Bài 1:Thêm đề: Cho tỉ lệ thức \(\dfrac{a}{b}=\dfrac{c}{d}\)

a, \(\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)

Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)

Ta có:

\(\left\{{}\begin{matrix}\dfrac{bk}{3bk+b}=\dfrac{bk}{b\left(3k+1\right)}=\dfrac{k}{3k+1}\\\dfrac{dk}{3dk+d}=\dfrac{dk}{d\left(3k+1\right)}=\dfrac{k}{3k+1}\end{matrix}\right.\)

\(\Rightarrow\dfrac{a}{3a+b}=\dfrac{c}{3c+d}\)

b, \(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\)

Ta có: \(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\)

Đặt \(\dfrac{a}{c}=\dfrac{b}{d}=k\Rightarrow\left\{{}\begin{matrix}a=ck\\b=dk\end{matrix}\right.\)

Ta có: \(\left\{{}\begin{matrix}\dfrac{\left(ck-dk\right)^2}{\left(c-d\right)^2}=\dfrac{c^2k^2-2cdk^2+d^2k^2}{c^2-2cd+d^2}=\dfrac{k^2\left(c^2-2cd+d^2\right)}{c^2-2cd+d^2}=k^2\\\dfrac{ck.dk}{c.d}=k^2\end{matrix}\right.\) \(\Rightarrow\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{ab}{cd}\) Chúc bạn học tốt!!!