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Khách
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N
2
30 tháng 10 2020
1. \(x^{10}=25x^8\Leftrightarrow x^{10}:x^8=25\Leftrightarrow x^2=25=5^2\Leftrightarrow x=5\)
2. \(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=\frac{2^{40}}{2^{30}}=2^{10}\)
30 tháng 10 2020
1)\(x^{10}=25x^8\)
\(\Rightarrow x^{10}:x^8=25\)
\(\Rightarrow x^2=5^2\)
\(\Rightarrow\orbr{\begin{cases}x=5\\x=-5\end{cases}}\)
2)\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}\)
12 tháng 7 2016
\(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}\)
\(M=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)
\(M=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}\)
\(M=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}\)
\(M=2^{10}\)
\(M=1024\)
PA
22 tháng 7 2016
\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\times\left(2^{20}+1\right)}{2^{30}\times\left(2^{20}+1\right)}=2^{10}=1024\)
Chúc bạn học tốt ^^
LD
1
HT
11 tháng 12 2016
\(\frac{8^{20}+4^{20}}{4^{25}+64^5}\)=\(\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)=\(\frac{2^{60}+2^{40}}{2^{50}+2^{30}}\)=\(\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}\)=\(\frac{2^{40}}{2^{30}}\)= 210
IM
19 tháng 7 2016
\(M=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}\)
19 tháng 7 2016
\(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}\)
= \(\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}\)
DD
5 tháng 4 2020
\(M=\frac{4^{20}.\left(2^{20}+1\right)}{4^{15}.\left(4^{10}+1\right)}\)
\(M=4^5\)
\(M=1024\)
Chúc bạn học tốt cho mik k nha.Thanks
31 tháng 8 2020
Bài 1 : \(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}=1024\)
Bài 2 : a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)=> \(x^8=x^7\)
=> \(x^8-x^7=0\)
=> \(x^7\left(x-1\right)=0\)
=> \(x-1=0\Rightarrow x=1\)(vì x7 = 0 => x = 0 mà x \(\ne\)0 nên loại)
b) \(x^{10}-25x^8=0\)
=> \(x^8\left(x^2-25\right)=0\)
=> x8 = 0 hoặc x2 - 25 = 0
=> x = 0 hoặc x2 = 25
=> x = 0 hoặc x = \(\pm\)5
Bài 3 : a) \(\left(2x+3\right)^2=\frac{9}{121}=\left(\pm\frac{3}{11}\right)^2\)
=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
b) \(\left(3x-1\right)^3=-\frac{8}{27}=\left(-\frac{2}{3}\right)^3\)
=> 3x - 1 = -2/3
=> 3x = 1/3
=> x = 1/3 : 3 = 1/9
XO
31 tháng 8 2020
1) Ta có \(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{30}+1\right)}=2^{10}=1024\)
2) a) \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\)
=> x8 = x7
=> x8 - x7 = 0
=> x7(x - 1) = 0
=> \(\orbr{\begin{cases}x^7=0\\x-1=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=1\end{cases}}\)
Vậy x \(\in\left\{0;1\right\}\)
b) x10 = 25x8
=> x10 - 25x8 = 0
=> x8(x2 - 25) = 0
=> \(\orbr{\begin{cases}x^8=0\\x^2-25=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=\pm5\end{cases}}\)
Vậy \(x\in\left\{0;5;-5\right\}\)
3) \(\left(2x+3\right)^2=\frac{9}{121}\)
=> \(\left(2x+3\right)^2=\left(\frac{3}{11}\right)^2\)
=> \(\orbr{\begin{cases}2x+3=\frac{3}{11}\\2x+3=-\frac{3}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}2x=\frac{-30}{11}\\2x=-\frac{36}{11}\end{cases}}\Rightarrow\orbr{\begin{cases}x=-\frac{15}{11}\\x=-\frac{18}{11}\end{cases}}\)
Vậy \(x\in\left\{-\frac{15}{11};-\frac{18}{11}\right\}\)
b) \(\left(3x-1\right)^3=-\frac{8}{27}\)
=> \(\left(3x-1\right)^3=\left(-\frac{2}{3}\right)^3\)
=> \(3x-1=-\frac{2}{3}\)
=> \(3x=\frac{1}{3}\)
=> \(x=\frac{1}{9}\)
Vậy \(x=\frac{1}{9}\)
\(2^5\).\(9^5\).\(2^8\).\(9^8\)
=(\(2^5\).\(2^8\)).(\(9^5\).\(9^8\))
=\(^{2^{13}}\).\(9^{13}\)
=\(^{2.9^{13}}\)
=\(18^{13}\)