Question

[Bùi Tiến Đạt]

Cho tỉ lệ thức: \(\frac{a}{b}=\frac{c}{d}\)(b,d khác 0). CMR:

a)   \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)                                                         b)   \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

Answer 1

[Huỳnh Quang Sang]

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=> a = bk,c = dk

Do đó \(\frac{2a+3b}{2a-3b}=\frac{2bk+3b}{2bk-3b}=\frac{b\left(2k+3\right)}{b\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(1)

\(\frac{2c+3d}{2c-3d}=\frac{2dk+3d}{2dk-3d}=\frac{d\left(2k+3\right)}{d\left(2k-3\right)}=\frac{2k+3}{2k-3}\)(2)

Từ (1) và (2) => \(\frac{2a+3b}{2a-3b}=\frac{2c+3d}{2c-3d}\)

b) Vì \(\frac{a}{b}=\frac{c}{d}\)=> \(\frac{a}{c}=\frac{b}{d}\)

=> \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}\)

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

\(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)

Answer 2

[Nobi Nobita]

a) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

Đặt \(\frac{a}{c}=\frac{b}{d}=k\left(k\ne0\right)\)\(\Rightarrow a=ck\); \(b=dk\)

Ta có: \(\frac{2a+3b}{2a-3b}=\frac{2.ck+3.dk}{2.ck-3.dk}=\frac{k\left(2c+3d\right)}{k\left(2c-3d\right)}=\frac{2c+3d}{2c-3d}\)( đpcm )

b) Từ \(\frac{a}{b}=\frac{c}{d}\)\(\Rightarrow\frac{a}{c}=\frac{b}{d}\)

\(\Rightarrow\left(\frac{a}{c}\right)^2=\left(\frac{b}{d}\right)^2=\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2-b^2}{c^2-d^2}\)

mà \(\left(\frac{a}{c}\right)^2=\frac{a}{c}.\frac{a}{c}=\frac{a}{c}.\frac{b}{d}=\frac{ab}{cd}\)

\(\Rightarrow\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\)( đpcm )