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Đặt \(\frac{b+c-a}{c}=\frac{a+b+c}{b}=\frac{b-c+a}{a}=k\)
\(\Rightarrow\hept{\begin{cases}b+c-a=ck\\a+b+c=bk\\b-c+a=ak\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}2b=k\left(a+c\right)\left(1\right)\\2c=k\left(b-a\right)\left(2\right)\\2b+2c=b\left(b+c\right)\Rightarrow k=2\end{cases}}\)
Thay k=2 vào (1) và (2) :
\(\hept{\begin{cases}2b=2\left(a+c\right)\\2c=2\left(b-a\right)\end{cases}\Rightarrow\hept{\begin{cases}b=a+c\\c=b-a\Rightarrow a=b-c\end{cases}}}\)
Vậy \(\frac{\left(b-a\right)\left(c+b\right)\left(a+c\right)}{abc}=\frac{\left(b-a\right)\left(c+b\right)\left(a+c\right)}{\left(b-c\right)\left(a+c\right)\left(b-a\right)}=\frac{b+c}{b-c}\)
\(Tacó\)
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=\frac{a+b+a+c+b+c-a-b-c}{a+b+c}=1\)
\(\Rightarrow a+b=2c;b+c=2a;c+a=2b\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}=\frac{2c.2c.2c}{c^3}=8\)
\(Taco:\)
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=\frac{a+b+a+c+b+c-a-b-c}{a+b+c}=1\)
\(\Rightarrow a+b=2c;b+c=2a;c+a=2b\)
\(\Leftrightarrow a=b=c\)
\(\Rightarrow\frac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}=\frac{2c.2c.2c}{c^3}=8\)
Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,
k nhé
Ta có: \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
Suy ra:
\(\frac{a}{b+c}=\frac{1}{2}\Rightarrow a=\frac{b+c}{2}=\frac{1}{2}\times\left(b+c\right)\)
\(\frac{b}{a+c}=\frac{1}{2}\Rightarrow b=\frac{a+c}{2}=\frac{1}{2}\times\left(a+c\right)\)
\(\frac{c}{a+b}=\frac{1}{2}\Rightarrow c=\frac{a+b}{2}=\frac{1}{2}\times\left(a+b\right)\)
Thay \(a=\frac{1}{2}\times\left(b+c\right)\); \(b=\frac{1}{2}\times\left(a+c\right)\); \(c=\frac{1}{2}\times\left(a+b\right)\) vào P ta được:
\(\frac{b+c}{\frac{1}{2}\times\left(b+c\right)}+\frac{c+a}{\frac{1}{2}\times\left(a+c\right)}+\frac{a+b}{\frac{1}{2}\times\left(a+b\right)}\)
\(=\frac{\text{ }1\text{ }}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}+\frac{1}{\frac{1}{2}}\)
\(=2+2+2=6\)
Vậy giá trị của P là 6
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(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
Chị tham khảo nhé
Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
l ike cho cái bạn chị tham khảo bài (:V
Vì \(a,b,c\ne0\)
\(\Rightarrow\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=2\)
\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=2+2+2=6\)
Ta có : \(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\)
=> \(\frac{a}{b+c}+1=\frac{b}{a+c}+1=\frac{c}{a+b}+1\)
=> \(\frac{a+b+c}{b+c}=\frac{a+b+c}{a+c}=\frac{a+b+c}{a+b}\)
Nếu a + b + c = 0
=> a + b = - c
=> b + c = - a
=> a + c = - b
Khi đó P = \(\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-1+\left(-1\right)+\left(-1\right)=-3\)
Nếu a + b + c \(\ne0\)
=> \(\frac{1}{b+c}=\frac{1}{a+c}=\frac{1}{a+b}\)
=> b + c = a + c = a + b
=> \(\hept{\begin{cases}b+c=a+c\\b+c=a+b\end{cases}\Rightarrow\hept{\begin{cases}a=b\\a=c\end{cases}}\Rightarrow a=b=c}\)
Khi đó P = \(\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}=2+2+2=6\)
=> P = 6
Vậy khi a + b + c = 0 => P = -3
khi a + b + c \(\ne0\) => P = 6
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(...=\frac{a+b-c+b+c-a+c+a-b}{c+a+b}=\frac{a+b+c}{a+b+c}=1\)( a, b, c khác 0 )
=> \(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}=1\)
=> \(\hept{\begin{cases}a+b-c=c\\b+c-a=a\\c+a-b=b\end{cases}}\Leftrightarrow\hept{\begin{cases}a+b=2c\\b+c=2a\\c+a=2b\end{cases}}\)
Thế vào P ta được :
\(P=\frac{2c}{a}\cdot\frac{2a}{b}\cdot\frac{2b}{c}=\frac{8abc}{abc}=8\)