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\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(T/C...)
Xét a+b+c=0
\(\Rightarrow a+b=-c,c+b=-a,a+c=-b\)
\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{-c}{b}\cdot\frac{-a}{c}\cdot\frac{-b}{a}=-1\)
Xét a+b+c\(\ne0\)
\(\Rightarrow a+b=2c,b+c=2a,c+a=2b\)
\(\Rightarrow\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{a+c}{a}=\frac{2c}{b}\cdot\frac{2a}{c}\cdot\frac{2b}{a}=8\)
Giải:
+) Xét a + b + c = 0
\(\Rightarrow-a=b+c\)
\(\Rightarrow-b=a+c\)
\(\Rightarrow-c=a+b\)
Ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{-c}{c}=\frac{-a}{a}=\frac{-b}{b}=-1\)
Lại có: \(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=-1\)
+) Xét \(a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Ta có:
\(M=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{a+b}{c}.\frac{b+c}{a}.\frac{c+a}{b}=2.2.2=8\)
Vậy M = -1 hoặc M = 8
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Leftrightarrow\)\(\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)
\(\Leftrightarrow\)\(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\frac{a+b}{a}.\frac{b+c}{b}.\frac{c+a}{c}\)
+) Nếu \(a+b+c=0\)
\(\Leftrightarrow\)\(\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow\)\(P=\frac{-c}{a}.\frac{-a}{b}.\frac{-b}{c}=\frac{-abc}{abc}=-1\)
+) Nếu \(a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{3\left(a+b+c\right)}{a+b+c}=3\)
Suy ra :
\(\frac{a+b+c}{c}=3\)\(\Leftrightarrow\)\(a+b=2c\)
\(\frac{a+b+c}{a}=3\)\(\Leftrightarrow\)\(b+c=2a\)
\(\frac{a+b+c}{b}=3\)\(\Leftrightarrow\)\(c+a=2b\)
\(\Rightarrow\)\(P=\frac{2c}{a}.\frac{2a}{b}.\frac{2b}{c}=\frac{8abc}{abc}=8\)
Vậy \(P=-1\) hoặc \(P=8\)
Chúc bạn học tốt ~
ta có: \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{c+a+b}.\)\(=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\hept{\begin{cases}\frac{a+b-c}{c}=1\\\frac{b+c-a}{a}=1\end{cases}\Rightarrow\hept{\begin{cases}a+b-c=c\\b+c-a=a\end{cases}}}\) => a+ c = a +b - c + b+c-a => a + c = 2b
tương tự như trên ta có: a + b = 2c; b + c = 2a
=> a=b=c
\(\Rightarrow P=\left(1+\frac{b}{a}\right).\left(1+\frac{c}{b}\right).\left(1+\frac{a}{c}\right)=\left(1+\frac{a}{a}\right).\left(1+\frac{c}{c}\right).\left(1+\frac{a}{a}\right)\)\(=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\) ( a,b,c khác 0 )
Áp dụng tc của dãy tỉ số bằng nhau ta cso:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b-c+b+c-a+c+a-b}{a+b+c}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}\)\(\Rightarrow\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}\)
Có: \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{c}{a}\right)\)
\(=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+1=\frac{b+c-a}{a}+1=\frac{c+a-b}{b}+1\)
\(\Rightarrow\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(T/C...)
Xét a+b+c=0
\(\Rightarrow a+b=-c,b+c=-a,c+a=-b\)
\(\Rightarrow P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\frac{a+b}{a}\cdot\frac{b+c}{b}\cdot\frac{c+a}{c}=\frac{-c}{a}\cdot\frac{-a}{b}\cdot\frac{-b}{c}=\frac{\left(-c\right)\left(-a\right)\left(-b\right)}{a\cdot b\cdot c}=-1\)
Xét a+b+c\(\ne0\Rightarrow a+b=2c,b+c=2a,c+a=2b\)
\(\Rightarrow P=\left(1+\frac{b}{a}\right)\left(1+\frac{c}{b}\right)\left(1+\frac{a}{c}\right)=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{a\cdot b\cdot c}=\frac{2c\cdot2a\cdot2b}{a\cdot b\cdot c}=8\)
Vậy P=8 hoặc P=-1
Ta có: a - b - c = 0
=> \(\hept{\begin{cases}a-c=b\\a-b=c\\-b-c=-a\end{cases}}\Rightarrow\hept{\begin{cases}a-c=b\\-\left(a-b\right)=-c\\-\left(b+c\right)=-a\end{cases}}\Rightarrow\hept{\begin{cases}a-c=b\\-a+b=-c\\b+c=a\end{cases}}\)
Lại có: \(P=\left(1-\frac{c}{a}\right)\left(1-\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\)
\(\Rightarrow P=\frac{a-c}{a}.\frac{b-a}{b}.\frac{c+b}{c}=\frac{b}{a}.\frac{-c}{b}.\frac{a}{c}=-1\)
Ta có:
(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c... (a+b+c)=(a+b+c)/(a+b+c)=1
=>(a+b-c)/c=1 => a+b-c=c =>a+b=2c (1)
Tương tự: (b+c-a)/a=1 =>b+c=2a (2)
(c+a-b)/b=1 =>c+a=2b (3)
Thế (1), (2), (3) vào P, ta có:
P=(a+b)/a . (b+c)/b .(a+c)/c=2c/a.2a/b.2b/c=2.2.2=8.
Từ giả thiết, suy ra:
(a+b-c)/c+2=(b+c-a)/a+2=(c+a-b)/b+2
<=> (a+b+c)/c=(b+c+a)/a=(c+a+b)/b
Xét 2 trường hợp:
Nếu a+b+c=0
=> (a+b)/a.(b+c)/b.(c+a)/c=((-c)(-a)(-b))/a...
Nếu a+b+c khác 0
=>a=b=c =>P=2.2.2=8
Vay :P=8
=>\(\frac{a-b+c}{2b}+1=\frac{c-a+b}{2a}+1=\frac{a-c+b}{2c}+1\)
\(\Rightarrow\frac{a+b+c}{2b}=\frac{a+b+c}{2a}=\frac{a+b+c}{2c}\)
*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
