Rút gọn : a) \(\dfrac{6^9.2^{10}+12^{10}}{2^{19}.27^3+15.4^9.9^4}\)
b) \(\dfrac{1+x^2+x^4+...+x^{22}+x^{24}+x^{26}}{1+x^4+x^8+...+x^{16}+x^{20}+x^{24}}\)
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
SS
25 tháng 6 2019
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8-6^8.20}\)
\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8-\left(2.3\right)^8.2^2.5}\)
\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8-2^{10}.3^8.5}\)
\(A=\frac{2^{10}.\left(3^8-3^9\right)}{2^{10}.3^8.\left(1-5\right)}=\frac{3^8-3^9}{3^8.\left(-4\right)}=\frac{3^8.\left(1-3\right)}{3^8.\left(-4\right)}=\frac{-2}{-4}=\frac{1}{2}\)
Vậy A = \(\frac{1}{2}\)
\(B=\frac{2^{19}.27^3+15.4^9.9^4}{6^9.2^{10}+12^{10}}\)
\(B=\frac{2^{19}.\left(3^3\right)^3+3.5.\left(2^2\right)^9.\left(3^2\right)^4}{\left(2.3\right)^9.2^{10}+\left(2^2.3\right)^{10}}\)
\(B=\frac{2^{19}.3^9+3.5.2^{18}.3^8}{2^9.3^9.2^{10}+2^{20}.3^{10}}\)
\(B=\frac{2^{19}.3^9+3^9.2^{18}.5}{2^{19}.3^9+2^{20}.3^{10}}\)
\(B=\frac{2^{18}.3^9.\left(2+5\right)}{2^{19}.3^9\left(1+2.3\right)}=\frac{7}{2.7}=\frac{1}{2}\)
Vậy B = \(\frac{1}{2}\)
KT
2
22 tháng 2 2021
Ta có: \(\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^{26}+x^{24}+x^{22}+...+x^2+1}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{26}+x^{22}+...+x^2\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{x^2\left(x^{24}+x^{20}+...+1\right)+\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)}\)
\(=\dfrac{x^{24}+x^{20}+x^{16}+...+x^4+1}{\left(x^{24}+x^{20}+x^{16}+...+1\right)\left(x^2+1\right)}\)
\(=\dfrac{1}{x^2+1}\)
OH
22 tháng 2 2021
x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1x24+x20+x16+...+x4+1x26+x24+x22+...+x2+1
=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1(x26+x22+...+x2)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)=x24+x20+x16+...+x4+1x2(x24+x20+...+1)+(x24+x20+x16+...+x4+1)
=x24+x20+x16+...+x4+1(x24+x20+x16+...+1)(x2+1)
FT
17 tháng 6 2017
BÀI 2
60%.x + 0,4.x + x:3= 2
\(\frac{60}{100}\)x + \(\frac{4}{10}\).x + x. \(\frac{1}{3}\)=2
\(\frac{3}{5}\).x + \(\frac{2}{5}\).x + x.\(\frac{1}{3}\)=2
(\(\frac{3}{5}\)+ \(\frac{2}{5}\)+ \(\frac{1}{3}\)) .x =2
\(\frac{4}{3}\).x =2
x = 2: \(\frac{4}{3}\)
x = \(\frac{3}{2}\)
Vậy x=\(\frac{3}{2}\)
k cho mik nha các bạn
17 tháng 6 2017
Bài 2 :
60%x + 0.4x + x : 3 = 2
\(x.\left(\frac{60}{100}+\frac{2}{5}+\frac{1}{3}\right)\)= 2
\(x.\frac{4}{3}\)= 2
\(x=2.\frac{3}{4}\)
\(x=1.5\)
NM
1
10 tháng 1 2018
Ta có:
\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)
Xét \(M=x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1\)
\(\Rightarrow x^4M=x^{28}+x^{24}+x^{20}+x^{16}+...+x^8+x^4\)
\(\Rightarrow x^4M-M=\left(x^{28}+x^{24}+x^{20}+...+x^8+x^4\right)-\left(x^{24}+x^{20}+x^{16}+...+x^4+1\right)\)
\(\Rightarrow\left(x^4-1\right)M=x^{28}-1\)
\(\Rightarrow M=\dfrac{x^{28}-1}{x^4-1}\)
Xét \(N=x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1\)
\(\Rightarrow x^2N=x^{28}+x^{26}+x^{24}+x^{20}+...+x^4+x^2\)
\(\Rightarrow x^2N-N=\left(x^{28}+x^{26}+x^{24}+...+x^4+x^2\right)-\left(x^{26}+x^{24}+x^{22}+...+x^2+1_{ }\right)\)
\(\Rightarrow\left(x^2-1\right)N=x^{28}-1\)
\(\Rightarrow N=\dfrac{x^{28}-1}{x^2-1}\)
Ta có:
\(\dfrac{x^{24}+x^{20}+x^{16}+x^{12}+...+x^4+1}{x^{26}+x^{24}+x^{22}+x^{20}+...+x^2+1}\)
\(=\dfrac{M}{N}=\dfrac{\dfrac{x^{28}-1}{x^4-1}}{\dfrac{x^{28}-1}{x^2-1}}\)
\(=\dfrac{x^{28}-1}{x^4-1}.\dfrac{x^2-1}{x^{28}-1}=\dfrac{x^2-1}{x^4-1}\)
\(=\dfrac{x^2-1}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{1}{x^2+1}\)
Chúc bạn học tốt!
TH
Thầy Hùng Olm
Manager
VIP
18 tháng 4 2023
A= \(\dfrac{10.11.\left(1+5.5+7.7\right)}{11.12.\left(1+5.5+7.7\right)}=\dfrac{10}{12}=\dfrac{5}{6}\)
a: \(=\dfrac{2^{19}\cdot3^9+2^{20}\cdot3^{10}}{2^{19}\cdot3^9+2^{18}\cdot3^9\cdot5}=\dfrac{2^{19}\cdot3^9\left(1+2\cdot3\right)}{2^{18}\cdot3^9\left(2+5\right)}=2\)