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\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.5}+...+\frac{1}{49.50}-\frac{1}{50.51}\)
\(=\frac{1}{2}-\frac{1}{50.51}\)
\(=\frac{1}{2}-\frac{1}{2550}=\frac{637}{1275}\)
Gọi A là tổng dãy phân số trên
Ta có :
\(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{49.50.51}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{49.50.51}\)
Ta thấy:
\(\frac{2}{1.2.3}=\frac{1}{1.2}-\frac{1}{2.3};\frac{2}{2.3.4}=\frac{1}{2.3}-\frac{1}{3.4};...;\frac{2}{49.50.51}=\frac{2}{49.50}-\frac{2}{50.51}\text{}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{49.50}-\frac{1}{50.51}\)
\(\Rightarrow2A=\frac{1}{1.2}-\frac{1}{50.51}\)
\(\Rightarrow2A=\frac{1}{2}-\frac{1}{2550}\)
\(\Rightarrow2A=\frac{1275}{2550}-\frac{1}{2550}\)
\(\Rightarrow2A=\frac{637}{1275}\Rightarrow A=\frac{637}{1275}:2=\frac{637}{2550}\)
Vậy tổng dãy phân số trên là :\(\frac{637}{2550}\)
Chúc bạn học tốt !!! :D
Đặt A=1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100
4A=(1.2.3+2.3.4+3.4.5+4.5.6+...+98.99.100)4
4A=1.2.3(4-0)+2.3.4(5-1)+3.4.5(6-2)+4.5.6(7-3)+...+98.99.100(101-97)
4A=1.2.3.4+2.3.4.5-1.2.3.4+3.4.5.6-2.3.4.5+4.5.6.7-3.4.5.6+...+98.99.100.101-97.98.99.100
4A=1.2.3.4-1.2.3.4+2.3.4.5-2.3.4.5+3.4.5.6-3.4.5.6+...+97.98.99.100-97.98.99.100+98.99.100.101
4A=98.99.100.101
=>A=98.99.100.101/4
đặt A=1.2.3+2.3.4+3.4.5+...+98.99.100
A.4=1.2.3.4+2.3.4.4+3.4.5.4+..+98.99.100.4
A.4=1.2.3(4-0)+2.3.4.(5-1)+3.4.5.(6-2)+......+98.99.100.(101-97)
a.4=1.2.3.4-0+2.3.4.5-2.3.4+3.4.5.6-2.3.4.5+...+98.99.100.101-97.98.99.100
A.4=98.99.100.101
A.4=97990200
A= 97990200:4
A= 24497550
Phải là phân số \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.99.100}\) chứ !
\(\frac{4}{7}x+\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{12.13.14}\right)=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{12.13}-\frac{1}{13.14}\right)\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\left(\frac{1}{1.2}-\frac{1}{13.14}\right)\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\left[\frac{1}{2}.\frac{45}{91}\right]=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x+\frac{45}{182}=\frac{39}{40}\)
\(\Leftrightarrow\frac{4}{7}x=\frac{39}{40}-\frac{45}{182}\Leftrightarrow\frac{4}{7}x=\frac{2649}{3640}\)
\(\Rightarrow x=\frac{2649}{3640}\div\frac{4}{7}=\frac{2649}{2080}\)
Vậy x = \(\frac{2649}{2080}\)
\(2C=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{98.99.100}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
\(=\dfrac{1}{1.2}-\dfrac{1}{99.100}=\dfrac{50.99-1}{100.99}=\dfrac{4949}{9900}\)
Lời giải:
\(2A=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+....+\frac{38-36}{36.37.38}+.\frac{39-37}{37.38.39}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{36.37}-\frac{1}{37.38}+\frac{1}{37.38}-\frac{1}{38.39}\)
\(=\frac{1}{1.2}-\frac{1}{38.39}=\frac{370}{741}\)
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+\frac{1}{4.5.6}+\frac{1}{5.6.7}+\frac{1}{6.7.8}+\frac{1}{7.8.9}+\frac{1}{8.9.10}\)
\(2A=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+\frac{2}{4.5.6}+\frac{2}{5.6.7}+\frac{2}{6.7.8}+\frac{2}{7.8.9}+\frac{2}{8.9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+\frac{1}{4.5}-\frac{1}{5.6}+...+\frac{1}{8.9}-\frac{1}{9.10}\)
\(2A=\frac{1}{1.2}-\frac{1}{9.10}=\frac{22}{45}\)
\(A=\frac{22}{45}:2=\frac{11}{45}\)
1/ 1.2.3 + 1/ 2.3.4 + 1/ 3.4.5+1/4.5.6+1/5.6.7+1/6.7.8+1/7.8.9+1/8.9.10
= 1 - 1/2 - 1/3 + 1/2 - 1/3 - 1/4 + 1/3 - 1/4 - 1/5 + 1/5 - 1/6 - 1/7 + 1/6 - 1/7 - 1/8 + 1/7 - 1/8 - 1/9 + 1/8 - 1/9 - 1/10
= 1 - 1/10
= 9/10
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{1999.2000}-\frac{1}{2000.2001}\right)\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{2000.2001}\right)=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{4002000}\right)=\frac{1}{2}\left(\frac{2000999}{4002000}\right)=\frac{2000999}{8004000}\)
A = 1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/1999.2000.2001
A = 1/2.(2/1.2.3 + 2/2.3.4 + 2/3.4.5 + 2/3.4.5 + ... + 2/1999.2000.2001)
A = 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + ... + 1/1999.2000 - 1/2000.2001)
A = 1/2.(1/1.2 - 1/2000.2001)
A = 1/2.(1/2 - 1/4002000)
Đến đây số to wa, bn tự lm típ
Chú ý: tính hiệu giữa: 1/1.2 - 1/2.3 = 3/1.2.3 - 1/1.2.3 = 2/1.2.3, nhân thêm 2 vào tử
Ủng hộ mk nha ^_-
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+..+\frac{1}{7.8.9}+\frac{1}{8.9.10}\)
\(B=2\times\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+..+\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2\times\left(1-\frac{1}{10}\right)\)
\(B=2\times\frac{9}{10}\)
\(B=\frac{9}{5}\)
\(B=2\times\left(\frac{1}{1\times2}-\frac{1}{2\times3}+\frac{1}{2\times3}-\frac{1}{3\times4}+\frac{1}{3\times4}-\frac{1}{4\times5}+..+\frac{1}{9\times10}\right)\)
\(B=2\times\left(\frac{1}{1\times2}-\frac{1}{9\times10}\right)\)
\(B=2\times\frac{22}{45}\)
\(B=\frac{44}{45}\)
Coi \(A=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(\Rightarrow2A=2x\left(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\right)\)
\(=\frac{2}{1.2.3}+\frac{2}{2.3.4}+\frac{2}{3.4.5}+...+\frac{2}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{4950}{9900}-\frac{1}{9900}\)
\(=\frac{4949}{9900}\)
các bn làm đúng rồi
tk mk nha
thnak