= 7/160

bn viết câu trả lời chi tiết đc ko, tính nhanh nếu có thể nha.

Lấy máy tính tính đi, câu này dễ mà còn hỏi cho bằng được nữa =="

\(1,\)

\(\dfrac{45^2.3^8.10^5}{5^5.3^7.18^5}\)

\(=\dfrac{3^4.5^2.3^8.2^5.5^5}{5^5.3^7.2^5.3^{10}}\)

\(=\dfrac{3^{12}.2^5.5^7}{5^5.3^{17}.2^5}\)

\(=\dfrac{1.5^2}{3^5.1}\)

\(=\dfrac{25}{243}\)

\(2,\)

\(\dfrac{4^5.9^4+2.6^9}{2^{10}.3^8+6^8.20}\)

\(=\dfrac{2^{10}.3^8+2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

\(=\dfrac{2^{10}.3^8+2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(=\dfrac{2^{10}.3^8.4}{2^{10}.3^8.6}\)

\(=\dfrac{2^{12}.3^8}{2^{11}.3^9}\)

\(=\dfrac{2}{3}\)

\(3,\)

\(\dfrac{15.3^{11}+4.27^4}{9^7}\)

\(=\dfrac{3.5.3^{11}+2^2.3^{12}}{3^{14}}\)

\(=\dfrac{5.3^{12}+2^2.3^{12}}{3^{14}}\)

\(=\dfrac{3^{12}\left(5+2^2\right)}{3^{14}}\)

\(=\dfrac{3^{12}.9}{3^{14}}\)

\(=\dfrac{3^{14}}{3^{14}}\)

\(=1\)

\(4,\)

\(\dfrac{4^7.2^8}{3.2^{15}.16^2-5^2\left(2^{10}\right)^2}\)

\(=\dfrac{2^{22}}{3.2^{23}-5^2.2^{20}}\)

\(=\dfrac{2^{22}}{2^{20}.\left(-1\right)}\)

\(=\dfrac{2^{22}}{-2^{20}}\)

\(=-4\)

* Mấy bài còn lại tương tự đấy bạn tự làm đi

Mình mỏi tay lắm rồi

P/s:khuyến khích tự làm,chỉ làm mẫu 1 câu:

1)\(\dfrac{45^2.3^8.10^5}{5^5.3^7.18^5}=\dfrac{\left(5.9\right)^2.3.3^7.\left(2.5\right)^5}{5^5.3^7.\left(2.9\right)^5}\)\(=\dfrac{5^2.9^2.3.3^7.2^5.5^5}{5^5.3^7.2^5.9^5}\)\(=\dfrac{5^2.9^2.3.1.1.1}{1.1.1.9^5}\)\(=\dfrac{5^2.9^2.3}{9^5}=\dfrac{5^2.9^2.3}{9^2.9^3}=\dfrac{5^2.3}{9^3}=\dfrac{75}{729}=\dfrac{25}{243}\)

\(S=\dfrac{2}{2\cdot6}+\dfrac{2}{6\cdot10}+...+\dfrac{2}{96\cdot100}\\ =\dfrac{1}{2}\left(\dfrac{4}{2\cdot6}+\dfrac{4}{6\cdot10}+...+\dfrac{4}{96\cdot100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{10}+...+\dfrac{1}{96}-\dfrac{1}{100}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{100}\right)=\dfrac{1}{2}\cdot\dfrac{49}{100}\\ =\dfrac{49}{100}\)

\(a,A=\dfrac{\dfrac{5}{4}+\dfrac{5}{5}+\dfrac{5}{7}-\dfrac{5}{11}}{\dfrac{10}{4}+\dfrac{10}{5}+\dfrac{10}{7}-\dfrac{10}{11}}\\ =\dfrac{5.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}{10.\left(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{7}-\dfrac{1}{11}\right)}\\ =\dfrac{5}{10}\\ =\dfrac{1}{2}\)

Vậy \(A=\dfrac{1}{2}\)

\(b,B=\dfrac{2+\dfrac{6}{5}-\dfrac{6}{7}-\dfrac{6}{11}}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =\dfrac{3.\left(\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}\right)}{\dfrac{2}{3}+\dfrac{2}{5}-\dfrac{2}{7}-\dfrac{2}{11}}\\ =3\)

Vậy \(B=3\)

Đoạn thẳng 1cm biểu diễn cho độ lớn của cường độ điện trường \(E = \frac{{{{10}^{ - 10}}}}{{6\pi {\varepsilon _0}}} = 0,6\)V/m.

Cường độ điện trường tại điểm cách Q một khoảng 2 cm:

\(E = \frac{{\left| Q \right|}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{\left| {{{6.10}^{ - 14}}} \right|}}{{4\pi 8,{{85.10}^{ - 12}}.0,{{02}^2}}} = 1,34\)V/m

Vectơ cường độ điện trường:

Cường độ điện trường tại điểm cách Q một khoảng 3 cm:

\(E = \frac{{\left| Q \right|}}{{4\pi {\varepsilon _0}{r^2}}} = \frac{{\left| {{{6.10}^{ - 14}}} \right|}}{{4\pi 8,{{85.10}^{ - 12}}.0,{{03}^2}}} = 0,6\)V/m

Vectơ cường độ điện trường:

\(e.\dfrac{7}{10}\cdot\dfrac{-3}{5}+\dfrac{7}{10}\cdot\dfrac{-2}{5}-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot\left[\left(\dfrac{-3}{5}\right)+\left(\dfrac{-2}{5}\right)\right]-\dfrac{3}{10}\)

\(=\dfrac{7}{10}\cdot1-\dfrac{3}{10}=\dfrac{4}{10}=\dfrac{2}{5}\)

\(f.\dfrac{-3}{7}\cdot\dfrac{5}{9}+\dfrac{4}{9}\cdot\dfrac{-3}{7}+2\dfrac{3}{7}\)

\(=\dfrac{-3}{7}\cdot\left(\dfrac{5}{9}+\dfrac{4}{9}\right)+\dfrac{17}{3}\)

\(=\dfrac{-3}{7}\cdot1+\dfrac{17}{3}=\dfrac{-9}{21}+\dfrac{119}{21}=\dfrac{110}{21}\)

\(g.\dfrac{5}{9}\cdot\dfrac{10}{17}+\dfrac{5}{9}\cdot\dfrac{9}{17}-\dfrac{5}{9}\cdot\dfrac{2}{17}\)

\(=\dfrac{5}{9}\cdot\left(\dfrac{10}{17}+\dfrac{9}{17}-\dfrac{2}{17}\right)\)

\(=\dfrac{5}{9}\cdot1=\dfrac{5}{9}\)