a)\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{5a+3b}{5a-3b}=\frac{5kb+3b}{5kb-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\)(1)
\(\frac{5c+3d}{5c-3d}=\frac{5dk+3d}{5dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\)(2)
từ (1)(2);\(\Rightarrow\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)(đfcm)
b)\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow a=bk;c=dk\)
\(\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7b^2.k^2+3b^2k}{11b^2k^2-8b^2}=\frac{b^2\left(7k^2+3k\right)}{b^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\)(1)
\(\frac{7c^2+3cd}{11c^2-8d^2}=\frac{7.k^2.d^2+3d^2.k}{11d^2.k^2-8d^2}=\frac{d^2\left(7k^2+3k\right)}{d^2\left(11k^2-8\right)}=\frac{7k^2+3k}{11k^2-8}\)(2)
từ(1)(2)\(\Rightarrow\frac{7a^2+3ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)(đfcm)
