a(y+z) = b(z+x) = c(x+y)

\(\frac{a\left(y+z\right)}{abc}=\frac{b\left(z+x\right)}{abc}=\frac{c\left(x+y\right)}{abc}\)

\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}\)

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

\(\frac{y+z}{bc}=\frac{z+x}{ac}=\frac{x+y}{ab}=\frac{\left(x+y\right)-\left(z+x\right)}{ab-ac}=\frac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\frac{\left(z+x\right)-\left(y+z\right)}{ac-bc}\)

\(\Rightarrow\frac{y-z}{a\left(b-c\right)}=\frac{z-x}{b\left(c-a\right)}=\frac{x-y}{c\left(a-b\right)}\)( đpcm )

Ta có: a.(y + z) = b.(x + z) = c.(x + y)

\(\Rightarrow\frac{a.\left(y+z\right)}{abc}=\frac{b.\left(x+z\right)}{abc}=\frac{c.\left(x+y\right)}{abc}\)

\(\Rightarrow\frac{y+z}{bc}=\frac{x+z}{ac}=\frac{x+y}{ab}\)

Áp dụng tính chất của dãy tỉ số = nhau ta có:

\(\frac{y+z}{bc}=\frac{x+z}{ac}=\frac{x+y}{ab}=\frac{\left(x+y\right)-\left(x+z\right)}{ab-ac}=\frac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\frac{\left(x+z\right)-\left(y+z\right)}{ac-bc}\)

                             \(=\frac{y-z}{a.\left(b-c\right)}=\frac{z-x}{b.\left(c-a\right)}=\frac{x-y}{c.\left(a-b\right)}\left(đpcm\right)\)

cho hỏi ai đây(trong lớp 7b)

\(\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\)

\(\Leftrightarrow\dfrac{a\left(y+z\right)}{abc}=\dfrac{b\left(z+x\right)}{abc}=\dfrac{c\left(x+y\right)}{abc}\)

\(\Leftrightarrow\dfrac{\left(x+y\right)-\left(z+x\right)}{ab-ac}=\dfrac{y-z}{a\left(b-c\right)}\)

\(\Leftrightarrow\dfrac{\left(y+z\right)-\left(x+y\right)}{bc-ab}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{\left(z+x\right)-\left(y+z\right)}{ac-bc}=\dfrac{x-y}{c\left(a-b\right)}\)

\(\Rightarrow\dfrac{y-z}{a\left(b-c\right)}=\dfrac{z-x}{b\left(c-a\right)}=\dfrac{x-y}{c\left(a-b\right)}\left(đpcm\right)\)

\(\hept{\begin{cases}\\\end{cases}^2_{ }}\)