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22 tháng 4 2018

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

9 tháng 5 2018

a , 2/8 - 2/13 + 2/13 - 2/18 + 2/18 - 2/23 + ... + 2/253 - 2/258

= 2/8 - 2/258

= 125/516

b , 1/6 - 1/10 + 1/10 - 1/14 + 1/14 - 1/18 + ... + 1/402 - 1/406

= 1/6 - 1/406

= 100/609

\(B=\dfrac{1}{6.10}+\dfrac{1}{10.14}+...+\dfrac{1}{402.406}\\ 4B=\dfrac{4}{6.10}+\dfrac{4}{10.14}+...+\dfrac{4}{402.406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{1}{402}-\dfrac{1}{406}\\ 4B=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}\\B=\dfrac{\dfrac{100}{609}}{4}=\dfrac{25}{609} \)

26 tháng 8 2017

\(B=\dfrac{1}{6}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{14}+...+\dfrac{ 1}{402}-\dfrac{1}{406}\)

\(=\dfrac{1}{6}-\dfrac{1}{406}=\dfrac{100}{609}.\)

12 tháng 6 2017

\(1,\)

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

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

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

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

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

\(2,\)

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

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

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

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

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

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

\(3,\)

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

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

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

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

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

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

\(=1\)

\(4,\)

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

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

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

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

\(=-4\)

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

Mình mỏi tay lắm rồi

12 tháng 6 2017

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

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

25 tháng 2 2018

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

25 tháng 2 2018

chuẩn

2 tháng 4 2017

\(S=\dfrac{5^2}{1.6}+\dfrac{5^2}{6.11}+\dfrac{5^2}{11.16}+...+\dfrac{5^2}{96.101}\\ S=\dfrac{25}{1.6}+\dfrac{25}{6.11}+\dfrac{25}{11.16}+...+\dfrac{25}{96.101}\\ S=5.\left(\dfrac{5}{1.6}+\dfrac{5}{6.11}+\dfrac{5}{11.16}+...+\dfrac{5}{96.101}\right)\\ S=5.\left(1-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{16}+...+\dfrac{1}{96}-\dfrac{1}{101}\right)\\ S=5.\left(1-\dfrac{1}{101}\right)\\ S=5.\dfrac{100}{101}\\ S=\dfrac{500}{101}\)

1 tháng 6 2017

S=500/101

Huỳnh Huyền Linh làm đúng rùi!

2 tháng 5 2017

Ta có :

\(S=3+\dfrac{3}{2}+\dfrac{3}{2^2}+..............+\dfrac{3}{2^9}\)

\(\Rightarrow2S=2\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+..............+\dfrac{3}{2^9}\right)\)

\(\Rightarrow2S=6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+.............+\dfrac{3}{2^8}\)

\(\Rightarrow2S-S=\left(6+3+\dfrac{3}{2}+\dfrac{3}{2^2}+.......+\dfrac{3}{2^8}\right)-\left(3+\dfrac{3}{2}+\dfrac{3}{2^2}+......+\dfrac{3}{2^9}\right)\)

\(\Rightarrow S=6-\dfrac{3}{2^9}\)

\(\Rightarrow S=6-\dfrac{3}{512}=\dfrac{3069}{512}\)

24 tháng 4 2022

4S=1+24+342+....+2014420134S=1+24+342+....+201442013

4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)4S−S=3S=1+24+342+....+201442013−(14+242+343+....+201442014)

3S=1+(24−14)+(342−242)+......+(201442013−201342013)−2014420143S=1+(24−14)+(342−242)+......+(201442013−201342013)−201442014

3S=1+14+142+143+.....+142013−2014420143S=1+14+142+143+.....+142013−201442014

đặt A=1+14+142+143+....+142023A=1+14+142+143+....+142023

4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)4A−A=4+1+14+142+.....+142022−(1+14+142+....+142023)

3A=4−1420233A=4−142023

A=43−13.42023A=43−13.42023

⇒3S=43−13.42023−201442024⇒3S=43−13.42023−201442024

⇒S=49−19.42023−20143.42024⇒S=49−19.42023−20143.42024

do 49<48=1249<48=12

⇒S=49−19.42023−20143.42024<48=12(đpcm)