cho a,b,c là ba số thực khác khác 0 tm đk \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
hãy tính \(b=\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\)
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27 tháng 11 2021
1.
Đặt \(\left(x;y;z\right)=\left(\dfrac{a}{a+b};\dfrac{b}{b+c};\dfrac{c}{c+a}\right)\Rightarrow\left\{{}\begin{matrix}1-x=\dfrac{b}{b+a}\\1-y=\dfrac{c}{b+c}\\1-z=\dfrac{a}{a+c}\end{matrix}\right.\)
\(\Rightarrow xyz=\dfrac{1}{8}\\ xyz=\left(1-x\right)\left(1-y\right)\left(1-z\right)\\ \Rightarrow xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)-xyz\\ \Rightarrow2xyz=1-\left(x+y+z\right)+\left(xy+yz+zx\right)=\dfrac{1}{4}\\ \Rightarrow x+y+z=\dfrac{3}{4}+xy+yz+zx\)
\(\RightarrowĐpcm\)
4 tháng 12 2021
Áp dụng t/c dtsbn ta có:
\(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}=\dfrac{a+b-c+b+c-a+c+a-b}{c+a+b}=\dfrac{a+b+c}{a+b+c}=1\)
\(\dfrac{a+b-c}{c}=1\Rightarrow a+b-c=c\Rightarrow a+b=2c\\ \dfrac{b+c-a}{a}=1\Rightarrow b+c-a=a\Rightarrow b+c=2a\\ \dfrac{c+a-b}{b}=1\Rightarrow c+a-b=b\Rightarrow c+a=2b\)
\(\left(1+\dfrac{b}{a}\right)\left(1+\dfrac{a}{c}\right)\left(1+\dfrac{c}{b}\right)\\ =\dfrac{\left(a+b\right)\left(a+c\right)\left(b+c\right)}{abc}\\ =\dfrac{2c.2b.2a}{abc}\\ =\dfrac{8abc}{abc}\\ =8\)
TT
20 tháng 12 2020
TH1 : a + b + c ≠ 0
Áp dụng t/c dãy tỉ số bằng nhau ta có
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b+b+c+a+c}{a+b+c}=2\)
\(\Rightarrow\left\{{}\begin{matrix}a+b=2c\\b+c=2a\\a+c=2b\end{matrix}\right.\)
Khi đó \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}=8\)
TH2 : a + b + c = 0
\(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\a+c=-b\\b+c=-a\end{matrix}\right.\)
Khi đó \(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\)
\(=\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{a+c}{a}=\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}=-1\)
20 tháng 12 2020
Xét 2 TH sau:
TH1: a+b+c=0
Khi đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =-1\)
TH2: a+b+c khác 0
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Suy ra: a+b=2c; b+c=2a; c+a=2b
Do đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)
20 tháng 12 2020
Xét 2 TH sau:
TH1: a+b+c=0
Khi đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{-c}{b}.\dfrac{-a}{c}.\dfrac{-b}{a}\\ =-1\)
TH2: a+b+c khác 0
Ta có:
\(\dfrac{a+b}{c}=\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\)
Suy ra: a+b=2c; b+c=2a; c+a=2b
Do đó:
\(M=\left(1+\dfrac{a}{b}\right)\left(1+\dfrac{b}{c}\right)\left(1+\dfrac{c}{a}\right)\\ =\dfrac{a+b}{b}.\dfrac{b+c}{c}.\dfrac{c+a}{a}\\ =\dfrac{2c}{b}.\dfrac{2a}{c}.\dfrac{2b}{a}\\ =8\)
Ta có : \(\dfrac{a+b-c}{c}=\dfrac{b+c-a}{a}=\dfrac{c+a-b}{b}\)
\(\Leftrightarrow\dfrac{a+b-c}{c}+2=\dfrac{b+c-a}{a}+2=\dfrac{c+a-b}{b}+2\)
\(\Leftrightarrow\dfrac{a+b+c}{c}=\dfrac{a+b+c}{a}=\dfrac{a+b+c}{b}\) (* )
Từ (*) => xảy ra 2 trường hợp : \(\left\{{}\begin{matrix}a=b=c\\a+b+c=0\end{matrix}\right.\)
Xét TH1 : Khi \(a=b=c.\)
\(b=\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)\left(1+\dfrac{a}{a}\right)=2.2.2=8\)
Xét TH2 : Khi \(a+b+c=0\) \(\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)
\(b=\left(\dfrac{a+b}{a}\right)\left(\dfrac{a+c}{c}\right)\left(\dfrac{b+c}{b}\right)=\left(\dfrac{-c}{a}\right)\left(\dfrac{-b}{c}\right)\left(\dfrac{-a}{b}\right)=-1.\)