Bài 3:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\left\{{}\begin{matrix}\dfrac{a}{c}=\dfrac{xa}{xc}=\dfrac{yb}{yd}\\\dfrac{a}{c}=\dfrac{za}{zc}=\dfrac{tb}{td}\end{matrix}\right.\)
\(\Rightarrow\dfrac{xa}{xc}=\dfrac{yb}{yd}=\dfrac{za}{zc}=\dfrac{tb}{td}\)
- APTCDTSBN, ta có:
\(\dfrac{xa}{xc}=\dfrac{yb}{yd}=\dfrac{za}{zc}=\dfrac{tb}{td}=\dfrac{xa+yb}{xc+yd}=\dfrac{za+tb}{zc+td}\)
\(\Rightarrow\dfrac{xa+yb}{za+tb}=\dfrac{xc+yd}{zc+td}\left(đpcm\right)\)
Bài 4:
\(\dfrac{a}{b}=\dfrac{c}{d}\Rightarrow\dfrac{a}{c}=\dfrac{b}{d}\Rightarrow\dfrac{a^2}{c^2}=\dfrac{a}{c}.\dfrac{b}{d}=\dfrac{ab}{cd}\left(1\right)\)
- ADTCDTSBN, ta có:
\(\)\(\dfrac{a^2}{c^2}=\dfrac{b^2}{d^2}=\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{a^2+b^2}{c^2+d^2}\left(2\right)\)
(1), (2) \(\Rightarrow\dfrac{ab}{cd}=\dfrac{a^2-b^2}{c^2-d^2}\)
- ADTCDTSBN, ta có:
\(\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}\)
\(\Rightarrow\dfrac{a^2}{c^2}=\left(\dfrac{a+b}{c+d}\right)^2\left(3\right)\)
(2), (3) \(\Rightarrow\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
