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 \(x=\frac{1}{9}\)

cảm ơn các bạn nhiều lắm nuôn.I love u

Bài 1:

a; (\(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\)) x \(\dfrac{3}{4}\) = \(\dfrac{1}{4}\)

     \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) : \(\dfrac{3}{4}\)

      \(\dfrac{1}{4}\)\(x\) - \(\dfrac{1}{8}\) = \(\dfrac{1}{4}\) x \(\dfrac{4}{3}\)

    \(\dfrac{1}{4}x\) - \(\dfrac{1}{8}\) =  \(\dfrac{1}{3}\)

      \(\dfrac{1}{4}x\) = \(\dfrac{1}{3}\) + \(\dfrac{1}{8}\)

       \(\dfrac{1}{4}\) \(x\)=  \(\dfrac{8}{24}\) + \(\dfrac{11}{24}\)

          \(\dfrac{1}{4}x=\dfrac{11}{24}\)

           \(x=\dfrac{11}{24}:\dfrac{1}{4}\)

           \(x=\dfrac{11}{24}\times4\)

           \(x=\dfrac{11}{6}\) 

b; \(\dfrac{12}{5}:x\) = \(\dfrac{14}{3}\) x \(\dfrac{4}{7}\)

     \(\dfrac{12}{5}\) : \(x\) = \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) : \(\dfrac{8}{3}\)

            \(x\) = \(\dfrac{12}{5}\) x \(\dfrac{3}{8}\)

             \(x\) = \(\dfrac{9}{10}\)

c; \(x+\dfrac{2}{3}\) = 8:4 - 1

   \(x+\dfrac{2}{3}\) = 2 - 1

   \(x+\dfrac{2}{3}\) = 1

  \(x=1-\dfrac{2}{3}\)

  \(x\) = \(\dfrac{1}{3}\)

\(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}\)

\(M=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}\)

\(M=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}\)

\(M=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}\)

\(M=2^{10}\)

\(M=1024\)

Tính :

a) \(\frac{8^{14}}{4^{12}}=\frac{\left(2^3\right)^{14}}{\left(2^2\right)^{12}}=\frac{2^{42}}{2^{24}}=2^{18}=262144.\)

b) \(\frac{120^3}{40^3}=\left(\frac{120}{40}\right)^3=3^3=27.\)

Tìm x:

b) \(x^2-0,25=0\)

\(\Rightarrow x^2=0+0,25\)

\(\Rightarrow x^2=0,25\)

\(\Rightarrow\left[{}\begin{matrix}x=0,5\\x=-0,5\end{matrix}\right.\)

Vậy \(x\in\left\{0,5;-0,5\right\}.\)

c) \(\frac{8}{2^x}=2\)

\(\Rightarrow2^x=8:2\)

\(\Rightarrow2^x=4\)

\(\Rightarrow2^x=2^2\)

\(\Rightarrow x=2\)

Vậy \(x=2.\)

Chúc bạn học tốt!

a, 2 mũ 17 phần 2 mũ 14

b,=30

mình chỉ làm được 2 câu thôi,chúc cậu học tốt!

\(\frac{8^{20}+4^{20}}{4^{25}+64^5}=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\times\left(2^{20}+1\right)}{2^{30}\times\left(2^{20}+1\right)}=2^{10}=1024\)

Chúc bạn học tốt ^^

a; 

c; \(\dfrac{1}{10}\) + \(\dfrac{4}{20}\) + \(\dfrac{9}{30}\)+\(\dfrac{16}{40}+\dfrac{25}{50}+\dfrac{36}{60}+\dfrac{49}{70}+\dfrac{64}{80}+\dfrac{81}{90}\)

= \(\dfrac{1}{10}+\dfrac{2}{10}+\dfrac{3}{10}+\dfrac{4}{10}+\dfrac{5}{10}+\dfrac{6}{10}+\dfrac{7}{10}+\dfrac{8}{10}+\dfrac{9}{10}\)

= \(\dfrac{1+2+3+4+5+6+7+8+9}{10}\)

=  \(\dfrac{\left(1+9\right)+\left(2+8\right)+\left(3+7\right)+\left(4+6\right)+5}{10}\)

= \(\dfrac{10+10+10+10+5}{10}\)

= \(\dfrac{\left(10+10+10+10\right)+5}{10}\)

= \(\dfrac{10\times4+5}{10}\)

= \(\dfrac{45}{10}\)

= \(\dfrac{9}{2}\)

\(M=\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}\)

\(M=\frac{8^{20}+4^{20}}{4^{25}+64^5}\)

     = \(\frac{\left(2^3\right)^{20}+\left(2^2\right)^{20}}{\left(2^2\right)^{25}+\left(2^6\right)^5}=\frac{2^{60}+2^{40}}{2^{50}+2^{30}}=\frac{2^{40}\left(2^{20}+1\right)}{2^{30}\left(2^{20}+1\right)}=2^{10}\)