Chia câu hỏi ra đk bn 

Vầy ai giải nổi khó hỉu lắm

Đây là bài thực hiện phép tính nha mấy bạn(tính nhanh nếu có thể)

\(\Leftrightarrow\left(\dfrac{2}{5}\right)^n\left(\dfrac{2}{5}-1\right)=\dfrac{-12}{125}\)

\(\Leftrightarrow\left(\dfrac{2}{5}\right)^n\left(\dfrac{-3}{5}\right)=\dfrac{-12}{125}\)

\(\Leftrightarrow\left(\dfrac{2}{5}\right)^n=\dfrac{4}{25}=\left(\dfrac{2}{5}\right)^2\)

\(\Rightarrow n=2\)

\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^3\cdot3\right)^4+8^4\cdot3^5}\)

\(=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^4+2^{12}\cdot3^5}\)

\(=\frac{2^{12}\cdot3^4\cdot\left(3-1\right)}{2^{12}\cdot3^4\cdot\left(3+1\right)}=\frac24=\frac12\)

\(\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(=\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\frac{5^{10}\cdot7^3\cdot\left(1-7\right)}{5^9\cdot7^3\left(1+2^3\right)}=\frac{-6}{9}=-\frac23\)

\(\frac{2^{12}\cdot3^5-4^6\cdot9^2}{\left(2^3\cdot3\right)^4+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{\left(125\cdot7\right)^3+5^9\cdot14^3}\)

\(=\frac12+\frac23=\frac76\)

\(A=\dfrac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\dfrac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot7^3\cdot2^3}\)

\(=\dfrac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\dfrac{5^{10}\cdot7^3\left(1-7\right)}{5^9\cdot7^3\cdot9}\)

\(=\dfrac{1}{6}-\dfrac{5\cdot\left(-6\right)}{9}=\dfrac{1}{6}+\dfrac{10}{3}=\dfrac{21}{6}=\dfrac{7}{2}\)

\(\frac{\left(5^4-5^3\right)}{125^4}-\frac{64}{125}\)

\(=\frac{\left(625-125\right)}{500}-\frac{64}{125}\)

\(=\frac{500}{500}-\frac{64}{125}\)

\(=0-0,51\)

\(=-0,51\)

a) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\left(-\dfrac{7}{12}\right)\cdot1\dfrac{2}{5}\)

\(\Rightarrow\dfrac{1}{6}x=\left(-\dfrac{7}{12}\right)\cdot\dfrac{7}{5}\)

\(\Rightarrow\dfrac{1}{6}x=-\dfrac{49}{60}\)

\(\Rightarrow x=-\dfrac{49}{60}:\dfrac{1}{6}\)

\(\Rightarrow x=-\dfrac{49}{10}\) 

b) \(\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\dfrac{9}{4}\)

\(\Rightarrow\left(\dfrac{1}{5}-\dfrac{3}{2}x\right)^2=\left(\pm\dfrac{3}{2}\right)^2\)

+) \(\dfrac{1}{5}-\dfrac{3}{2}x=\dfrac{3}{2}\)

\(\Rightarrow\dfrac{3}{2}x=\dfrac{1}{5}-\dfrac{3}{2}\)

\(\Rightarrow\dfrac{3}{2}x=-\dfrac{13}{10}\)

\(\Rightarrow x=-\dfrac{13}{10}:\dfrac{3}{2}\)

\(\Rightarrow x=-\dfrac{13}{15}\)

+) \(\left(1,25-\dfrac{4}{5}x\right)^3=-125\)

\(\Rightarrow\left(\dfrac{5}{4}-\dfrac{4}{5}x\right)^3=\left(-5\right)^3\)

\(\Rightarrow\dfrac{5}{4}-\dfrac{4}{5}x=-5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{5}{4}+5\)

\(\Rightarrow\dfrac{4}{5}x=\dfrac{25}{4}\)

\(\Rightarrow x=\dfrac{25}{4}:\dfrac{4}{5}\)

\(\Rightarrow x=\dfrac{125}{16}\)

a, \(\dfrac{2}{3}\)\(x\) - \(\dfrac{1}{2}\)\(x\) = (- \(\dfrac{7}{12}\)). 1\(\dfrac{2}{5}\)

    \(x\).(\(\dfrac{2}{3}\) - \(\dfrac{1}{2}\)) = (- \(\dfrac{7}{12}\)) . \(\dfrac{7}{5}\)

    \(x\). \(\dfrac{1}{6}\) = - \(\dfrac{49}{60}\)

    \(x\)      = - \(\dfrac{49}{60}\).6

    \(x\)      = -\(\dfrac{49}{10}\)