Question

Cho TLT \(\frac{a}{b}=\frac{c}{d}\) hãy CTR

a) \(\frac{a}{b}=\frac{a+c}{b+d}\)

b) \(\frac{a-b}{b}=\frac{c-d}{d}\)

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

Answer 1

Đặt \(\frac{a}{b}=\frac{c}{d}=k\) => \(\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

a) Khi đó, ta có:

 +) \(\frac{bk}{b}=k\)

+) \(\frac{bk+dk}{b+d}=\frac{k\left(b+d\right)}{b+d}=k\)

=> \(\frac{a}{b}=\frac{a+c}{b+d}\)

b) Ta có:

 +) \(\frac{bk-b}{b}=\frac{b\left(k-1\right)}{b}=k-1\)

 +) \(\frac{dk-d}{d}=\frac{d\left(k-1\right)}{d}=k-1\)

=> \(\frac{a-b}{b}=\frac{c-d}{d}\)

Answer 2

c) Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)

Do đó \(\frac{ac}{bd}=\frac{bk.dk}{bd}=k^2\)(1)

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

Từ (1) và (2) suy ra \(\frac{ac}{bd}=\frac{a^2+c^2}{b^2+d^2}\left(=k^2\right)\left(đpcm\right)\)