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1.
a) \(-\frac{8}{27}=-\left(\frac{2}{3}\right)^3\)
b) \(\frac{81}{625}=\left(\frac{3}{5}\right)^4\)
2.
a) 27.81=2187=37
b) sai đề
1. \(\frac{x^7}{81}=27\Leftrightarrow x^7=2187\)
\(\Leftrightarrow x^7=3^7\Leftrightarrow x=3\)
2. \(\left(x^4\right)^2=\frac{x^{12}}{x^5}\Leftrightarrow x^8=x^7\)
\(\Leftrightarrow x^8-x^7=0\Leftrightarrow x^7\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^7=0\\x-1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy,...
3.\(x^{10}=25x^8\Leftrightarrow x^{10}-25x^8=0\)
\(\Leftrightarrow x^8\left(x^2-25\right)=0\Leftrightarrow x^8\left(x+5\right)\left(x-5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x^8=0\\x+5=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-5\\x=5\end{matrix}\right.\)
4. \(\left(3x-1\right)^3=\frac{-8}{27}\Leftrightarrow\left(3x-1\right)^3=\left(\frac{-2}{3}\right)^3\)
\(\Leftrightarrow3x-1=\frac{-2}{3}\Leftrightarrow3x=\frac{1}{3}\)
\(\Leftrightarrow x=\frac{1}{9}\)
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
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 \(x=\frac{1}{9}\)
\(a,7^6+7^5-7^4⋮55\)
\(7^4\left(7^2+7-1\right)⋮55\)
\(7^4\times55⋮55\left(dpcm\right)\)
\(8^{12}-2^{33}-2^{30}\)
\(=8^{12}-\left(2^3\right)^{11}-\left(2^3\right)^{10}\)
\(=8^{12}-8^{11}-8^{10}\)
\(=8^{10}\left(8^2-8-1\right)\)
\(=8^{10}\times55⋮55\left(dpcm\right)\)
a) 420 x 810 = (42)10 x 810 = 810 x 810 = 820
b) 8110x278 = ( 34 )10 x ( 33 )10 =340 x 330= 370
c) ) a12 : a8 = a4