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Áp dụng tính chất dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}=\dfrac{\left(a+b+c\right)-\left(a-b+c\right)}{\left(a+b-c\right)-\left(a-b-c\right)}=\dfrac{a+b+c-a+b-c}{a+b-c-a+b+c}=\dfrac{2b}{2b}=1\)
\(\Rightarrow\dfrac{a+b+c}{a+b-c}=1\) \(\Rightarrow a+b+c=1\times\left(a+b-c\right)\) \(\Rightarrow a+b+c=a+b-c\) \(\Rightarrow\left(a+b+c\right)-\left(a+b-c\right)=0\) \(\Rightarrow a+b+c-a-b+c=0\) \(\Rightarrow2c=0\) \(\Rightarrow c=0\div2\) \(\Rightarrow c=0\)
Vậy \(c=0\).
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
b) Ta có (a+b+c+d)(a-b-c-d)=(a-b+c-d)(a+b-c-d) với dạng a.d = b.c
\(\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+b+c+d}{a+b-c-d}=\frac{a-b-c-d}{a-b+c-d}\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\left(1\right)\)
\(\frac{a}{b}=\frac{c}{d}=\frac{a-c}{b-d}\left(2\right)\)
Từ (1) và (2) => \(\frac{\left(a+b+c+d\right)=\left(a-b-c-d\right)}{\left(a+b-c-d\right)=\left(a-b+c-d\right)}\Rightarrow\frac{a+b+c+d}{a+b-c-d}=\frac{a-b-c-d}{a-b+c-d}\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
Từ \(\frac{a+b-c}{c}=\frac{a-b+c}{b}=\frac{-a+b+c}{a}\)
\(\Rightarrow\frac{a+b-c}{c}+2=\frac{a-b+c}{b}+2=\frac{-a+b+c}{a}+2\)
\(\Rightarrow\frac{a+b+c}{c}=\frac{a+b+c}{b}=\frac{a+b+c}{a}\)
Nếu a + b + c = 0
=> a + b = - c
=> b + c = - a
=> c + a = - b
Khi đó \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}=\frac{-a.\left(-b\right).\left(-c\right)}{abc}=-\frac{abc}{abc}=-1\)
Nếu \(a+b+c\ne0\)
\(\Rightarrow\frac{1}{c}=\frac{1}{b}=\frac{1}{a}\)
\(\Rightarrow a=b=c\)
Khi đó \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}=\frac{2a.2b.2c}{abc}=\frac{8.abc}{abc}=8\)
Vậy nếu a + b + c = 0 thì \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}=-1\)
nếu a + b + c \(\ne\)0 thì \(\frac{\left(a+b\right).\left(b+c\right).\left(c+a\right)}{abc}=8\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{a}{b}=\frac{c}{d}\)
\(\Leftrightarrow\frac{b}{a}=\frac{d}{c}\Leftrightarrow\frac{b}{a}+1=\frac{d}{c}+1\Leftrightarrow\frac{a+b}{a}=\frac{c+d}{c}\) (1)
\(\Rightarrow\frac{a}{a+b}=\frac{c}{c+d}\)
\(\frac{a}{b}=\frac{c}{d}\Leftrightarrow\frac{b}{a}=\frac{d}{c}\Leftrightarrow1-\frac{b}{a}=1-\frac{d}{c}\)
\(\Leftrightarrow\frac{a-b}{a}=\frac{c-d}{c}\Leftrightarrow\frac{a}{a-b}=\frac{c}{c-d}\) (2)
Nhân vế (1) và (2) lại ta được:
\(\frac{a+b}{a}\cdot\frac{a}{a-b}=\frac{c+d}{c}\cdot\frac{c}{c-d}\Rightarrow\frac{a+b}{a-b}=\frac{c+d}{c-d}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a+b+c}{a+b-c}=\dfrac{a-b+c}{a-b-c}=\dfrac{\left(a+b+c\right)-\left(a-b+c\right)}{\left(a+b-c\right)-\left(a-b-c\right)}=\dfrac{2b}{2b}=1\)
\(\Rightarrow\dfrac{a+b+c}{a+b-c}=1\)
\(\Rightarrow a+b+c=a+b-c\)
\(\Rightarrow c+c=a-a+b-b\)
\(\Rightarrow2c=0\\ \Rightarrow c=0\)