đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)\(\Rightarrow\left\{{}\begin{matrix}a=bk\\c=dk\end{matrix}\right.\)
a) thay \(a=bk;c=dk\) ta có
\(\dfrac{5a+3b}{5a-3b}=\dfrac{5bk+3b}{5bk-3b}=\dfrac{b\left(5k+3\right)}{b\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(1)
\(\dfrac{5c+3d}{5c-3d}=\dfrac{5dk+3d}{5dk-3d}=\dfrac{d\left(5k+3\right)}{d\left(5k-3\right)}=\dfrac{5k+3}{5k-3}\)(2)
từ (1);(2)\(\Rightarrow\dfrac{5a+3b}{5a-3b}=\dfrac{5c+3d}{5c-3d}\)
b) thay \(a=bk;c=dk\) ta có
\(\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7(bk)^2+3bkb}{11(bk)^2-8b^2}=\dfrac{7b^2k^2+3b^2k}{11b^2k^2-8b^2}\)
\(=\dfrac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(3)
\(\dfrac{7c^2+3cd}{11c^2-8d^2}=\dfrac{7\left(dk\right)^2+3dkd}{11\left(dk\right)^2-8d^2}=\dfrac{7d^2k^2+3d^2k}{11d^2k^2-8d^2}\)
\(=\dfrac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\dfrac{7k^2+3k}{11k^2-8}\)(4)
từ (3);(4)\(\Rightarrow\dfrac{7a^2+3ab}{11a^2-8b^2}=\dfrac{7c^2+3cd}{11c^2-8d^2}\)
