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Ta có: a+b+c=0a+b+c=0
\Rightarrow b+a=-c⇒b+a=−c
\Rightarrow c+b=-a⇒c+b=−a
\Rightarrow a+c=-b⇒a+c=−b
Ta có: A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)A=(1+
b
a
)(1+
c
b
)(1+
a
c
)
\Rightarrow A=\left(\frac{b+a}{b}\right)\left(\frac{c+b}{c}\right)\left(\frac{a+c}{a}\right)⇒A=(
b
b+a
)(
c
c+b
)(
a
a+c
)
\Rightarrow A=\left(\frac{-c}{b}\right)\left(\frac{-a}{c}\right)\left(\frac{-b}{a}\right)⇒A=(
b
−c
)(
c
−a
)(
a
−b
)
\Rightarrow A=-1⇒A=−1
\(\frac{b+c-a}{a}+\frac{2a}{a}=\frac{a+c-b}{b}+\frac{2b}{b}=\frac{a+b-c}{c}+\frac{2c}{c}\)
\(\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\)
=> a=b=c
A=(1+1)(1+1)(1+1) = 2.2.2 =8
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Đặt \(\hept{\begin{cases}a-b=x\\b-c=y\\c-a=z\end{cases}}\)
Thế vào bài toán trở thành
Cho: \(\frac{x+z}{xz}+\frac{x+y}{xy}+\frac{y+z}{yz}=2013\left(1\right)\)
Tính \(M=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Từ (1) ta có
\(\left(1\right)\Leftrightarrow\frac{xy+yz+zx+yz+xy+zx}{xyz}=2013\)
\(\Leftrightarrow\frac{2\left(xy+yz+zx\right)}{xyz}=2013\)
\(\Leftrightarrow\frac{xy+yz+zx}{xyz}=\frac{2013}{2}\)
Ta lại có
\(M=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{xy+yz+zx}{xyz}=\frac{2013}{2}\)
\(\frac{b-c}{\left(a-b\right)\left(a-c\right)}+\frac{c-a}{\left(b-a\right)\left(b-c\right)}+\frac{a-b}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{\left(a-c\right)-\left(a-b\right)}{\left(a-b\right)\left(a-c\right)}+\frac{\left(b-a\right)-\left(b-c\right)}{\left(b-a\right)\left(b-c\right)}+\frac{\left(c-b\right)-\left(c-a\right)}{\left(c-b\right)\left(c-a\right)}\)
\(=\frac{1}{a-b}-\frac{1}{a-c}+\frac{1}{b-c}-\frac{1}{b-a}+\frac{1}{c-a}-\frac{1}{c-b}\)
\(=2\left(\frac{1}{a-b}+\frac{1}{b-c}+\frac{1}{c-a}\right)=2013\)
\(\Rightarrow M=\frac{2013}{2}\)
=>\(\frac{a-b+c}{2b}+1=\frac{c-a+b}{2a}+1=\frac{a-c+b}{2c}+1\)
\(\Rightarrow\frac{a+b+c}{2b}=\frac{a+b+c}{2a}=\frac{a+b+c}{2c}\)
*TH1: nếu a+b+c=0 => a+b=-c; b+c=-a; c+a=-b
=>P=\(\left(\frac{b+c}{b}\right)\left(\frac{a+b}{a}\right)\left(\frac{c+a}{c}\right)\)
=\(\frac{-a}{b}.\frac{-c}{a}.\frac{-b}{c}=\frac{-\left(a.b.c\right)}{a.b.c}=-1\)
*TH2: Nếu a+b+c khác 0: thì a=b=c
Khi đó P=2.2.2=8
Vậy P= -1 hoặc 8
\(\frac{a-b+c}{2b}=\frac{c-a+b}{2a}=\frac{a-c+b}{2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
=> 2a-2b+2c=2b <=> a+c=2b. Chia cả 2 vế cho c ta được: \(1+\frac{a}{c}=\frac{2b}{c}\)
Tương tự: \(1+\frac{c}{b}=\frac{2a}{b}\) và \(1+\frac{b}{a}=\frac{2c}{a}\)
=> \(\left(1+\frac{c}{b}\right)\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)=\frac{2a}{b}.\frac{2c}{a}.\frac{2b}{c}=\frac{8.abc}{abc}=8\)
Đáp số: 8
\(A=\left(\frac{a+b}{b}\right).\left(\frac{b+c}{c}\right).\left(\frac{a+c}{a}\right)\)
Vì \(a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\a+c=-b\end{cases}}\)
\(\Rightarrow A=\frac{-c}{b}.\left(\frac{-a}{c}\right).\left(\frac{-b}{a}\right)\)
\(\Rightarrow A=-1\)
toán nâng cao ak bn