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Theo T/C dãy tỉ số bằng nhau
\(\frac{a+b}{c}=\frac{b+c}{a}=\frac{c+a}{b}=\frac{a+b+b+c+c+a}{a+b+c}=\frac{2\left(a+b+c\right)}{a+b+c}=2\)
\(\frac{a+b}{c}=2\Rightarrow a+b=2c\)
Tương tự ta có
\(b+c=2a\)
\(c+a=2b\)
Xét \(P=\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\left(\frac{a+b}{b}\right)\left(\frac{b+c}{c}\right)\left(\frac{c+a}{a}\right)\)
\(P=\frac{\left(a+b\right)\left(b+c\right)\left(c+a\right)}{abc}=\frac{2a\cdot2b\cdot2c}{abc}=8\)
Câu 2 :
\(x-y=7\)
\(\Rightarrow x=7+y\)
*)
\(B=\dfrac{3\left(7+y\right)-7}{2\left(7+y\right)+y}-\dfrac{3y+7}{2y+7+y}\)
\(=\dfrac{21+3y-7}{14+3y}-\dfrac{3y+7}{3y+7}\)
\(=\dfrac{14y+3y}{14y+3y}-1\)
\(=1-1\)
\(=0\)
Vậy B = 0
2/ Ta có :
\(B=\dfrac{3x-7}{2x+y}-\dfrac{3y+7}{2y+x}\)
\(=\dfrac{3x-\left(x-y\right)}{2x+y}-\dfrac{3y+\left(x-y\right)}{2y+x}\)
\(=\dfrac{3x-x+y}{2y+x}-\dfrac{3y+x-y}{2y+x}\)
\(=\dfrac{2x+y}{2x+y}-\dfrac{2y+x}{2y+x}\)
\(=1-1=0\)
\(P=\dfrac{a}{b+c}+\dfrac{b}{a+c}+\dfrac{c}{a+b}\\ \Rightarrow P+3=\left(\dfrac{a}{b+c}+1\right)+\left(\dfrac{b}{a+c}+1\right)+\left(\dfrac{c}{a+b}+1\right)\\ \Rightarrow P+3=\dfrac{a+b+c}{b+c}+\dfrac{a+b+c}{a+c}+\dfrac{a+b+c}{a+b}\\ =\left(a+b+c\right)\left(\dfrac{1}{b+c}+\dfrac{1}{a+c}+\dfrac{1}{a+b}\right)=2018.\dfrac{2021}{4034}=1011.000992\\ \Rightarrow P=1008.000992\)
Lời giải:
TH1: $a+b+c=0$
Khi đó: \(a+b=-c; b+c=-a; c+a=-b\)
\(\Rightarrow M=\frac{(-c)(-a)(-b)}{abc}=\frac{-abc}{abc}=-1\)
TH2: \(a+b+c\neq 0\)
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}=\frac{a+b-c+a-b+c-a+b+c}{c+b+a}=\frac{a+b+c}{a+b+c}=1\)
\(\Rightarrow \left\{\begin{matrix} a+b-c=c\\ a-b+c=b\\ -a+b+c=a\end{matrix}\right.\Rightarrow a+b=2c; a+c=2b; b+c=2a\)
\(\Rightarrow M=\frac{2c.2a.2b}{abc}=\frac{8abc}{abc}=8\)
Ta có:
\(\dfrac{a+b-c}{c}=\dfrac{a-b+c}{b}=\dfrac{-a+b+c}{a}\)
\(=\dfrac{a+b-c+a-b+c-a+b+c}{a+b+c}\)
\(=\dfrac{a+b+c}{a+b+c}=1\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{a+b-c}{c}=1\\\dfrac{a-b+c}{b}=1\\\dfrac{-a+b+c}{a}=1\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b-c=c\\a-b+c=b\\-a+b+c=a\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\a+c=2b\\b+c=2a\end{matrix}\right.\)(1)
Thay (1) vào M ta được
\(M=\dfrac{2c.2a.2b}{abc}=\dfrac{8abc}{abc}=8\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có:
\(\frac{2a+b+c+d}{a}=\frac{a+2b+c+d}{b}=\frac{a+b+2c+d}{c}=\frac{a+b+c+2d}{d}=\frac{2a+b+c+d-a-2b-c-d}{a-b}=1\)
\(\Rightarrow\left\{\begin{matrix}-a=b+c+d\\-b=a+c+d\\-c=b+c+d\\-d=a+b+c\end{matrix}\right.\)\(\Rightarrow a=b=c=d\)
\(M=\frac{a+b}{c+d}+\frac{b+c}{a+d}+\frac{c+d}{a+b}+\frac{a+d}{b+c}\)
\(\Rightarrow M=\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}+\frac{a+a}{a+a}\)
\(\Rightarrow M=1+1+1+1\)
\(\Rightarrow M=4\)
Vậy \(M=4\)