\(\text{a) }\dfrac{5^4\cdot20^4}{25^5\cdot4^5}=\dfrac{5^4\cdot\left(5\cdot4\right)^4}{\left(5^2\right)^5\cdot4^5}=\dfrac{5^4\cdot5^4\cdot4^4}{5^{10}\cdot4^5}=\dfrac{5^8\cdot4^4}{5^{10}\cdot4^5}=\dfrac{1}{5^2\cdot4}=\dfrac{1}{25\cdot4}=\dfrac{1}{100}\)

\(\text{b) }\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}=\dfrac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{2^7\cdot3^6}{2^5\cdot2^6\cdot3^5}=\dfrac{2^7\cdot3^6}{2^{11}\cdot3^5}=\dfrac{3}{2^4}=\dfrac{3}{16}\)

\(\text{c) }\dfrac{45^{10}\cdot5^{20}}{75^5}=\dfrac{\left(5\cdot9\right)^{10}\cdot5^{20}}{\left(25\cdot3\right)^5}=\dfrac{5^{10}\cdot9^{10}\cdot5^{20}}{25^5\cdot3^5}=\dfrac{5^{10}\cdot5^{20}\cdot\left(3^2\right)^{10}}{\left(5^2\right)^5\cdot3^5}=\dfrac{5^{30}\cdot3^{20}}{5^{10}\cdot3^5}=5^{20}\cdot3^{15}\)

\(\text{d) }\left(0.8\right)^5=\left(\dfrac{8}{10}\right)^5=\left(\dfrac{4}{5}\right)^5=\dfrac{4^5}{5^5}=\dfrac{64}{3125}\)

\(\text{e) }\dfrac{2^{15}\cdot9^4}{6^6\cdot8^3}=\dfrac{2^{15}\cdot\left(3^2\right)^4}{\left(2\cdot3\right)^6\cdot\left(2^3\right)^3}=\dfrac{2^{15}\cdot3^8}{2^6\cdot3^6\cdot2^9}=\dfrac{2^{15}\cdot3^8}{2^6\cdot2^9\cdot3^6}=\dfrac{2^{15}\cdot3^8}{2^{15}\cdot3^6}=3^2=9\)

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

Vì \(x⋮18;15;12\) nên \(\Rightarrow x\in BC_{\left(18;15;12\right)}\) \(\left(1\right)\)

\(\text{Ta có : }18=2\cdot3^2\\ 15=3\cdot5\\ 12+2^2\cdot3\\ \Rightarrow BCNN_{\left(18;15;12\right)}=3^2\cdot5\cdot2^2=180\\ \Rightarrow BC_{\left(18;15;12\right)}=\left\{180;360;540;...\right\}\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra : \(x\in\left\{180;360;540;...\right\}\)

Mà \(200\le x\le500\)

\(\Rightarrow x=360\)

Vậy \(x=360\)

Cho phép mình làm bài này theo cái đề do bạn Nguyễn Bảo Trung chữa lại nhé Nguyễn Lê Ngọc Thanh

\(A=\dfrac{11\cdot13\cdot15+33\cdot39\cdot45+55\cdot65\cdot75+99\cdot117\cdot135}{13\cdot15\cdot17+39\cdot45\cdot51+65\cdot75\cdot85+117\cdot135\cdot153}\text{ và }B=\dfrac{1111}{1717}\\ \)

\(\text{Ta có : }\)

\(A=\dfrac{11\cdot13\cdot15+33\cdot39\cdot45+55\cdot65\cdot75+99\cdot117\cdot135}{13\cdot15\cdot17+39\cdot45\cdot51+65\cdot75\cdot85+117\cdot135\cdot153}\\ A=\dfrac{11\cdot13\cdot15+3\cdot11\cdot13\cdot15+5\cdot11\cdot13\cdot15+9\cdot11\cdot13\cdot15}{13\cdot15\cdot17+3\cdot13\cdot15\cdot17+5\cdot13\cdot15\cdot17+9\cdot13\cdot15\cdot17}\\ A=\dfrac{11\cdot13\cdot15\left(1+3+5+9\right)}{13\cdot15\cdot17\left(1+3+5+9\right)}\\ A=\dfrac{11}{17}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left(1\right)\)

\(B=\dfrac{1111}{1717}\\ B=\dfrac{11\cdot101}{17\cdot101}\\ B=\dfrac{11}{17}\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\left(2\right)\)

Từ \(\left(1\right)\) và \(\left(2\right)\) suy ra :

\(A=B=\dfrac{11}{17}\)

\(\Rightarrow A=B\)

Vậy \(A=B\)

\(\text{Ta có : }\\ \dfrac{ }{ab}_+\dfrac{ }{ba}=\left(10a+b\right)+\left(10b+a\right)\\ =10a+b+10b+a\\ =\left(10a+a\right)+\left(b+10b\right)\\ =11a+11b\\ =11\left(a+b\right)\text{ }⋮\text{ }11\\ \text{ }\Rightarrow\dfrac{ }{ab}_+\dfrac{ }{ba}\text{ }⋮\text{ }11\text{ }\left(ĐPCM\right)\\ \text{ Vậy }\dfrac{ }{ab}_+\dfrac{ }{ba}\text{ }⋮\text{ }11\)

\(71\cdot64-32\cdot\left(-7\right)-32\cdot\left(-11\right)\\ =71\cdot2\cdot32-32\cdot\left(-7\right)-32\cdot\left(-11\right)\\ =32\left[142-\left(-7\right)-\left(-11\right)\right]\\ =32\cdot160\\ =5120\)

\(14x+36-37x=48-16x\\ \Leftrightarrow\left(14x-37x\right)+36=48-16x\\ \Leftrightarrow-23x+36=48-16x\\ \Leftrightarrow-23x-16x=48-36\\ \Leftrightarrow-7x=12\\ \Leftrightarrow x=-\dfrac{12}{7}\)