Đặt \(\dfrac{a}{b}=\dfrac{c}{d}=k\)
=>a=bk; c=dk
a: \(\dfrac{2a+11b}{2a-5b}=\dfrac{2bk+11b}{2bk-5b}=\dfrac{b\left(2k+11\right)}{b\left(2k-5\right)}=\dfrac{2k+11}{2k-5}\)
\(\dfrac{2c+11d}{2c-5d}=\dfrac{2dk+11d}{2dk-5d}=\dfrac{d\left(2k+11\right)}{d\left(2k-5\right)}=\dfrac{2k+11}{2k-5}\)
Do đó: \(\dfrac{2a+11b}{2a-5b}=\dfrac{2c+11d}{2c-5d}\)
b: \(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\left[\dfrac{b\left(k+1\right)}{d\left(k+1\right)}\right]^2=\left(\dfrac{b}{d}\right)^2\)
\(\dfrac{a^2+b^2}{c^2+d^2}=\dfrac{\left(bk\right)^2+b^2}{\left(dk\right)^2+d^2}=\dfrac{b^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}=\dfrac{b^2}{d^2}=\left(\dfrac{b}{d}\right)^2\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2+b^2}{c^2+d^2}\)
c: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2\cdot k}{d^2k}=\dfrac{b^2}{d^2}\)
\(\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}=\dfrac{\left(bk-b\right)^2}{\left(dk-d\right)^2}=\dfrac{b^2\cdot\left(k-1\right)^2}{d^2\cdot\left(k-1\right)^2}=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\dfrac{\left(a-b\right)^2}{\left(c-d\right)^2}\)
