Question

Cho \(a+c=2b\) và 2bd = c(b+d), (b,d \(\ne0\))

CMR: \(\dfrac{a}{b}=\dfrac{c}{d}\)

Rút gọn:

\(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\) và \(\dfrac{5\left(3.7^{15}-19.17^{14}\right)}{7^{14}+3.7^{15}}\)

Answer 1

Bài 1 :

Ta có :

\(a+c=2b\left(1\right)\)

\(2bd=c\left(b+d\right)\left(2\right)\)

Thay \(\left(1\right)\) vào \(\left(2\right)\) ta được :

\(\left(a+c\right)d=c\left(b+d\right)\)

\(\Leftrightarrow ad+cd=cb+cd\)

\(\Leftrightarrow ad=cb\)

\(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\rightarrowđpcm\)

Answer 2

Bài 2 :

\(a,\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)

\(=\dfrac{5^{21}\left(2.5-9\right)}{5^{20}}\)

\(=5\left(10-9\right)\)

\(=5\)

b, \(\dfrac{5\left(3.7^{15}-19.17^{14}\right)}{7^{14}+3.7^{15}}\)

\(=\dfrac{5.2.7^{14}}{10.7^{15}}\)

\(=\dfrac{1}{7}\)