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\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{b}{a}=1;\frac{a}{c}=1;\frac{c}{b}=1\)
\(\Rightarrow B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Câu hỏi của Chu Hoàng THủy Tiên - Toán lớp 7 - Học toán với OnlineMath
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)
<=> \(\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1\)
<=> \(\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
<=> a + b + c = 0 hoặc a = b = c.
Th1: a + b + c = 0
=> a + b = - c ; a + c = -b ; b + c = -a.
Thế vào P :
\(P=\left(1+\frac{a}{b}\right)\cdot\left(1+\frac{b}{c}\right)\cdot\left(1+\frac{c}{a}\right)\)
\(=\left(\frac{a+b}{b}\right)\cdot\left(\frac{b+c}{c}\right)\cdot\left(\frac{c+a}{a}\right)\)
\(=-\frac{c}{b}.\frac{\left(-a\right)}{c}.\frac{\left(-b\right)}{a}=-1\)
TH2: a = b = c. THế vào P
\(P=\left(1+1\right).\left(1+1\right).\left(1+1\right)=8\)
Vậy: P = -1 nếu a + b + c = 0
hoặc P = 8 nếu a = b = c.
\(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}.\frac{b+c}{c}.\frac{c+a}{a}\)
Ta có: \(\frac{a+b}{c}=\frac{b+c}{a}=\frac{a+c}{b}\)\(\Rightarrow\frac{a+b}{c}+1=\frac{b+c}{a}+1=\frac{a+c}{b}+1=\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
TH1: Nếu \(a+b+c=0\)\(\Rightarrow\hept{\begin{cases}a+b=-c\\b+c=-a\\c+a=-b\end{cases}}\)
\(\Rightarrow P=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=\frac{\left(-a\right).\left(-b\right).\left(-c\right)}{abc}=-1\)
TH2: Nếu \(a+b+c\ne0\)\(\Rightarrow a=b=c\)
\(\Rightarrow\hept{\begin{cases}a+b=2b\\b+c=2c\\c+a=2a\end{cases}}\)\(\Rightarrow P=\frac{2b}{b}.\frac{2c}{c}.\frac{2a}{a}=2.2.2=8\)
Vậy \(P=-1\)hoặc \(P=8\)
Ta có:
(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(c+a+b)=0/(c+a+b)=0
=> a+b-c=0 =>a+b=c
b+c-a=0 =>b+c=a
c+a-b=0 =>c+a=b
=>B=(a+b)/a.(c+a)/c.(b+c)/b
=c/a.b/c.a/b=1
TK!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Ta có:
(a+b-c)/c=(b+c-a)/a=(c+a-b)/b=(a+b-c+b+c-a+c+a-b)/(c+a+b)=0/(c+a+b)=0
=> a+b-c=0 =>a+b=c
b+c-a=0 =>b+c=a
c+a-b=0 =>c+a=b
=>B=(a+b)/a.(c+a)/c.(b+c)/b
=c/a.b/c.a/b=1
Áp dụng tính chất hãy tỉ số bằng nhau ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow a+b=2c;b+c=2a;a+c=2b\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{b}{a}=\frac{a}{c}=\frac{c}{b}=1\)
\(\Rightarrow B=2.2.2=8\)
ta có: \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{a-a+a+b+b-b-c+c+c}{c+a+b}=\frac{a+b+c}{c+a+b}=1\)
nếu a+b+c =0
=> a =0-b-c => a = -(b+c)
b = 0-a-c => b = -(a+c)
c = 0-a-b => c = -(a+b)
thay vào \(B=\left(1+\frac{-\left(a+c\right)}{a}\right).\left(1+\frac{-\left(b+c\right)}{c}\right).\left(1+\frac{-\left(a+b\right)}{b}\right)\)
\(B=\left(\frac{a-\left(a+c\right)}{a}\right).\left(\frac{c-\left(b-c\right)}{c}\right).\left(\frac{b-\left(a+b\right)}{b}\right)\)
\(B=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}\)
\(B=-1\)
nếu a+b+c khác 0
mà \(\frac{a+b+c}{c+a+b}=\frac{a}{c}=\frac{b}{a}=\frac{c}{b}=1\Rightarrow a=b=c\)
=> \(B=\left(1+\frac{b}{a}\right).\left(1+\frac{a}{c}\right).\left(1+\frac{c}{b}\right)\)
\(B=\left(1+1\right).\left(1+1\right).\left(1+1\right)\)
\(B=2.2.2\)
\(B=8\)
KL: B= -1 hoặc B=8
Chúc bn học tốt !!!!
Áp dụng tính chất của dãy tỉ số = nhau ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{\left(a+b-c\right)+\left(b+c-a\right)+\left(c+a-b\right)}{c+a+b}=\frac{a+b+c}{a+b+c}=1\) (1)
Xét 2 trường hợp:
- TH1: a + b + c = 0 \(\Rightarrow\begin{cases}a+b=-c\\a+c=-b\\b+c=-a\end{cases}\)
\(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)\)
\(P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}\)
\(P=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=-1\)
- TH2: a + b + c \(\ne\) 0
Từ (1) \(\Rightarrow\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=1\)
\(\Rightarrow\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}\)\(\Rightarrow\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}\)
\(P=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=8\)
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
=\(\frac{a+b-c+b+c-a+c+a-b}{a+b+c}\)=\(\frac{a+b+c}{a+b+c}\)=1
=>\(\frac{a+b-c}{c}=1\)
a+b-c=c
2c=a+b
=>\(\frac{b+c-a}{a}=1\)
b+c-a=a
2a=b+c
=>\(\frac{c+a-b}{b}=1\)
c+a-b=b
=>c+a=2b
ta co \(P=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(\frac{a+b}{a}\right)\left(\frac{a+c}{c}\right)\left(\frac{c+b}{b}\right)\)
=\(\frac{2c}{a}.\frac{2b}{c}.\frac{2a}{b}=2.2.2=8\)
Giải:
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
+) Xét \(a+b+c=0\Rightarrow\left\{\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
\(B=\left(1+\frac{b}{a}\right).\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{-c}{a}.\frac{-b}{c}.\frac{-a}{b}=-1\)
+) Xét \(a+b+c\ne0\)
\(\left\{\begin{matrix}\frac{a+b-c}{c}=2\\\frac{b+c-a}{a}=2\\\frac{c+a-b}{b}=2\end{matrix}\right.\Rightarrow\left\{\begin{matrix}a+b=3c\\b+c=3a\\a+c=3b\end{matrix}\right.\)
\(B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\frac{a+b}{a}.\frac{a+c}{c}.\frac{b+c}{b}=\frac{3c}{a}.\frac{3b}{c}.\frac{3a}{b}\)
\(=3.3.3=27\)
Vậy B = -1 hoặc B = 27