Question

Cho \(\frac{a}{b}\)=\(\frac{c}{d}\) Chứng minh rằng :a) \(\frac{ab}{cd}\)=\(\frac{a^2-b^2}{c^2-d^2}\)

                                              b) \(\left(\frac{a+b}{c+d}\right)^2\)= \(\frac{a^2+b^2}{c^2+d^2}\)

Answer 1

\(\frac{a}{b}=\frac{c}{d}\Rightarrow ad=bc=X\)

a/

 \(\frac{ab}{cd}=\frac{a^2-b^2}{c^2-d^2}\Leftrightarrow ab\left(c^2-d^2\right)=cd\left(a^2-b^2\right)\)\(\Leftrightarrow abc^2-abd^2=cda^2-cdb^2\Leftrightarrow a.c.X-b.d.X=a.c.X-b.d.X\)

Do đẳng thức cuối cùng đúng nên ta có \(\frac{ad}{bc}=\frac{a^2-b^2}{c^2-d^2}\) đúng

b/ \(\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{c}=\frac{b}{d}\Rightarrow\frac{a^2}{c^2}=\frac{b^2}{d^2}\)

Ta có: \(\frac{a^2}{c^2}=\frac{b^2}{d^2}=\frac{a^2+b^2}{c^2+d^2}=\frac{a^2-b^2}{c^2-d^2}=\frac{ab}{cd}=\frac{2ab}{2cd}=\frac{a^2+b^2+2ab}{c^2+d^2+2cd}=\frac{\left(a+b\right)^2}{\left(c+d\right)^2}\)