Question

cho \(\frac{a}{b}\)=  \(\frac{c}{d}\)chứng minh rằng:

a) \(\frac{a}{3a+b}=\frac{c}{3c+d}\)

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

c) \(\frac{a.b}{c.d}=\frac{a^2-b^2}{c^2-d^2}\)

Answer 1

Ta có: \(\frac{a}{b}=\frac{c}{d}\)

=> ad = bc 

=> 3ac + ad = 3ac + bc

=> a(3c + d) = c(3a + b)

=> \(\frac{a}{3a+b}=\frac{c}{3a+d}\) (ĐPCM)

Answer 2

b) Ta có:

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

đặt \(\frac{a}{c}=k\Rightarrow\frac{b}{d}=k\)

=> a = c.k; b = d.k

=> a² = c².k²; b² = d².k²

=> \(\frac{a^2+c^2}{b^2+d^2}=\frac{\left(c^2.k^2\right)+c^2}{\left(d^2.k^2\right)+d^2}\)= \(\frac{c^2\left(k^2+1\right)}{d^2\left(k^2+1\right)}\)=\(\frac{c^2}{d^2}=\frac{a^2}{b^2}=\frac{ac}{bd}\)

=> ĐPCM

Answer 3

Câu c tương tự câu b