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\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)\(\Rightarrow b+c=2a\)
\(\Rightarrow a+c=2b\)
\(\Rightarrow a+b=2c\)
\(D=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)
\(D=\frac{2a}{a}=\frac{2b}{b}=\frac{2c}{c}\)
\(D=2+2+2\)
\(D=6\)
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
=>b+c=2a
=>a+c=2b
=>a+b=2c
\(D=\frac{b+c}{a}+\frac{a+c}{b}=\frac{a+b}{c}\)
\(D=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}\)
\(D=2+2+2\)
D=6
Vậy D=6
^...^ 
^_^
Vì \(a,b,c\ne0\) nên:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow\hept{\begin{cases}b+c=2a\\a+c=2b\\a+b=2c\end{cases}}\)
\(\Rightarrow D=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}=2+2+2=6\)
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}=\frac{a+b+c}{b+c+a+c+a+b}=\frac{a+b+c}{2a+2b+2c}=\frac{a+b+c}{2\left(a+b+c\right)}=\frac{1}{2}\)
\(\Rightarrow b+c=2a\)
\(\Rightarrow a+c=2b\)
\(\Rightarrow a+b=2c\)
\(D=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}\)
\(D=\frac{2a}{a}+\frac{2b}{b}+\frac{2c}{c}\)
\(D=2+2+2\)
\(D=6\)
Vậy \(D=6\)
Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính $A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$A=(1+ab )+(1+bc )+(1+ca )
Cho a, b, c khác 0 thoả mãn a+b+c=0. Tính $A=\left(1+\frac{a}{b}\right)+\left(1+\frac{b}{c}\right)+\left(1+\frac{c}{a}\right)$A=(1+ab )+(1+bc )+(1+ca )
Khó quá do anh thien
Bạn tham khảo câu hỏi tương tự.
Câu hỏi của Đào Thị Lan Nhi - Toán lớp 7 - Học trực tuyến OLM
vì a+b+c=0 => a+b= -c; b+c=-a; c+a=-b
(1+a/b)(1+b/c)(1+c/a)
=(a+b/b)(b+c/c)(a+c/a)
= (-c/b)(-a/c)(-b/a)
=-1
Thay a = -2 ; b = 1 ; c = 1 ( vì -2 + 1 + 1 = 0 )
Ta có : \(A=\left(1+\frac{-2}{1}\right)\left(1+\frac{1}{1}\right)\left(1+\frac{1}{-2}\right)\)
\(A=-1.2..\frac{1}{2}\)
\(A=-1\)
\(1\)
+ TH1 : \(a+b+c=0\Rightarrow\frac{a+b+c}{2}=0\)
\(\Rightarrow\hept{\begin{cases}a+b-2=0\\b+c+1=0\\c+a+1=0\end{cases}}\)\(\Rightarrow\hept{\begin{cases}a+b+c=c+2=0\\a+b+c=a-1=0\\a+b+c=b-1=0\end{cases}}\)\
\(\Rightarrow\hept{\begin{cases}a=1\\b=1\\c=-2\end{cases}}\left(TM\right)\)
+ TH2 : \(a+b+c\ne0\)
\(\frac{a+b-2}{c}=\frac{b+c+1}{a}=\frac{c+a+1}{b}\)\(=\frac{2\left(a+b+c\right)}{a+b+c}=2\) ( Theo tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\hept{\begin{cases}a+b-2=2c\\b+c+1=2a\\c+a+1=2b\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}a+b+c=3c+2\\a+b+c=3a-1\\a+b+c=3b-1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}3c+2=4\\3a-1=4\\3b-1=4\end{cases}}\) \(\left(do\frac{a+b+c}{2}=2\Rightarrow a+b+c=4\right)\)
\(\Rightarrow\hept{\begin{cases}a=b=\frac{5}{3}\\c=\frac{2}{3}\end{cases}\left(TM\right)}\)
Vậy \(\hept{\begin{cases}a=b=1\\c=-2\end{cases}}\) hoặc \(\hept{\begin{cases}a=b=\frac{5}{3}\\c=\frac{2}{3}\end{cases}}\)
Ta có: a+b+c=0 => a+b=-c;b+c=-a;a+c=-b
=>\(A=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{b+a}{b}.\frac{c+b}{c}.\frac{a+c}{a}=\frac{-c}{b}.\frac{-a}{c}.\frac{-b}{a}=-1\)
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
Ta có:
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Leftrightarrow\)
\(\frac{b+c}{a}=\frac{a+c}{b}=\frac{a+b}{c}=\frac{b+c+a+c+a+b}{a+b+c}=2\)
\(\Rightarrow P=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}=3.2=6\)
bài này có 2 trường hợp nhé =))
\(\frac{a}{b+c}=\frac{b}{a+c}=\frac{c}{a+b}\Rightarrow1+\frac{a}{b+c}=1+\frac{b}{a+c}=1+\frac{c}{a+b}\)
\(\Rightarrow\frac{a+b+c}{b+c}=\frac{a+b+c}{a+c}=\frac{a+b+c}{a+b}\)
\(TH1:a+b+c=0\)
\(\Rightarrow\hept{\begin{cases}b+c=-a\\a+c=-b\\a+b=-c\end{cases}\Rightarrow P=\frac{-a}{a}+\frac{-b}{b}+\frac{-c}{c}=-3}\)
\(TH2:a+b+c\ne0\)
\(\Rightarrow\hept{\begin{cases}b+c=a+c\Rightarrow a=b\\a+c=a+b\Rightarrow c=b\\a+b=b+c\Rightarrow a=c\end{cases}\Rightarrow a=b=c}\)
\(\Rightarrow P=\frac{a+a}{a}+\frac{b+b}{b}+\frac{c+c}{c}=2.3=6\)
Vậy P=-3 hay P=6