Question

[Nguyễn Duy Sơn]

CHO BIẾT \(\frac{AB}{CD}=\frac{A'B'}{C'D'}=\frac{4}{5}\).TÌM CÁC TỈ SỐ:

A,\(\frac{AB+CD}{CD}\)

B,\(\frac{A'B'}{C'D'-A'B'}\)

C,\(\frac{AB}{A'B'}\)BIẾT \(3\cdot CD=C'D'\)

Answer 1

[Thám Tử THCS Nguyễn Hiếu]

a) \(\frac{AB+CD}{CD}=\frac{AB}{CD}+1\)Hay \(\frac{AB+CD}{CD}=\frac{4}{5}+1=\frac{9}{5}\)

b) \(\frac{C'D'-A'B'}{A'B'}=\frac{C'D'}{A'B'}-1\)Hay \(\frac{C'D'-A'B'}{A'B'}=\frac{5}{4}-1=\frac{1}{4}\)

=> \(\frac{A'B'}{C'D'-A'B'}=4\)

c) Ta có: 3CD = C'D' => \(\frac{CD}{C'D'}=\frac{1}{3}\)

Mà \(\frac{CD}{C'D'}=\frac{AB}{A'B'}\) nên \(\frac{AB}{A'B'}=\frac{1}{3}\)