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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 \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)\)
\(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{a+c-b}{b}+2=\dfrac{b+c-a}{a}+2\)
\(\Rightarrow\dfrac{a+b-c}{c}+\dfrac{2c}{2}=\dfrac{a+c-b}{b}+\dfrac{2b}{b}=\dfrac{b+c-a}{a}+\dfrac{2a}{a}\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+c+b}{b}=\dfrac{b+c+a}{a}\)
\(\Rightarrow a=b=c\) 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 :
\(A=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\\ \Rightarrow A=\dfrac{3c\cdot3a\cdot3b}{abc}=\dfrac{27abc}{abc}=27\)
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:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
Đặt \(\frac{a}{2013}=\frac{b}{2014}=\frac{c}{2015}=k\Rightarrow\hept{\begin{cases}a=2013k\\b=2014k\\c=2015k\end{cases}}\)
Ta có: 4(a - b)(b - c) = 4(2013k - 2014k)(2014k - 2015k) = 4(-k)(-k) = 4k2 (1)
(c - a)2 = (2015k - 2013k)2 = (2k)2 = 4k2 (2)
Từ (1) và (2) ta có đpcm
Đặt a2013 =b2014 =c2015 =k⇒{
| a=2013k |
| b=2014k |
| c=2015k |
Ta có: 4(a - b)(b - c) = 4(2013k - 2014k)(2014k - 2015k) = 4(-k)(-k) = 4k2 (1)
(c - a)2 = (2015k - 2013k)2 = (2k)2 = 4k2 (2)
Từ (1) và (2) ta có đpcm
Với: \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Khi đó: \(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{-abc}{abc}=-1\)
Với \(a+b+c\ne0\) áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)
Khi đó: \(P=\dfrac{8abc}{abc}=8\)
TH1: Với \(a+b+c\ne0\)
Áp dụng tính chất của dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)=> a + b = 2c ; a + c = 2b và b + c = 2a
\(\Rightarrow P=\dfrac{2c.2a.2b}{abc}=\dfrac{8abc}{abc}=8\)
TH2: Với a + b + c = 0
=> a + b = -c ; a + c = -b và c + b = -a
\(\Rightarrow P=\dfrac{\left(-c\right)\left(-a\right)\left(-b\right)}{abc}=-1\)
Vậy P = 8 hoặc P = -1
+) Nếu \(a+b+c=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
\(\Leftrightarrow P=\left(\dfrac{a}{b}+1\right)\left(\dfrac{b}{c}+1\right)\left(\dfrac{c}{a}+1\right)\)
\(=\left(\dfrac{a}{b}+\dfrac{b}{b}\right)\left(\dfrac{b}{c}+\dfrac{c}{c}\right)\left(\dfrac{c}{a}+\dfrac{a}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)
\(=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\)
\(=-1\)
+) Nếu \(a+b+c\ne0\)
Theo t.c dãy tỉ số bằng nhau ta có
\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}=\dfrac{a+b+c}{2\left(a+b+c\right)}=\dfrac{1}{2}\)
Vậy ...
Áp dụng t/c dtsbn:
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b+c}{a+b+c}=1\\ \Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\Rightarrow a=b=c\)
\(\Rightarrow P=\dfrac{\left(a+a\right)\left(a+a\right)\left(a+a\right)}{a\cdot a\cdot a}=\dfrac{8a^3}{a^3}=8\)
\(\dfrac{a+b-c}{c}=\dfrac{a+c-b}{b}=\dfrac{b+c-a}{a}=\dfrac{a+b-c+a+c-b+b+c-a}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a+c-b=b\\b+c-a=a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)
\(P=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\dfrac{2a.2b.2c}{abc}=8\)