\(\dfrac{7}{10^2+6\cdot10^2}=\dfrac{7}{100+6\cdot100}=\dfrac{7}{100+600}=\dfrac{7}{700}=\dfrac{1}{100}\)

\(=\dfrac{7}{10^2.\left(6+1\right)}=\dfrac{7}{100.7}=\dfrac{7}{700}=\dfrac{1}{100}\)

\(\dfrac{7}{10^2+6.10^2}\)=\(\dfrac{7}{100+6.100}=\dfrac{7}{700}=\dfrac{1}{100}\)

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

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

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:

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

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 \(B=3\)

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

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

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

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

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

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

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

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

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