Cho tỉ lệ thức\(\frac{a}{b}=\frac{c}{d}\). Chứng minh :
d) \(\frac{a+c}{b+d}=\frac{a-c}{b-d}\) f) \(\frac{\left(a+b\right)^2}{\left(c+d\right)^2}=\frac{ab}{cd}\) h) \(\frac{7a^2+ab}{11a^2-8b^2}=\frac{7c^2+3cd}{11c^2-8d^2}\)
e) \(\frac{3a+5b}{2a-7b}=\frac{3c+5d}{2c-7d}\) g) \(\left(\frac{a+b}{c+d}\right)^2=\frac{a^2+b^2}{c^2+d^2}\) i) \(\left(\frac{a-b}{c-d}\right)^{2000}=\frac{a^{2000}+b^{2000}}{c^{2000}+d^{2000}}\)
Đặt \(\frac{a}{b}\)=\(\frac{c}{d}\)=k \(\hept{\begin{cases}a=kb\\c=kd\end{cases}}\)
Ta có: \(\frac{a+c}{b+d}\)= \(\frac{kb+kd}{b+d}\)=\(\frac{k\left(b+d\right)}{b+d}\)=k (1)
\(\frac{a-c}{b-d}\)= \(\frac{kb-kd}{b-d}\)=\(\frac{k\left(b-d\right)}{b-d}\)=k (2)
Từ (1) và (2) =>\(\frac{a+c}{b+d}\)=\(\frac{a-c}{b-d}\)
