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![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(M=\frac{bc}{a}+\frac{ac}{b}+\frac{ab}{c}=\frac{abc}{a^2}+\frac{abc}{b^2}+\frac{abc}{c^2}=abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)=8.\frac{3}{4}=6\)
Vậy M = 6
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt: \(\frac{a}{3}=\frac{b}{4}=\frac{c}{5}=k\)
Suy ra: a=3k, b=4k, c=5k
a.b.c=480 suy ra 3k.4k.5k=480
suy ra: 60.k^3=480
k^3=480:60=8
Vậy k=2
Thay vào ta có:a=6, b=8,c=10
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt\(\frac{a}{3}=\frac{b}{12}=\frac{c}{5}\)= k => a= 3k; b= 12k;c=5k
a.b.c = 22,5 => 3k.12k.5k = 22,5 = 180k3 = 22,5 => k3 = 0,125 => k = 0,5
Do đó:\(\frac{a}{3}=0,5=>a=1,5\)
\(\frac{b}{12}=0,5=>b=6\)
\(\frac{c}{5}=0,5=>c=2,5\)
Vậy...
Đặt \(\frac{a}{3}=\frac{b}{12}=\frac{c}{5}\)= k => a = 3k ; b = 12k ; c = 5k
a.b.c = 22,5 => 3k.12k.5k = 22,5 => 180k3 = 22,5 => k3 = 0,125 => k= 0,5
Do đó : \(\frac{a}{3}=0,5\Rightarrow a=1,5\)
\(\frac{b}{12}=0,5\Rightarrow b=6\)
\(\frac{c}{5}=0,5\Rightarrow c=2,5\)
Vậy ...
![](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)
tớ bik
tick đi tớ làm cho
chưa học thì đừng cmt!!!