1: Đặt \(\frac{a}{b}=\frac{c}{d}=k\)
=>a=bk; c=dk
\(\frac{5a+3b}{5a-3b}=\frac{5\cdot bk+3b}{5bk-3b}=\frac{b\left(5k+3\right)}{b\left(5k-3\right)}=\frac{5k+3}{5k-3}\)
\(\frac{5c+3d}{5c-3d}=\frac{5\cdot dk+3d}{5\cdot dk-3d}=\frac{d\left(5k+3\right)}{d\left(5k-3\right)}=\frac{5k+3}{5k-3}\)
Do đó: \(\frac{5a+3b}{5a-3b}=\frac{5c+3d}{5c-3d}\)
2: \(\frac{2a-5b}{2a+5b}=\frac{2\cdot bk-5b}{2bk+5b}=\frac{b\left(2k-5\right)}{b\left(2k+5\right)}=\frac{2k-5}{2k+5}\)
\(\frac{2c-5d}{2c+5d}=\frac{2\cdot dk-5d}{2\cdot dk+5d}=\frac{d\left(2k-5\right)}{d\left(2k+5\right)}=\frac{2k-5}{2k+5}\)
Do đó: \(\frac{2a-5b}{2a+5b}=\frac{2c-5d}{2c+5d}\)
5: \(\frac{a^2+b^2}{c^2+d^2}=\frac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\frac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\frac{b^2}{d^2}\)
\(\frac{ab}{cd}=\frac{bk\cdot b}{dk\cdot d}=\frac{b^2}{d^2}\)
\(\left(\frac{a-b}{c-d}\right)^2=\left(\frac{bk-b}{dk-d}\right)^2=\left\lbrack\frac{b\left(k-1\right)}{d\left(k-1\right)}\right\rbrack^2=\left(\frac{b}{d}\right)^2=\frac{b^2}{d^2}\)
\(\left(\frac{2a+b}{2c+d}\right)^2=\left(\frac{2\cdot bk+b}{2\cdot dk+d}\right)^2=\left\lbrack\frac{b\left(2k+1\right)}{d\left(2k+1\right)}\right\rbrack^2=\left(\frac{b}{d}\right)^2=\frac{b^2}{d^2}\)
Do đó: \(\frac{a^2+b^2}{c^2+d^2}=\frac{ab}{cd}=\left(\frac{a-b}{c-d}\right)^2=\left(\frac{2a+b}{2c+d}\right)^2\)
