Question

Cho a,b,c \(\ne0\), đôi một khác nhau thoả mãn:

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

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

Answer 1

Đề sai hay sao á, k rút gọn được.

fix: \(a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)\)

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

Lời giải:

Từ \(a\left(y+z\right)=b\left(z+x\right)=c\left(x+y\right)\)

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

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

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

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