Question

a) \(\frac{5a-3b}{3a+2b}=\frac{5c-3d}{3c+2d}\)                     b)\(\frac{ac}{bd}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

cho \(\frac{a}{b}=\frac{c}{d}\) (a,b,c,d khác 0)

chứng minh ta có tỉ lệ thức trên.

Answer 1

b)\(\frac{ac}{bd}=\frac{bkdk}{bd}=k.k=k^2\)

\(\frac{\left(a+c\right)^2}{\left(b+d\right)^2}=\frac{\left(bk+dk\right)^2}{\left(b+d\right)^2}=\frac{\left[k\left(b+d\right)\right]^2}{\left(b+d\right)^2}=\frac{k^2.\left(b+d\right)^2}{\left(b+d\right)^2}=k^2\)

=> \(\frac{ac}{bd}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

Answer 2

Đặt k ( với k khác 0 , thuộc Z ) sao cho \(\frac{a}{b}=\frac{c}{d}=k\) => \(a=kb\)  /  \(c=dk\) .

a) Thế vào \(\frac{5a-b}{3a+2b}\) , ta có \(\frac{5kb-3b}{3kb+2b}\)\(=\frac{b\left(5k-3\right)}{b\left(3k+2\right)}\)\(=\frac{5k-3}{3k+2}\)  /  \(\frac{5c-3d}{3c+2d}=\frac{5dk-3d}{3dk-2d}=\frac{d\left(5k-3\right)}{d\left(3k+2\right)}=\frac{\left(5k+3\right)}{\left(3k+2\right)}\)

=> VT = VP