Proste sumy II
: 8 lut 2024, o 18:17
Jeśli \(\displaystyle{ \frac{a-b}{c-d}=2}\) i \(\displaystyle{ \frac{a-c}{b-d}= 3, }\) to obliczyć \(\displaystyle{ \frac{a-d}{b-c}. }\)
\(\displaystyle{ 3=\frac{a-c}{b-d}=\frac{a-b+b-c}{b-c+c-d}=\frac{2(c-d)+(b-c)}{(b-c)+(c-d)}}\),
skąd
\(\displaystyle{ c-d=-2(b-c)}\)
\(\displaystyle{ \frac{a-d}{b-c}=\frac{a-b+b-c+c-d}{b-c}=\frac{3(c-d)+b-c}{b-c}=\frac{-5(b-c)}{b-c}=-5}\)