Question

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

 

Answer 1

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)

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

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

Answer 2

Ai muốn kết bn ko!

Tiện thể tk mình luôn nha!

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