1/ CMR nếu a(y+z)=b(z+x)=c(x+y) (1). Trong đó a,b,c là các số khác nhau và khác 0 thì:
\(\dfrac{y-z}{a\left(b-c\right)}\)=\(\dfrac{z-x}{b\left(c-a\right)}\)=\(\dfrac{x-y}{c\left(a-b\right)}\)
2/cho \(\dfrac{bz-cz}{a}\)=\(\dfrac{cx-az}{b}\)=\(\dfrac{ay-bx}{c}\)
CM: \(\dfrac{x}{a}\)=\(\dfrac{y}{b}\)=\(\dfrac{z}{c}\)
3/biết \(\dfrac{a}{b}\)+\(\dfrac{b}{c}\)=1 và \(\dfrac{b}{b}\)+\(\dfrac{c}{c}\)=1 CMR abc + a'b'c' = 0
Từ \(a\left(y+z\right)=b\left(z+x\right)\), áp dụng t/c dãy tỉ số bằng nhau ta được
\(\dfrac{z+x}{a}=\dfrac{y+z}{b}=\dfrac{z+x-y-z}{a-b}=\dfrac{x-y}{a-b}\)
\(\Rightarrow\dfrac{z+x}{a}.\dfrac{1}{c}=\dfrac{y+z}{b}.\dfrac{1}{c}=\dfrac{x-y}{c\left(a-b\right)}\)(1)
Tương tự : từ \(b\left(z+x\right)=c\left(x+y\right)\)
\(\Rightarrow\dfrac{z+x}{c}=\dfrac{x+y}{b}=\dfrac{z+x-x-y}{c-b}=\dfrac{y-z}{c-b}\)\(\Rightarrow\dfrac{z+x}{c}.\dfrac{1}{a}=\dfrac{x+y}{b}.\dfrac{1}{a}=\dfrac{y-z}{c-b}.\dfrac{1}{a}\)
\(\Rightarrow\dfrac{z+x}{ac}=\dfrac{x+y}{ab}=\dfrac{y-z}{a\left(c-b\right)}\)(2)
từ \(a\left(y+z\right)=c\left(x+y\right)\)
\(\Rightarrow\dfrac{y+z}{c}=\dfrac{x+y}{a}=\dfrac{y+z-x-y}{c-a}=\dfrac{z-x}{c-a}\)\(\Rightarrow\dfrac{y+z}{c}.\dfrac{1}{b}=\dfrac{x+y}{a}.\dfrac{1}{b}=\dfrac{z-x}{c-a}.\dfrac{1}{b}\)
\(\Rightarrow\dfrac{y+z}{bc}=\dfrac{x+y}{ab}=\dfrac{z-x}{b\left(c-a\right)}\)(3)
Kết hợi (1);(2)(3) => ĐPCM
tik mik nha !!!
