2/3*5 + 2/3*7 + 2/3*9 + 2/3*11 + 2/3*13 + 2/3*15
2/1*2 + 2/2*3 + ....................... + 2/9*10
3/1*2 + 3/2*3 + ..........................+ 3/9*10
Giup mk theo kiểu dễ hiểu và đơn giản nhất nhé !
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
\(A=\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{13.15}\)
\(A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{13}-\frac{1}{15}\)
\(A=\frac{1}{3}-\frac{1}{15}=\frac{4}{15}\)
\(B=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{9.10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(B=2.\left(\frac{1}{1}-\frac{1}{10}\right)=2.\frac{9}{10}\)
\(B=\frac{9}{5}\)
Ta có 1/2*3=1/2-1/3;
1/3*4=1/3-1/4
......................(tương tự với các số khác)
1/149*150=1/149-1/150
=>A=1/2-1/3+1/3-1/4+1/4-1/5+...-1/149+1/149-1/150=1/2-1/150
A=75/150-1/150=74/150=37/75
Vậy A= 37/75
`@` `\text {Ans}`
`\downarrow`
`a.`
`A=(1/2-7/13-1/3)+(-6/13+1/2+1 1/3)`
`= 1/2 - 7/13 - 1/3 - 6/13 + 1/2 + 1 1/3`
`= (1/2 + 1/2) + (-7/13 - 6/13) + (-1/3 + 1 1/3) `
`= 1 - 1 + 1`
`= 1`
`b.`
`B=0,75+2/5+(1/9-1 1/2+5/4)`
`= 3/4 + 2/5 + 1/9 - 3/2 + 5/4`
`= (3/4+5/4)+ 1/9 + 2/5 - 3/2`
`= 2 + 1/9 - 11/10`
`= 19/9 - 11/10`
`= 91/90`
`c.`
`(-5/9).3/11+(-13/18).3/11`
`= 3/11*[(-5/9) + (-13/18)]`
`= 3/11*(-23/18)`
`= -23/66`
`d.`
`(-2/3).3/11+(-16/9).3/11`
`= 3/11* [(-2/3) + (-16/9)]`
`= 3/11*(-22/9)`
`= -2/3`
`e.`
`(-1/4).(-2/13)-7/24.(-2/13)`
`= (-2/13)*(-1/4-7/24)`
`= (-2/13)*(-13/24)`
`= 1/12`
`f.`
`(-1/27).3/7+(5/9).(-3/7)`
`= 3/7*(-1/27 - 5/9)`
`= 3/7*(-16/27)`
`= -16/63`
`g.`
`(-1/5+3/7):2/11+(-4/5+4/7):2/11`
`=[(-1/5+3/7)+(-4/5+4/7)] \div 2/11`
`= (-1/5+3/7 - 4/5 + 4/7) \div 2/11`
`= [(-1/5-4/5)+(3/7+4/7)] \div 2/11`
`= (-1+1) \div 2/11`
`= 0 \div 2/11 = 0`
a: \(=11+\dfrac{3}{13}-2-\dfrac{4}{7}-5-\dfrac{3}{13}=4-\dfrac{4}{7}=\dfrac{24}{7}\)
b: \(=\dfrac{11}{2}\cdot\dfrac{15}{4}=\dfrac{165}{8}\)
c: \(=10+\dfrac{2}{9}+2+\dfrac{3}{5}-6-\dfrac{2}{9}=6+\dfrac{3}{5}=\dfrac{33}{5}\)
d: \(=6+\dfrac{4}{9}+3+\dfrac{7}{11}-4-\dfrac{4}{9}=5+\dfrac{7}{11}=\dfrac{62}{11}\)
c: Ta có: \(\dfrac{5}{3}+\dfrac{5}{3\cdot5}+\dfrac{5}{5\cdot7}+...+\dfrac{5}{101\cdot103}\)
\(=\dfrac{5}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{101\cdot103}\right)\)
\(=\dfrac{5}{2}\left(1-\dfrac{1}{103}\right)\)
\(=\dfrac{5}{2}\cdot\dfrac{102}{103}\)
\(=\dfrac{255}{103}\)
\(\frac{2}{3}\times5+\frac{2}{3}\times7+\frac{2}{3}\times9+\frac{2}{3}\times11+\frac{2}{3}\times13+\frac{2}{3}\times15\)
=\(\frac{2}{3}\)\(\times\)( 5+7+9+11+15)
= \(\frac{2}{3}\)\(\times\)47
= \(\frac{94}{3}\)
Có friend forever II Lê Tiến Đạt giải rồi nhé nên đừng bắt tui giải nữa ( chuồn là thượng sách)