\(\dfrac{a}{b}=\dfrac{c}{d}\rightarrow a=n.c,b=n.d\)
Thay vào biểu thức \(\dfrac{a+c}{b+d}\), ta có:
\(\dfrac{a+c}{b+d}=\)\(\dfrac{a+na}{b+nb}=\dfrac{a.1+n.a}{b.1+n.b}=\dfrac{a.\left(1+n\right)}{b.\left(1+n\right)}\)\(=\dfrac{a.\left(n+1\right):\left(n+1\right)}{b.\left(n+1\right):\left(n+1\right)}=\dfrac{a}{b}\)
Vậy \(\dfrac{a+c}{b+d}=\dfrac{a}{b}\)
Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\). \(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
Ta có:
\(\dfrac{a+c}{b+d}=\dfrac{bk+dk}{b+d}=\dfrac{k\left(b+d\right)}{b+d}=k\)
Mà \(\dfrac{a}{b}=k\) nên \(\dfrac{a}{b}=\dfrac{a+c}{b+d}=k\)
\(\Rightarrowđpcm\)
Làm dùm : https://hoc24.vn/hoi-dap/question/512875.html
