Question

Cho \(\frac{a}{b}=\frac{c}{d}\)(b,d ko bằng 0). CM rằng \(\frac{ac}{bd}=\frac{a^2-c^2}{b^2-d^2}\)

Answer 1

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

áp dụng t/c dãy tỉ số bằng nhau ta có:

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

\(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a^2}{b^2}=\frac{a}{b}\cdot\frac{a}{b}=\frac{a}{b}\cdot\frac{c}{d}=\frac{ac}{bd}\left(dpcm\right)\)

Answer 2

Có \(\frac{a}{b}=\frac{c}{d}\left(b,d\ne0\right)\)

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

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

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

Answer 3

Sửa lại tí:

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

Answer 4

Siêu sao bóng đá sai hết rồi, sửa lại cũng sai 

\(\frac{a}{b}\ne\frac{a^2}{b^2}\)

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

Answer 5

Đặt \(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=kb\\c=kd\end{cases}}\). Thay vào,ta có:

\(\frac{ac}{bd}=\frac{kb.kd}{bd}=\frac{bdk^2}{1.bd}=k^2\) (1)

\(\frac{a^2-c^2}{b^2-d^2}=\frac{k^2b^2-k^2d^2}{b^2-d^2}=\frac{k^2\left(b^2-d^2\right)}{1\left(b^2-d^2\right)}=k^2\) (2)

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