Question

Cho \(\frac{a}{b}=\frac{c}{d}\) . CMR:

a) \(\frac{a}{b}=\frac{11a+9c}{11b+9d}\)            ;       b)    \(\frac{3a^2+5c^2}{3b^2+5d^2}=\frac{\left(a+c\right)^2}{\left(b+d\right)^2}\)

Answer 1

a) Ta có:

\(\frac{a}{b}=\frac{11a}{11b}\) và \(\frac{c}{d}=\frac{9c}{9d}\)

Mà \(\frac{a}{b}=\frac{c}{d}\) nên suy ra \(\frac{a}{b}=\frac{11a}{11b}=\frac{9c}{9d}\)

=> \(\frac{a}{b}=\frac{11a+9c}{11b+9d}\)

Answer 2

\(\text{a) Ta có: }\)

\(\frac{a}{b}=\frac{11a}{11b}\)\(\text{và }\)\(\frac{c}{d}=\frac{9c}{9d}\)

\(\text{Ma dau bai cho}\) \(\frac{a}{b}=\frac{c}{d}\) \(\Rightarrow\)\(\frac{a}{b}=\frac{11a}{11b}=\frac{9c}{9d}\)

\(\text{Vay }\)\(\frac{a}{b}=\frac{11a+9c}{11b+9d}\)

Answer 3

dựa vào công thức 2 phân số bằng nhau

a)\(\frac{a}{b}=\frac{11a+9c}{11b+9d}\) 

do \(\frac{a}{b}=\frac{c}{d}\)nên \(\frac{11a+9c}{11b+9d}=\frac{a\times\left(11+9\right)}{b\times\left(11+9\right)}\)

\(\Rightarrow\)tự làm tiếp đi