Bn ghi thiếu đề, mk sửa lại:
Cho \(\dfrac{a}{b}=\dfrac{c}{d}\). CMR: \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\)
Giải:
Đặt: \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{\left[b\left(k-1\right)\right]^2}{\left[d\left(k-1\right)\right]^2}=\dfrac{b^2}{d^2}\) (1)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}\) (2)
Từ (1) và (2) \(\Rightarrow\) \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{a^2+b^2}{c^2+d^2}\).
