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\(A=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{95\cdot98}\)
\(A=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{95\cdot98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\cdot\frac{48}{98}\)
\(A=\frac{16}{98}=\frac{8}{49}\)
\(B=\frac{2}{1\cdot4}+\frac{2}{4\cdot7}+\frac{2}{7\cdot10}+...+\frac{2}{97\cdot100}\)
\(B=2\left(\frac{1}{1\cdot4}+\frac{1}{4\cdot7}+\frac{1}{7\cdot10}+...+\frac{1}{97\cdot100}\right)\)
\(B=2\left[\frac{1}{3}\left(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\right)\right]\)
\(B=2\left[\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\right)\right]\)
\(B=2\left[\frac{1}{3}\left(1-\frac{1}{100}\right)\right]\)
\(B=2\left[\frac{1}{3}\cdot\frac{99}{100}\right]\)
\(B=2\cdot\frac{33}{100}\)
\(B=\frac{33}{50}\)
A = \(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
3A = 3/2.5 + 3/5.8 + 3/8.11 + ... + 3/92.95 + 3/95.98
3A = 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 + ... + 1/92 - 1/95 + 1/95 - 1/98
3A = 1/2 - 1/98
3A = 24/49
A = 24/49 : 3
A = 72/49
B = 2/1.4 + 2/4.7 + 2/7.10 + ... + 2/97.100
3/2B = 3/1.4 + 3/4.7 + 3/7.10 + ... + 3/97.100
3/2B = 1/1 - 1/4 + 1/4 - 1/7 + 1/7 - 1/10 + .... + 1/97 - 1/100
3/2B = 1 - 1/100
3/2B = 99/100
B = 99/100 : 3/2
B = 33/50
Bài nhìn vô muốn xỉu rồi ='((
1. a) \(\frac{2}{1.4}+\frac{2}{4.7}+\frac{2}{7.10}+...+\frac{2}{91.94}+\frac{2}{94.97}\)
\(=\frac{2}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{91.94}+\frac{3}{94.97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\right)\)
\(=\frac{2}{3}\left(1-\frac{1}{97}\right)=\frac{2}{3}.\frac{96}{97}=\frac{64}{97}\)
b) Bạn tự làm, làm nữa chắc xỉu =((( Khi nào rảnh mình sẽ làm, nếu bạn cần
2 )
a) \(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{x\left(x+2\right)}=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow\frac{1}{2}\left(1-\frac{1}{x+2}\right)=\frac{1005}{2011}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{1005}{2011}:2=\frac{1005}{4022}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{1005}{4022}=\frac{3017}{4020+2}\)
\(\Rightarrow x=4020\)
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+............+\frac{1}{92.95}+\frac{1}{95.98}\)
\(A=\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+..........+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-.............-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\left(\frac{49}{98}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{48}{98}\)
\(A=\frac{8}{49}\)
Vậy A = \(\frac{8}{49}\)
Phân tích: 1/2.5 = 1/2 - 1/5
1/5.8 = 1/5 - 1/8
1/8.11 = 1/8 - 1/11
...
1/92.95 = 1/92 - 1/95
1/95.98 = 1/95 - 1/98
Ta có: 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
3 = 3/2.5 + 3/5.8 + 3/8.11 + ...+ 3/92.95 + 3/95.98
3 = 1 - 1/2 + 1/2 - 1/5 + 1/5 - 1/8 + 1/8 - 1/11 +...+ 1/92 - 1/95 + 1/95 - 1/98
= 1 - 1/98
= 97/98 : 3 = 97/98 x 1/3 = (tự tính)
a) \(\frac{5.4^{15}.9^9-4.3^{20}.8^9}{5.2^9.6^{19}-7.2^{29}.27^6}\)
\(=\frac{5.2^{30}.3^{18}-2^2.2^{27}.3^{20}}{5.2^9.2^{19}.3^{19}-7.2^{29}.3^{18}}\)
\(=\frac{2^{29}.3^{18}\left(5.2-3^2\right)}{2^{18}.3^{18}\left(5.3-7.2\right)}\)
\(=\frac{2.1}{1}=2\)
c)1*(1/2-1/3+1/3-1/4+.....+1/91-1/94)
1/2-1/94 ban tu tinh nhe
d)1*(1/1-1/4+1/4-1/7+......+1/91-1/94)
1-1/94 ban tu tinh nhe
tk nha
a) \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)
\(\Leftrightarrow\frac{n+1}{n\left(n+1\right)}-\frac{n}{n\left(n+1\right)}\Leftrightarrow\frac{n+1-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
b) \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)
\(\Leftrightarrow\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{n+3-n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)
c,d dễ bn tách ra rồi trừ đi
A=1/3x(1/2x5+1/5x8+......+1/95x98)
A=1/3x(1/2-1/5+1/5-1/8+.........+1/95-1/98)
A=1/3x(1/2-1/98)
A=1/3x24/49
A=8/49
A =\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
A = \(\frac{1.3}{2.5.3}+\frac{1.3}{5.8.3}+\frac{1.3}{8.11.3}+...+\frac{1.3}{92.95.3}+\frac{1.3}{95.98.3}\)
A = \(\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
A =\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
A =\(\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
A =\(\frac{1}{3}.\frac{97}{98}\)
A =\(\frac{97}{294}\)
1) Ta có: A=\(\frac{1}{3}\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{x\left(x+3\right)}\right)=\)
=\(\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{x}-\frac{1}{x+3}\right)=\)
=\(\frac{1}{3}\left(1-\frac{1}{x+3}\right)=\frac{1}{3}.\frac{x+2}{x+3}=\frac{125}{376}\)
<=> \(\frac{x+2}{x+3}=\frac{375}{376}\)<=> 376(x+2)=375(x+3) <=> 376x+752=375x+1125 => X=373
1.
a) \(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{97\cdot100}\\ A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{97}-\frac{1}{100}\\ A=1-\frac{1}{100}=\frac{99}{100}\)
b) Sửa đề: B = 1/2.5 + 1/5.8 + 1/8.11 + ...
\(B=\frac{1}{2\cdot5}+\frac{1}{5\cdot8}+\frac{1}{8\cdot11}+...+\frac{1}{92\cdot95}+\frac{1}{95\cdot98}\\ B=\frac{1}{3}\left(\frac{3}{2\cdot5}+\frac{3}{5\cdot8}+\frac{3}{8\cdot11}+...+\frac{3}{92\cdot95}+\frac{3}{95\cdot98}\right)\\ B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\\ B=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{98}\right)\\ B=\frac{1}{6}-\frac{1}{294}\\ B=\frac{49}{294}-\frac{1}{294}=\frac{48}{294}=\frac{8}{49}\)
2.
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2000}\\ \frac{2}{6}+\frac{2}{12}+\frac{2}{20}+...+\frac{2}{n\left(n+1\right)}=\frac{1999}{2000}\\ 2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{n\left(n+1\right)}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}\right)=\frac{1999}{2000}\\ 2\left(\frac{1}{2}-\frac{1}{n+1}\right)=\frac{1999}{2000}\\ \frac{1}{2}-\frac{1}{n+1}=\frac{1999}{2000}:2\\ \frac{1}{2}-\frac{1}{n+1}=\frac{1999}{4000}\\ \frac{1}{2}-\frac{1999}{4000}=\frac{1}{n+1}\\ \frac{1}{n+1}=\frac{1}{4000}\\ \Rightarrow n+1=4000\\ \Rightarrow n=3999\)
Vậy n = 3999