b: \(ac=b^2;bd=c^2\)
=>\(\dfrac{a}{b}=\dfrac{b}{c};\dfrac{b}{c}=\dfrac{c}{d}\)
=>\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=k\)
=>\(c=dk;b=ck=dk\cdot k=dk^2;a=bk=dk^3\)
\(\dfrac{a}{d}=\dfrac{dk^3}{d}=k^3\)
\(\left(\dfrac{2a+3b-c}{2b+3c-d}\right)^3=\left(\dfrac{2\cdot dk^3+3\cdot dk^2-dk}{2\cdot dk^2+3\cdot dk-d}\right)^3\)
\(=\left(\dfrac{dk\left(2k^2+3k-1\right)}{d\left(2k^2+3k-1\right)}\right)^3=k^3\)
=>\(\dfrac{a}{d}=\left(\dfrac{2a+3b-c}{2b+3c-d}\right)^3\)
a: 
Đúng 1
Bình luận (0)
