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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:
\(M=\left(1+\dfrac{a}{b}\right)\cdot\left(1+\dfrac{b}{c}\right)\cdot\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{a+b}{b}\right)\cdot\left(\dfrac{b+c}{a}\right)\cdot\left(\dfrac{c+a}{c}\right)\)
\(=\dfrac{-c}{b}\cdot\dfrac{-a}{c}\cdot\dfrac{-b}{a}\)
\(=\left(-1\right)\)
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 \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)\)
\(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(=\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\)
\(=\dfrac{\left(a+b+b+c+c+a\right)-\left(c+a+b\right)}{a+b+c}\)
\(=\dfrac{2a+2b+2c-a-b-c}{a+b+c}=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-c}{c}=1\\\dfrac{b+c-a}{a}=1\\\dfrac{c+a-b}{b}=1\end{matrix}\right.\)
\(PHUCDZ=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{c}{b}\right)\left(1+\dfrac{a}{c}\right)\)
\(PHUCDZ=\left(\dfrac{b+c-a}{a}+\dfrac{b}{a}\right)\left(\dfrac{c+a-b}{b}+\dfrac{c}{b}\right)\left(\dfrac{a+b-c}{c}+\dfrac{a}{c}\right)\)
\(PHUCDZ=\dfrac{b+c-a+b}{a}.\dfrac{c+a-b+c}{b}.\dfrac{a+b-c+a}{c}\)
\(PHUCDZ=\dfrac{2b+c-a}{a}.\dfrac{2c+a-b}{b}.\dfrac{2a+b-c}{c}\)
\(PHUCDZ=\dfrac{\left(2b+c-a\right)\left(2c+a-b\right)\left(2a+b-c\right)}{abc}\)
\(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Ta có: \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
Hay \(A=\dfrac{-abc}{abc}=-1\)
\(a+b+c=0\\ \Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(A=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =\dfrac{-abc}{abc}=-1\)
Có chỗ khác sp nha!!!
Lời giải \(B=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)=\dfrac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}\)
Ta có: \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(\Rightarrow\dfrac{a+b-c}{c}+2=\dfrac{b+c-a}{a}+2=\dfrac{c+a-b}{b}+2\)
\(\Rightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\)
\(\Rightarrow\left[{}\begin{matrix}a+b+c=0\\a=b=c\end{matrix}\right.\)
Khi \(a+b+c=0\Leftrightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\Leftrightarrow B=\dfrac{-abc}{abc}=-1\)
Khi \(a=b=c\Leftrightarrow B=\dfrac{8abc}{abc}=8\)
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 $(*)$)
Đề thiếu điều kiện \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\) nữa đấy
Ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(=\dfrac{a+b-c+b+c-a+c+a-b}{a+b+c}\)
\(=\dfrac{a+b+c}{a+b+c}\)
\(=1\)
Với \(\dfrac{a+b-c}{c}=1\)
\(\Rightarrow a+b-c=c\)
\(\Rightarrow a+b=2c\left(1\right)\)
Với \(\dfrac{b+c-a}{a}=1\)
\(\Rightarrow b+c-a=a\)
\(\Rightarrow b+c=2a\left(2\right)\)
Với \(\dfrac{c+a-b}{b}=1\)
\(\Rightarrow c+a-b=b\)
\(\Rightarrow c+a=2b\left(3\right)\)
Ta lại có:
\(\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\left(\dfrac{b}{b}+\dfrac{a}{b}\right)\left(\dfrac{c}{c}+\dfrac{b}{c}\right)\left(\dfrac{a}{a}+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}\)
Thay (1) , (2) và (3) vào ta được
\(=\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\)
\(=\dfrac{8abc}{abc}\)
\(=8\)