cho a,b,c tỉ lệ nghịch với -x+y+z,x-y+z,x+y-z.cmr x,y,z tỉ lệ thuận với a(b+c),b(a+c),c(b+a)

0 bình luận về “cho a,b,c tỉ lệ nghịch với -x+y+z,x-y+z,x+y-z.cmr x,y,z tỉ lệ thuận với a(b+c),b(a+c),c(b+a)”

1. Giải thích các bước giải:

   $\rightarrow (-x+y+z)a=(x-y+z)b=(x+y-z)c$

   $\rightarrow \dfrac{-x+y+z}{\dfrac{1}{a}}= \dfrac{x-y+z}{\dfrac{1}{b}}= \dfrac{x+y-z}{\dfrac{1}{c}}=\dfrac{-x+y+z+x-y+z+x+y-z}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}$

   $\rightarrow \dfrac{-x+y+z}{\dfrac{1}{a}}= \dfrac{x-y+z}{\dfrac{1}{b}}= \dfrac{x+y-z}{\dfrac{1}{c}}=\dfrac{-x+y+z+x-y+z+x+y-z}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}$

   $\rightarrow \dfrac{-x+y+z}{\dfrac{1}{a}}= \dfrac{x-y+z}{\dfrac{1}{b}}= \dfrac{x+y-z}{\dfrac{1}{c}}=\dfrac{x+y+z}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}$

   $\rightarrow \dfrac{-x+y+z+x-y+z}{\dfrac{1}{a}+\dfrac{1}{b}}= \dfrac{x-y+z+x+y-z}{\dfrac{1}{b}+\dfrac{1}{c}}= \dfrac{x+y-z-x+y+z}{\dfrac{1}{c}+\dfrac{1}{a}}=\dfrac{x+y+z}{\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}}$

   $\rightarrow \dfrac{2z}{\dfrac{1}{a}+\dfrac{1}{b}}= \dfrac{2x}{\dfrac{1}{b}+\dfrac{1}{c}}= \dfrac{2y}{\dfrac{1}{c}+\dfrac{1}{a}}$

   $\rightarrow \dfrac{z}{\dfrac{a+b}{ab}}= \dfrac{x}{\dfrac{b+c}{bc}}= \dfrac{y}{\dfrac{c+a}{ca}}$

   $\rightarrow \dfrac{z}{\dfrac{c(a+b)}{abc}}= \dfrac{x}{\dfrac{a(b+c)}{abc}}= \dfrac{y}{\dfrac{b(c+a)}{cab}}$

   $\rightarrow \dfrac{z}{c(a+b)}= \dfrac{x}{a(b+c)}= \dfrac{y}{b(c+a)}$

   $\rightarrow \dfrac{x}{a(b+c)}= \dfrac{y}{b(c+a)}=\dfrac{z}{c(a+b)}$

   $\rightarrow x,y,z$ tỉ lệ thuận với $a(b+c), b(c+a), c(a+b)\rightarrow đpcm$

   Bình luận