{2c}{2}=\dfrac{a+c-b}{b}+\dfrac{2b}{b}=\dfrac{b+c-a}{a}+\dfrac{2a}{a}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+c+b}{b}=\dfrac{b+c+a}{a}\)
\(\Rightarrow a=b=c\) Thay vào A ta được :
\(A=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a.a.a}=\dfrac{2a.2a.2a}{a^3}=\dfrac{8.a^3}{a^3}=8\)
Áp dụng tính chất dãy tỉ số bằng nhau, ta có :
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{c+b+a}=\dfrac{2a+2b+2c}{a+b+c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-c}{c}=2\\\dfrac{a+c-b}{b}=2\\\dfrac{b+c-a}{a}=2\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=2c\\a+c-b=2b\\b+c-a=2a\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b=2c+c\\a+c=2b+b\\b+c=2a+a\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}a+b=3c\\a+c=3b\\b+c=3a\end{matrix}\right.\)
Ta có :
\(A=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\\ \Rightarrow A=\dfrac{3c\cdot3a\cdot3b}{abc}=\dfrac{27abc}{abc}=27\)