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\(\frac{2}{1\cdot2}+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+\frac{2}{5\cdot6}\\ =2\left(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\right)\\ =2\left(\frac{2-1}{1\cdot2}+\frac{3-2}{2\cdot3}+\frac{4-3}{3\cdot4}+\frac{5-4}{4\cdot5}+\frac{6-5}{5\cdot6}\right)\\ =2\left(\frac{2}{1\cdot2}-\frac{1}{1\cdot2}+\frac{3}{2\cdot3}-\frac{2}{2\cdot3}+\frac{4}{3\cdot4}-\frac{3}{3\cdot4}+\frac{5}{4\cdot5}-\frac{4}{4\cdot5}+\frac{6}{5\cdot6}-\frac{5}{5\cdot6}\right)\\ =2\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...-\frac{1}{6}\right)\\ =2\left(1-\frac{1}{6}\right)\\ =2\cdot\frac{5}{6}=\frac{10}{6}\)
Chúc bạn học tốt nha
.
Ng ta năm nay mới lên lớp 6, dùng x là đúng r, ngày trc chúng mik cx vậy mà.
1/1x2+1/2x3+...+1/49x50
=1-1/2+1/2-1/3+.....+1/49-1/50
=1-1/50(1)
Ta co 1(2)
So sanh (1) voi (2) ta thay 1-1/50<1
=>1/1x2+...+1/49x50<1
(Phuong phap khu)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}\)
=\(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{49}-\frac{1}{50}\)
=\(\frac{1}{1}-\frac{1}{50}=\frac{50}{50}-\frac{1}{50}=\frac{49}{50}<1\)
Vậy \(\frac{49}{50}<1\)
B = \(\frac{3^2}{2.4}+\frac{3^2}{4.6}+\frac{3^2}{6.8}+...+\frac{3^2}{198.200}\)
B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}\right)+\frac{3^2}{2}.\left(\frac{1}{4}-\frac{1}{6}\right)+\frac{3^2}{2}.\left(\frac{1}{6}-\frac{1}{8}\right)+...+\frac{3^2}{2}.\left(\frac{1}{198}-\frac{1}{200}\right)\)
B = \(\frac{3^2}{2}.\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+\frac{1}{6}-\frac{1}{8}+...+\frac{1}{198}-\frac{1}{200}\right)\)
B = \(\frac{9}{2}.\left(\frac{1}{2}-\frac{1}{200}\right)\)
B = \(\frac{9}{2}.\frac{99}{200}\)
B = \(\frac{891}{400}\)
D = 1 x 2 + 2 x 3 + 3 x 4 + 4 x 5 + ... + 48 x 49
3D = 1 x 2 x 3 + 2 x 3 x 3 + 3 x 4 x 3 + 4 x 5 x 3 + ... + 48 x 49 x 3
3D = 1 x 2 x 3 + 2 x 3 x ( 4 - 1 ) + 3 x 4 x ( 5 - 2 ) + 4 x 5 x ( 6 - 3 ) + ... + 48 x 49 x ( 50 - 47 )
3D = 1 x 2 x 3 + 2 x 3 x 4 - 1 x 2 x 3 + 3 x 4 x 5 - 2 x 3 x 4 + 4 x 5 x 6 - 3 x 4 x 5 + ... + 48 x 49 x 50 - 47 x 48 x 49
3D = 48 x 49 x 50
D = ( 48 x 49 x 50 ) : 3
D = 39200
E = 12 + 22 + 32 + ... + 482
E = 1 x 1 + 2 x 2 + 3 x 3 + ... + 48 x 48
E = 1 x ( 2 - 1 ) + 2 x ( 3 - 1 ) + 3 x ( 4 - 1 ) + ... + 48 x ( 49 - 1 )
E = 1 x 2 - 1 + 2 x 3 - 2 + 3 x 4 - 3 + ... + 48 x 49 - 49
E = ( 1 x 2 + 2 x 3 + 3 x 4 + ... + 48 x 49 ) - ( 1 + 2 + 3 + ... + 49 )
Ta tính được vế trong ngoặc thứ nhất là 39200 , còn vế trong ngoặc thứ hai là 1225
thay vào ta được :
E = 39200 - 1225
E = 37975
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\)
\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)
\(\Rightarrow2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}\right)\)
\(\Rightarrow A=1-\frac{1}{2^{100}}\)
a) \(\frac{3}{40}+\frac{5}{3}+\frac{7}{60}=\frac{9}{120}+\frac{200}{120}+\frac{14}{120}=\frac{223}{120}\)
b) \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{19.20}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
a) \(\frac{2^{12}x3^5-4^6.9^2}{\left(2^2x3\right)^6+8^4x3^5}=\frac{2^{12}x3^5+\left(2^2\right)^6x\left(3^2\right)^2}{2^{12}x3^6+\left(2^3\right)^4x3^5}\)
\(=\frac{2^{12}x3^5-2^{12}x3^4}{2^{12}x3^6+2^{12}x3^5}=\frac{2^{12}x3^4x\left(3-1\right)}{2^{12}x3^5x\left(3+1\right)}\)
\(=\frac{2}{3.4}=\frac{1}{3.2}=\frac{1}{6}\)
b) \(\frac{1}{9x10}-\frac{1}{8x9}-\frac{1}{7x8}-\frac{1}{6x7}-\frac{1}{5x6}-\frac{1}{4x5}-\frac{1}{3x4}-\frac{1}{2x3}-\frac{1}{1x2}\)
\(=-\left(\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{7x8}+\frac{1}{8x9}+\frac{1}{9x10}\right)\)
\(=-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=-\left(1-\frac{1}{10}\right)\)
\(=\frac{-9}{10}\)
sorry bn nha! mk ko bk lm phần c
\(A=\frac{1}{6.10}+\frac{1}{10.14}+\frac{1}{14.18}+\frac{1}{18.22}+\frac{1}{22.26}+\frac{1}{26.30}\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{10}+\frac{1}{10}-\frac{1}{14}+\frac{1}{14}-\frac{1}{18}+\frac{1}{18}-\frac{1}{22}+\frac{1}{22}-\frac{1}{26}+\frac{1}{26}-\frac{1}{30}\right)\)
\(=\frac{1}{4}.\left(\frac{1}{6}-\frac{1}{30}\right)=\frac{1}{4}.\frac{2}{15}=\frac{1}{30}\)
\(B=\frac{5}{2.3}+\frac{5}{3.4}+\frac{5}{4.5}+...+\frac{5}{8.9}\)\(=5.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{8.9}\right)\) \(=5.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{8}-\frac{1}{9}\right)\)
\(=5.\left(\frac{1}{2}-\frac{1}{9}\right)=5.\frac{7}{18}=\frac{35}{18}\)
\(C=\left(\frac{7^2}{2.9}+\frac{7^2}{9.16}+....+\frac{7^2}{65.72}\right):\left(\frac{1}{3}-\frac{7}{36}\right)\)
\(=7.\left(\frac{7}{2.9}+\frac{7}{9.16}+...+\frac{7}{65.72}\right):\frac{5}{36}\) \(=7.\left(\frac{1}{2}-\frac{1}{9}+\frac{1}{9}-\frac{1}{16}+...+\frac{1}{65}-\frac{1}{72}\right):\frac{5}{36}\)'
\(=7.\left(\frac{1}{2}-\frac{1}{72}\right):\frac{5}{36}=7.\frac{35}{72}:\frac{5}{36}=\frac{49}{2}\)
\(D=\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{37.38.39}+\frac{2}{38.39.40}\)
\(=2.\left(\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{37.38.39}+\frac{1}{38.39.40}\right)\)
\(=2.\frac{1}{2}.\left(\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{37.38}-\frac{1}{38.39}+\frac{1}{38.39}-\frac{1}{39.40}\right)\)
\(=\frac{1}{2.3}-\frac{1}{39.40}=\frac{259}{1560}\)
\(E=\frac{202202}{1212}+\frac{202202}{2020}+\frac{202202}{3030}+\frac{202202}{4242}+\frac{202202}{5656}\)
\(=202202.\left(\frac{1}{3.4.101}+\frac{1}{4.5.101}+\frac{1}{5.6.101}+\frac{1}{6.7.101}+\frac{1}{7.8.101}\right)\)
\(=2002.\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\right)\)
\(=2002.\left(\frac{1}{3}-\frac{1}{8}\right)=2002.\frac{5}{24}=\frac{5005}{12}\)
\(\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
\(=\frac{1}{1}-\frac{1}{6}\)
\(=\frac{5}{6}\)
\(\frac{1}{1.2}\)\(+\)\(\frac{1}{2.3}\)\(+\)\(\frac{1}{3.4}\)\(+\)\(\frac{1}{4.5}\)\(+\)\(\frac{1}{5.6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{2}\)\(+\)\(\frac{1}{2}\)\(-\)\(\frac{1}{3}\)\(+\)\(\frac{1}{3}\)\(-\)\(\frac{1}{4}\)\(+\)\(\frac{1}{4}\)\(-\)\(\frac{1}{5}\)\(+\)\(\frac{1}{5}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{1}{1}\)\(-\)\(\frac{1}{6}\)
\(=\)\(\frac{5}{6}\)
Hok tốt
c)
\(C=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{19.21}\)
\(=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{19}-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{21}\right)\)
\(=\frac{1}{2}.\frac{20}{21}\)
\(=\frac{10}{21}\)
\(A\)= \(\frac{1}{3.4}+\frac{1}{4.5}+..+\frac{1}{49.50}=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{49}-\frac{1}{50}=\)\(\frac{1}{3}-\frac{1}{50}=\frac{50}{150}-\frac{3}{150}=\frac{47}{150}\)
đề có sai không đó bạn làm gì tích 2 số tự nhiên liên mà = 900 chớ