\(\dfrac{a}{b+c}=\dfrac{b}{c+a}=\dfrac{c}{a+b}\\ \Leftrightarrow\dfrac{b+c}{a}=\dfrac{c+a}{b}=\dfrac{a+b}{c}=\dfrac{2\left(a+b+c\right)}{a+b+c}=2\\ \Leftrightarrow\left\{{}\begin{matrix}b+c=2a\left(1\right)\\c+a=2b\left(2\right)\\a+b=2c\left(3\right)\end{matrix}\right.\\ \Leftrightarrow\left(1\right)-\left(2\right)=b-a=2a-2b\Leftrightarrow a-b=0\Leftrightarrow a=b\\ \left(2\right)-\left(3\right)=c-b=2b-2c\Leftrightarrow b-c=0\Leftrightarrow b=c\\ \left(3\right)-\left(1\right)=a-c=2c-2a\Leftrightarrow a-c=0\Leftrightarrow a=c\)

Vậy \(a=b=c\)

Áp dụng BĐT Am-Gm ta được:

\(\dfrac{ab}{c}+\dfrac{bc}{a}\ge2\sqrt{\dfrac{ab^2c}{ca}}=2b^2\)

\(\dfrac{bc}{a}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{abc^2}{ab}}=2c^2\)

\(\dfrac{ab}{c}+\dfrac{ac}{b}\ge2\sqrt{\dfrac{a^2bc}{bc}}=2a^2\)

\(\Rightarrow\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}\ge a^2+b^2+c^2=1\)

Vậy giá trị nhỏ nhất của \(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ac}{b}=1\)

thay \(a^2=b.c\)vào biểu thức, ta có:

\(\frac{b.c+c^2}{b^2+b.c}=\frac{c.\left(c+b\right)}{b.\left(b+c\right)}=\frac{c}{b}\)

a)

\(c=\dfrac{a\cdot b\cdot c}{a\cdot b}=\dfrac{35}{-35}=-1\\ a=\dfrac{a\cdot b\cdot c}{b\cdot c}=\dfrac{35}{7}=5\\ b=\dfrac{b\cdot c}{c}=\dfrac{7}{-1}=-7\)

Vậy ...

b)

\(d=\dfrac{a\cdot b\cdot c\cdot d}{a\cdot b\cdot c}=\dfrac{120}{-30}=-4\\ c=\dfrac{a\cdot b\cdot c}{a\cdot b}=\dfrac{-30}{-6}=5\\ a=\dfrac{a\cdot b\cdot c}{b\cdot c}=\dfrac{-30}{-15}=2\\ b=\dfrac{a\cdot b}{a}=\dfrac{-6}{2}=-3\)

Vậy ...

c)

\(a+b+b+c+c+a=-1+1+6\\ 2a+2b+2c=6\\ 2\left(a+b+c\right)=6\\ a+b+c=3\\ a=\left(a+b+c\right)-\left(b+c\right)=3-1=2\\ b=\left(a+b+c\right)-\left(a+c\right)=3-6=-3\\ c=\left(a+b+c\right)-\left(a+b\right)=3-\left(-1\right)=4\)

Vậy ...

\(\frac{b}{a+b}=\frac{c}{b+c}=\frac{a}{a+c}\Rightarrow\frac{a+b}{b}=\frac{b+c}{c}=\frac{a+c}{a}\)

\(\Leftrightarrow\frac{a}{b}+1=\frac{b}{c}+1=\frac{c}{a}+1\)mà\(a,b,c>0\Rightarrow a+b+c\ne0\)

\(\Rightarrow\frac{a}{b}=\frac{b}{c}=\frac{c}{a}=\frac{a+b+c}{b+c+a}=1\Rightarrow a=b=c\)

\(\Rightarrow M=\frac{ab+bc+ac}{a^2+b^2+c^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2}=1\)

a² = b.c \(\Rightarrow\)\(\frac{a}{b}\)= \(\frac{c}{a}\)

Đặt \(\frac{a}{b}\)= \(\frac{c}{a}\) = k ( k \(\in\)Z) 

\(\Rightarrow\)a = b.k

          c = a.k

Ta có:

\(\frac{a+b}{a-b}\)= \(\frac{b.k+b}{b.k-b}\)= \(\frac{b.\left(k+1\right)}{b.\left(k-1\right)}\)= \(\frac{k+1}{k-1}\)(1)

\(\frac{c+a}{c-a}\)= \(\frac{a.k+a}{a.k-a}\)= \(\frac{a.\left(k+1\right)}{a.\left(k-1\right)}\)= \(\frac{k+1}{k-1}\)(2)

Từ (1) và (2) suy ra \(\frac{a+b}{a-b}\)= \(\frac{c+a}{c-a}\)