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\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{a+b+3c}{c}=\frac{\left(3a+b+c\right)+\left(a+3b+c\right)+\left(a+b+3c\right)}{a+b+c}\)
\(=\frac{5\left(a+b+c\right)}{a+b+c}=5\)
\(\Rightarrow\frac{3a+b+c}{a}=5\Rightarrow3a+b+c=5a\Rightarrow b+c=2a\)
Tương tự ta có : \(a+c=2b;a+b=2c\)
\(\Rightarrow B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)\)
\(=\frac{a+b}{b}.\frac{b+c}{c}.\frac{a+c}{a}=\frac{2c}{b}.\frac{2a}{c}.\frac{2b}{a}\)
\(=\frac{8abc}{abc}=8\)
Ta có:
\(\frac{3a+b+c}{a}=\frac{a+3b+c}{b}=\frac{3c+b+a}{c}\)
\(\Rightarrow3+\frac{b+c}{a}=3+\frac{a+c}{b}=3+\frac{b+a}{c}\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{b+a}{c}=\frac{2\left(a+b+c\right)}{a+b+c}\)
TH1:\(a+b+c=0\)\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\a+c=-b\end{matrix}\right.\)
Thay vào B, ta có:
\(B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=-1\)
TH2:\(a+b+c\ne0\)
\(\Rightarrow\frac{b+c}{a}=\frac{a+c}{b}=\frac{b+a}{c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}b+c=2a\\a+c=2b\\b+a=2c\end{matrix}\right.\)
Thay vào B, ta có:
\(B=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=8\)
Vậy \(\left[{}\begin{matrix}B=-1\\B=8\end{matrix}\right.\)
\(\frac{2a+a+b+c}{a}=\frac{2b+a+b+c}{b}=\frac{2c+a+b+c}{c}\)
\(\Rightarrow2+\frac{a+b+c}{a}=2+\frac{a+b+c}{b}=2+\frac{a+b+c}{c}\)
\(\Rightarrow\frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}\Rightarrow a=b=c\)
\(\Rightarrow B=\left(1+1\right)\left(1+1\right)\left(1+1\right)\)
Câu hỏi của Chu Hoàng THủy Tiên - Toán lớp 7 - Học toán với OnlineMath
\(\frac{a+b-c}{c}=\frac{b+c-a}{a}=\frac{c+a-b}{b}\)
\(\Rightarrow\frac{a+b-c}{c}+2=\frac{b+c-a}{a}+2=\frac{c+a-b}{b}+2\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{a}=\frac{a+b+c}{b}\)
\(\Rightarrow a=b=c\)
\(\Rightarrow\frac{b}{a}=1;\frac{a}{c}=1;\frac{c}{b}=1\)
\(\Rightarrow B=\left(1+\frac{b}{a}\right)\left(1+\frac{a}{c}\right)\left(1+\frac{c}{b}\right)=\left(1+1\right)\left(1+1\right)\left(1+1\right)=8\)
Có: \(\frac{3a+b+2c}{2a+c}=\frac{a+3b+c}{2b}=\frac{a+2b+2c}{b+c}\)
\(\Rightarrow\frac{a+b+c+2a+c}{2a+c}=\frac{a+b+c+2b}{2b}=\frac{a+b+c+b+c}{b+c}\)
\(\Rightarrow\frac{a+b+c}{2a+c}+1=\frac{a+b+c}{2b}+1=\frac{a+b+c}{b+c}+1\)
\(\Rightarrow\frac{a+b+c}{2a+c}=\frac{a+b+c}{2b}=\frac{a+b+c}{b+c}\)
\(\Rightarrow2a+c=2b=b+c\)
\(\Rightarrow\hept{\begin{cases}c=b\\a=\frac{1}{2}b\end{cases}}\)
Thay vào biểu thức trên , ta được:
\(P=\)\(\frac{\left(\frac{1}{2}b+b\right)\left(b+b\right)\left(b+\frac{1}{2}b\right)}{\frac{1}{2}b.b.b}=9\)
Vậy \(P=9\)
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{c+a+b}=2\)(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
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{2b+c-a}{a}=\frac{2c-b+a}{b}=\frac{2a+b-c}{c}=\frac{2a+2b+2c}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
Do đó :
\(\frac{2b+c-a}{a}=2\)\(\Rightarrow\)\(c=3a-2b\)\(;\)\(2b=3a-c\)\(\left(1\right)\)
\(\frac{2c-b+a}{b}=2\)\(\Rightarrow\)\(a=3b-2c\)\(;\)\(2c=3b-a\)\(\left(2\right)\)
\(\frac{2a+b-c}{c}=2\)\(\Rightarrow\)\(b=3c-2a\)\(;\)\(2a=3c-b\)\(\left(3\right)\)
Thay (1), (2) và (3) vào \(P=\frac{\left(3a-2b\right)\left(3b-2c\right)\left(3c-2a\right)}{\left(3a-c\right)\left(3b-a\right)\left(3c-b\right)}\) ta được :
\(P=\frac{c.a.b}{2b.2c.2a}=\frac{abc}{8abc}=\frac{1}{8}\)
Vậy \(P=\frac{1}{8}\)
Chúc bạn học tốt ~
Phùng Minh Quân sai nha nếu a+b+c = 0 thì a+b+c / 2(a+b+c) thì nó không bằng 1/2 đc mà nó bằng 0
Giải toán trên mạng - Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath