\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}=\dfrac{3y}{3b}=\dfrac{2z}{2c}=\dfrac{x-3y+2z}{a-3b+2c}=4\)

\(\frac{a}{x}=\frac{b}{y}=\frac{c}{z}=-4\)

\(\Rightarrow\frac{-a}{x}=\frac{-b}{y}=\frac{-c}{z}=4\)

\(=\frac{-a}{x}=\frac{3b}{-3y}=\frac{-2c}{2z}=\frac{-a+3b-2c}{x-3y+2z}=4\)

cho dãy tỉ số bằng nhau: 3a+b+2c/2a+c=a+3b+c/2b=a+2b+2c/b+c. tính giá trị biểu thức (a+b)(b+c)(c+a)/abc, với các mẫu số khác 0. Cái này cũng khó, nếu sai thì mong bạn thông cảm! 

1.

\(\dfrac{3a+b+2c}{2a+c}=\dfrac{a+3b+c}{2b}=\dfrac{a+2b+2c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c+2a+c}{2a+c}=\dfrac{a+b+c+2b}{2b}=\dfrac{a+b+c+b+c}{b+c}\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}+1=\dfrac{a+b+c}{2b}+1=\dfrac{a+b+c}{b+c}+1\)

\(\Leftrightarrow\dfrac{a+b+c}{2a+c}=\dfrac{a+b+c}{2b}=\dfrac{a+b+c}{b+c}\)

TH1: \(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a+b=-c\\b+c=-a\\c+a=-b\end{matrix}\right.\)

\(\Rightarrow A=\dfrac{\left(-c\right).\left(-a\right).\left(-b\right)}{abc}=-1\)

TH2: \(a+b+c\ne0\)

\(\Rightarrow\dfrac{1}{2a+c}=\dfrac{1}{2b}=\dfrac{1}{b+c}\)

\(\Rightarrow\left\{{}\begin{matrix}2a+c=b+c\\2b=b+c\\\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}2a=b\\b=c\end{matrix}\right.\) \(\Rightarrow2a=b=c\)

\(\Rightarrow P=\dfrac{\left(a+2a\right)\left(2a+2a\right)\left(2a+a\right)}{a.2a.2a}=9\)

Bài 2:

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x+3y-z}{z}=\frac{y+3z-x}{x}=\frac{z+3x-y}{y}=\frac{x+3y-z+y+3z-x+z+3x-y}{x+y+z}=\frac{3x+3y+3z}{x+y+z}=3\)

=>x+3y-z=3z; y+3z-x=3x; z+3x-y=3y

=>x+3y=4z; y+3z=4x; z+3x=4y

\(P=\left(\frac{x}{y}+3\right)\left(\frac{y}{z}+3\right)\left(\frac{z}{x}+3\right)\)

\(=\frac{x+3y}{y}\cdot\frac{y+3z}{z}\cdot\frac{z+3x}{x}\)

\(=\frac{4z}{y}\cdot\frac{4x}{z}\cdot\frac{4y}{x}=4\cdot4\cdot4=64\)

Bài 2:

Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:

\(\frac{x+3y-z}{z}=\frac{y+3z-x}{x}=\frac{z+3x-y}{y}=\frac{x+3y-z+y+3z-x+z+3x-y}{x+y+z}=\frac{3x+3y+3z}{x+y+z}=3\)

=>x+3y-z=3z; y+3z-x=3x; z+3x-y=3y

=>x+3y=4z; y+3z=4x; z+3x=4y

\(P=\left(\frac{x}{y}+3\right)\left(\frac{y}{z}+3\right)\left(\frac{z}{x}+3\right)\)

\(=\frac{x+3y}{y}\cdot\frac{y+3z}{z}\cdot\frac{z+3x}{x}\)

\(=\frac{4z}{y}\cdot\frac{4x}{z}\cdot\frac{4y}{x}=4\cdot4\cdot4=64\)