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\(A=\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5\left(3+1\right)}=\frac{2}{3.4}=\frac{1}{6}\)
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=\(\frac{2^{12}.3^5+2^{12}.3^4}{2^{12}.3^6+2^{12}.3^3}\)
=\(\frac{2^{12}\left(3^5+3^4\right)}{2^{12}\left(3^6+3^3\right)}\)
\(=\frac{324}{756}\)
=\(\frac{3}{7}\)
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\(\frac{2^{12}.3^5-4^6.81}{\left(2^2.3\right)^6+8^4.3^5}\)
\(=\frac{2^{12}.3^5-2^{12}.3^4}{2^{12}.3^6+2^{12}.3^5}\)
\(=\frac{2^{12}.\left(3^5-3^4\right)}{2^{12}.\left(3^6+3^5\right)}\)
\(=\frac{3^5-3^4}{3^6+3^5}=\frac{3^4.\left(3-1\right)}{3^5\left(3+1\right)}\)
\(=\frac{3^4.2}{3^5.4}=\frac{3^4.2}{3^4.3.4}=\frac{2}{12}=\frac{1}{6}\)
P/s: Hoq chắc ạ (: Ms lp 6 lm đại
\(\frac{x}{2}=\frac{y}{3}\)
\(\Leftrightarrow\frac{x}{8}=\frac{y}{12}\)(1)
\(\frac{y}{4}=\frac{z}{5}\)
\(\Leftrightarrow\frac{y}{12}=\frac{z}{15}\)(2)
Từ (1) (2)
\(\Rightarrow\frac{x}{8}=\frac{y}{12}=\frac{z}{15}=\frac{x+y-z}{8+12-15}=\frac{10}{5}=2\)
\(\Rightarrow\hept{\begin{cases}x=2.8\\y=2.12\\z=2.15\end{cases}\Rightarrow}\hept{\begin{cases}x=16\\y=24\\z=30\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\frac{2^5\cdot2^{12}\cdot2^6}{2^{24}}=\frac{2^{23}}{2^{24}}=\frac{1}{2}\)
Các phần kia tương tự, à bạn đăng 1 2 câu hỏi 1 lần thôi, đăng nhiều quá ko ai trả lời đâu
@-@
Thế à ! Vậy bạn hãy nhấp vào https://h.vn/hoi-dap/question/646555.html?pos=1792187 mà xem
Bài giải
\(\frac{2^{12}\cdot3^5-4^6\cdot81}{\left(2^2\cdot6\right)^6+8^4\cdot3^5}=\frac{2^{12}\cdot3^5-\left(2^2\right)^6\cdot3^4}{2^{12}\cdot6^6+\left(2^3\right)^4\cdot3^5}=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot2^6\cdot3^6+2^{12}\cdot3^5}=\frac{2^{12}\cdot3^4\left(3-1\right)}{2^{12}\cdot3^4\left(2^6\cdot3^2+3\right)}\)
\(=\frac{2}{64\cdot9+3}=\frac{2}{576+3}=\frac{2}{579}\)