Bài 7 (Tớ không ghi lại đề)
a)Nếu $\dfrac{x}{a+2b+c}=\dfrac{y}{2a+b-c}=\dfrac{z}{4a-4a+c}$
$=>\dfrac{a+2b+c}{x}=\dfrac{2a+b-c}{y}=\dfrac{4a-4a+c}{z} (1)$
$\text{Áp dụng tính chất dãy tỉ số bằng nhau}$
Từ $(1)=>\dfrac{a+2b+c}{x}=\dfrac{2.(2a+b-c)}{y}=\dfrac{4a-4a+c}{z}=\dfrac{(a+2b+c)+2.(2a-b+c)+4a-4b+c}{x+2y+z}=\dfrac{9a}{x+2y+z}$
$+)\dfrac{2.(a+2b+c)}{2x}=\dfrac{2a+b-c}{y}=\dfrac{4a-4a+c}{z}=\dfrac{2a+4b+2c+2a+b-c-4a+4b-c}{2x+y-z}=\dfrac{9b}{2x+y-z}$
$+)\dfrac{4.(a+2b+c)}{4x}=\dfrac{4.(2a+b-c)}{4y}=\dfrac{4a-4a+c}{z}=\dfrac{4a+8b+4c-8a-4b+4c+4a-4b+c}{4x-4y+z}=\dfrac{9c}{4x-y+z}$
$=>\dfrac{9c}{4x-y+z}=\dfrac{9b}{2x+y-z}=\dfrac{9a}{x+2y+z}$
$=>\dfrac{c}{4x-y+z}=\dfrac{b}{2x+y-z}=\dfrac{a}{x+2y+z}$
b)
$\text{Áp dụng tính chất dãy tỉ số bằng nhau:}$
$\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{a+b+c}{b+c+d}$
$=>(\dfrac{a+b+c}{b+c+d})^{3}=\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{d}=\dfrac{a}{d}$
Vậy $(\dfrac{a+b+c}{b+c+d})^{3}=\dfrac{a}{d}$
