Question

cho tỉ lệ thức  a/b=c/d chứng minh rằng a^2+ac/c^2-ac=b^2+bd/d^2-bd

Answer 1

Đặt \(\frac{a}{b}=\frac{c}{d}=k\)

=>a=bk; c=dk

\(\frac{a^2+ac}{c^2-ac}=\frac{\left(bk\right)^2+bk\cdot dk}{\left(dk\right)^2-bk\cdot dk}=\frac{b^2k^2+bd\cdot k^2}{d^2k^2-bd\cdot k^2}=\frac{b\cdot k^2\left(b+d\right)}{d\cdot k^2\left(d-b\right)}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)

\(\frac{b^2+bd}{d^2-bd}=\frac{b\left(b+d\right)}{d\left(d-b\right)}\)

Do đó: \(\frac{a^2+ac}{c^2-ac}=\frac{b^2+bd}{d^2-bd}\)