Bạn cần bài nào trước nhỉ? Nếu cần tất cả thì bạn tách từng bài ra để mọi người giúp nhé!
3
a
\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{1999.2000}\\ =\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\\ =1-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
b
\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+....+\dfrac{1}{100.103}\\ =\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+....+\dfrac{3}{100.103}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+....+\dfrac{1}{100}-\dfrac{1}{103}\right)\\ =\dfrac{1}{3}\left(1-\dfrac{1}{103}\right)\\ =\dfrac{1}{3}.\dfrac{102}{103}\\ =\dfrac{34}{103}\)
c
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-...-\dfrac{1}{6}-\dfrac{1}{2}\\ =\dfrac{8}{9}+\dfrac{1}{9}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{7}+...+\dfrac{1}{3}-\dfrac{1}{2}+\dfrac{1}{2}\\ =1\)
`@` `\text {Ans}`
`\downarrow`
Bài tính toán thì mình nghĩ bạn có thể làm được nên mình làm 2 bài sau nhé!
`2,`
`a)`
\(\dfrac{17}{b}-\left(x-\dfrac{7}{b}\right)=\dfrac{7}{4}\)
`=>`\(\dfrac{17}{b}-x+\dfrac{7}{b}=\dfrac{7}{4}\)
`=>`\(\dfrac{17}{b}+\dfrac{7}{b}-x=\dfrac{7}{4}\)
`=>`\(\dfrac{24}{b}-x=\dfrac{7}{4}\)
`=>`\(x=\dfrac{24}{b}-\dfrac{7}{4}\)
`=>`\(x=\dfrac{89}{4b}\)
Vậy, `x = 89/4b`
`b)`
\(\dfrac{4}{3}+1,25-x=2,25\)
`=>`\(\dfrac{19}{12}-x=2,25\)
`=>`\(x=\dfrac{19}{12}-2,25\)
`=>`\(x=-\dfrac{2}{3}\)
Vậy, `x = -2/3`
`c)`
\(2x-3=x+\dfrac{1}{2}\)
`=> 2x - 3 - x - 1/2 = 0`
`=> (2x - x) - (3 + 1/2) = 0`
`=> x - 7/2 = 0`
`=> x = 7/2`
Vậy, `x = 7/2`
`d)`
\(4x-2x+1=3-\dfrac{1}{3}+x\)
`=> 2x + 1 - 3 + 1/3 - x = 0`
`=> (2x - x) + (1 - 3 + 1/3) = 0`
`=> x - 5/3 = 0`
`=> x = 5/3`
Vậy, `x = 5/3`
`3,`
Bạn trên làm rồi, mình làm lại phần này chi tiết hơn nếu bạn cần nhé.
`a)`
\(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{1999\cdot2000}\)
`=`\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{1999}-\dfrac{1}{2000}\)
`=`\(1-\left(\dfrac{1}{2}-\dfrac{1}{2}\right)-\left(\dfrac{1}{3}-\dfrac{1}{3}\right)-...-\dfrac{1}{2000}\)
`=`\(1-\dfrac{1}{2000}=\dfrac{1999}{2000}\)
`b)`
\(\dfrac{1}{1\cdot4}+\dfrac{1}{4\cdot7}+\dfrac{1}{7\cdot10}+...+\dfrac{1}{100\cdot103}\)
`=`\(\dfrac{1}{3}\cdot\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{100\cdot103}\right)\)
`=`\(\dfrac{1}{3}\cdot\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\right)\)
`=`\(\dfrac{1}{3}\cdot\left(1-\dfrac{1}{103}\right)\)
`=`\(\dfrac{1}{3}\cdot\dfrac{102}{103}=\dfrac{34}{103}\)
`c)`
`8/9 - 1/72 - 1/56 - 1/42 - ... - 1/6 - 1/2`
`= 8/9 - (1/2 + 1/6 + ... + 1/42 + 1/56 + 1/72)`
`= 8/9 -`\(\left(\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{6\cdot7}+\dfrac{1}{7\cdot8}+\dfrac{1}{8\cdot9}\right)\)
`= 8/9 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/7 - 1/8 + 1/8 - 1/9)`
`= 8/9 - (1 - 1/9)`
`= 8/9 - 8/9`
`= 0`
4 và 5