(505+495)+(327+673)

=1000+1000

=2000

\(327 + 495+673+505 = (327+673)+(495+505) = 1000+1000 = 2000\)

B

mik đang cần ak

B

a) - ( -25) - ( 202 + 125 ) + 402 = 100

b) - (-22) - ( 150 + 22 ) - 50 = (-200)

c) ( - 505 ) - (- 154 + 495 ) - ( - 846 ) = 990

d) ( 35 - 814 ) - ( 186 - 65 ) = (-1030)

hơi lệch

\(1,A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\\ 2A=\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{11}+...+\dfrac{1}{99}-\dfrac{1}{103}\\ 2A=\dfrac{1}{3}-\dfrac{1}{103}=\dfrac{100}{309}\\ A=\dfrac{100}{309}\cdot\dfrac{1}{2}=\dfrac{50}{309}\)

\(2,A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\\ A=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{9\cdot10}\right)\\ A=7\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)\\ A=7\left(1-\dfrac{1}{10}\right)=7\cdot\dfrac{9}{10}=\dfrac{63}{10}\)

Bài 1: 

Ta có: \(A=\dfrac{2}{3\cdot7}+\dfrac{2}{7\cdot11}+\dfrac{2}{11\cdot15}+...+\dfrac{2}{99\cdot103}\)

\(=\dfrac{1}{2}\left(\dfrac{4}{3\cdot7}+\dfrac{4}{7\cdot11}+\dfrac{4}{11\cdot15}+...+\dfrac{4}{99\cdot103}\right)\)

\(=\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{103}\right)\)

\(=\dfrac{1}{2}\cdot\dfrac{100}{309}=\dfrac{50}{309}\)

Bài 2: 

Ta có: \(A=\dfrac{7}{2}+\dfrac{7}{6}+\dfrac{7}{12}+\dfrac{7}{20}+\dfrac{7}{30}+\dfrac{7}{42}+\dfrac{7}{56}+\dfrac{7}{72}+\dfrac{7}{90}\)

\(=7\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dfrac{1}{4\cdot5}+\dfrac{1}{5\cdot6}+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}+\dfrac{1}{9\cdot10}\right)\)

\(=7\left(1-\dfrac{1}{10}\right)\)

\(=\dfrac{63}{10}\)

10¹⁰ và 48⁵⁰⁵

10¹⁰ = 2¹⁰.5¹⁰

48^{505 =} 2⁵⁰⁵.24⁵⁰⁵

Do 2⁵⁰⁵ > 2¹⁰

     24⁵⁰⁵ > 5¹⁰

Nên 48⁵⁰⁵ > 10¹⁰

Nhớ li ke mk nhé

10 lũy thừa 10 lớn hơn 48 lũy thừa 505 nếu sai đừng k mình

10^10 và 48^505

10^10=2^10.5^10

48^505=2^505.24^505

vì 2^505>2^10

24^505>5^10

=> 48^505>10^10

\(\frac{1}{1\times10}+\frac{1}{2\times15}+\frac{1}{3\times20}+...+\frac{1}{98\times495}+\frac{1}{99\times500}\)

\(=\frac{1}{1\times2\times5}+\frac{1}{2\times3\times5}+\frac{1}{3\times4\times5}+...+\frac{1}{98\times99\times5}+\frac{1}{99\times100\times5}\)

\(=\frac{1}{5}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{100}\right)=\frac{1}{5}\times\frac{99}{100}=\frac{99}{500}\)

\(\frac{1}{1\times10}+\frac{1}{2\times15}+\frac{1}{3\times20}+...+\frac{1}{98\times495}+\frac{1}{99\times500}\)

\(=\frac{1}{1\times2\times5}+\frac{1}{2\times3\times5}+\frac{1}{3\times4\times5}+...+\frac{1}{98\times90\times5}+\frac{1}{90\times100\times5}\)

\(=\frac{1}{5}\times\left(\frac{1}{1\times2}+\frac{1}{2\times3}+\frac{1}{3\times4}+...+\frac{1}{98\times99}+\frac{1}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(\frac{2-1}{1\times2}+\frac{3-2}{2\times3}+...+\frac{99-98}{98\times99}+\frac{100-99}{99\times100}\right)\)

\(=\frac{1}{5}\times\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{5}\times\left(1-\frac{1}{100}\right)=\frac{99}{500}\)

cảm ơn bạn Xyz và bạn Đoàn Đức Hà nhìu (^-^) thank you !!!!!!!!!!!!!!!!!