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 \(\dfrac{5\left(3.7^{15}-19.17^{14}\right)}{7^{14}+3.7^{15}}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
NA
2
R
24 tháng 1 2022
Đặt P= \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\) : \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
Có : \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)
= \(\dfrac{\left(2.5-9\right).5^{21}}{\left(5^2\right)^{10}}\)= \(\dfrac{\left(10-9\right).5^{21}}{5^{20}}\)=\(\dfrac{5^{21}}{5^{20}}\)= 5 (1)
Có: \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
= \(\dfrac{5.\left[7^{14}.\left(3.7-19\right)\right]}{\left[7^{15}.\left(3+7\right)\right]}\)=\(\dfrac{5.7^{14}.2}{7^{15}.10}\)=\(\dfrac{10.7^{14}}{7^{15}.10}\)=\(\dfrac{1}{7}\) (2)
Từ (1) và (2) suy ra:
A= 5:\(\dfrac{1}{7}\)=5.7=35
Vậy A=35 hay \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\):\(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)= 35
21 tháng 6 2022
Câu 2:
\(B=\dfrac{5^{21}\cdot\left(2\cdot5-9\right)}{5^{20}}\cdot\dfrac{7^{15}\left(7+3\right)}{15\cdot7^{15}-95\cdot7^{14}}\)
\(=\dfrac{5\cdot1}{1}\cdot\dfrac{7^{15}\cdot10}{7^{14}\cdot\left(15\cdot7-95\right)}\)
\(=5\cdot\dfrac{7\cdot10}{105-95}=5\cdot7=35\)
1 tháng 12 2018
Bài làm
\(C=\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(C=\frac{2.5^{21}.5-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{14}.7-19.7^{14}\right)}{7^{16}.7+3.7^{15}}\)
\(C=\frac{5^{21}.\left(2-9\right).5}{\left(5.5\right)^{10}}:\frac{5.[7^{15}.\left(3-19\right)].7}{7^{15}.\left(3+1\right).7}\)
\(C=\frac{5^{21}.\left(-7\right).5}{5^{10}.5^{10}}:\frac{5.7^{15}.\left(-16\right).7}{7^{15}.4.7}\)
\(C=\frac{5^{21}.\left(-35\right)}{5^{10}.5^{10}}:\frac{7^{15}.\left(-112\right).5}{7^{15}.28}\)
\(C=5.\left(-35\right):\frac{7^{15}.560}{7^{15}.28}\)
\(C=5.\left(-35\right):\frac{1.560}{1.28}\)
\(C=5.\left(-35\right):20\)
\(C=5.\left(-35\right).\frac{1}{20}\)
\(C=-\frac{175}{20}\)
\(C=-\frac{35}{4}\)
Vậy biểu thức trên \(C=\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)bằng \(C=-\frac{35}{4}\)
# Chúc bạn học tốt #
DH
11 tháng 2 2020
\(\frac{2.5^{22}-9.5^{21}}{25^{10}}:\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}\)
\(=\frac{5^{21}.\left(2.5-9\right)}{5^{20}}:\frac{5.7^{14}\left(3.7-19\right)}{7^{15}\left(7+3\right)}\)
\(=5:\frac{5.2}{7.10}\)
\(=5:\frac{1}{7}\)
\(=35\)
10 tháng 7 2018
Đặt A= \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\) : \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
Có : \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\)
= \(\dfrac{\left(2.5-9\right).5^{21}}{\left(5^2\right)^{10}}\)= \(\dfrac{\left(10-9\right).5^{21}}{5^{20}}\)=\(\dfrac{5^{21}}{5^{20}}\)= 5 (1)
Có: \(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)
= \(\dfrac{5.\left[7^{14}.\left(3.7-19\right)\right]}{\left[7^{15}.\left(3+7\right)\right]}\)=\(\dfrac{5.7^{14}.2}{7^{15}.10}\)=\(\dfrac{10.7^{14}}{7^{15}.10}\)=\(\dfrac{1}{7}\) (2)
Từ (1) và (2) suy ra:
A= 5:\(\dfrac{1}{7}\)=5.7=35
Vậy A=35 hay \(\dfrac{2.5^{22}-9.5^{21}}{25^{10}}\):\(\dfrac{5.\left(3.7^{15}-19.7^{14}\right)}{\left(7^{16}+3.7^{15}\right)}\)= 35
Chúc học tốt nhé
LH
20 tháng 8 2016
\(\frac{5\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.\left[7^{14}\left(3.7-19\right)\right]}{7^{15}\left(7+3\right)}=\frac{5.7^{14}.2}{7^{15}.10}=\frac{2}{7.2}=\frac{2}{14}=\frac{1}{7}\)
20 tháng 8 2016
a) \(\frac{2.5^{22}-9.5^{21}}{25^{10}}=\frac{5^{21}.\left(2.5-9\right)}{\left(5^2\right)^{10}}=\frac{5^{21}.1}{5^{20}}=5\)
b) \(\frac{5.\left(3.7^{15}-19.7^{14}\right)}{7^{16}+3.7^{15}}=\frac{5.7^{14}.\left(3.7-19\right)}{7^{15}.\left(7+3\right)}=\frac{5.7^{14}.2}{7^{15}.10}=\frac{1}{7}\)
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 :
\(\left(a+c\right)d=c\left(b+d\right)\)
\(\Leftrightarrow ad+cd=cb+cd\)
\(\Leftrightarrow ad=cb\)
\(\Leftrightarrow\dfrac{a}{b}=\dfrac{c}{d}\rightarrowđpcm\)
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, \(\dfrac{5\left(3.7^{15}-19.17^{14}\right)}{7^{14}+3.7^{15}}\)
\(=\dfrac{5.2.7^{14}}{10.7^{15}}\)
\(=\dfrac{1}{7}\)