Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Bài 1:
\(A=\frac{5}{3.6}+\frac{5}{6.9}+....+\frac{5}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{3}{3.6}+\frac{3}{6.9}+....+\frac{3}{96.99}\)
\(\Rightarrow\frac{3}{5}A=\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
\(\Rightarrow A=\frac{32}{99}\div\frac{3}{5}=\frac{160}{297}\)
Bái 2:
\(B=\frac{2}{3.7}+\frac{2}{7.11}+...+\frac{2}{99.103}\)
\(\Rightarrow2B=\frac{4}{3.7}+\frac{4}{7.11}+....+\frac{4}{99.103}\)
\(\Rightarrow2B=\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+....+\frac{1}{99}-\frac{1}{103}\)
\(=\frac{1}{3}-\frac{1}{103}=\frac{100}{309}\)
\(\Rightarrow B=\frac{100}{309}\div2=\frac{50}{309}\)
Bài 1:
Ta có:
\(\frac{5}{n.\left(n+3\right)}=\frac{5}{3}.\frac{3}{n.\left(n+3\right)}=\frac{5}{3}.\frac{\left(n+3\right)-n}{n.\left(n+3\right)}=\frac{5}{3}.\left[\frac{n+3}{n.\left(n+3\right)}-\frac{n}{n\left(n+3\right)}\right]\)\(=\frac{5}{3}\left(\frac{1}{n}-\frac{1}{n+3}\right)\)
\(\frac{5}{3.6}+\frac{5}{6.9}+\frac{5}{9.12}+...+\frac{5}{96.99}=\frac{5}{3}\left(\frac{1}{3}-\frac{1}{6}+\frac{1}{6}-\frac{1}{9}+...+\frac{1}{96}-\frac{1}{99}\right)\)
a) SSH : (101 - 2) : 1 + 1 = 100
=> Tổng : \(\frac{\left(2+101\right)\cdot100}{2}=5150\)
b) SSH : (201 - 101) : 2 + 1 = 51
=> Tổng : \(\frac{\left(101+201\right)\cdot51}{2}=7701\)
c) SSH : (293 - 5) : 3 + 1 = 97
=> Tổng : \(\frac{\left(5+293\right)\cdot97}{2}=14453\)
P/S : Đề bài là gì ?? '-'
1)
\(\left(a\right)37+397+3997+39997\)
\(=40-3+400-3+4000-3+40000-3\)
\(=\left(40+400+4000+40000\right)-\left(3+3+3+3\right)\)
\(=44440-12=44428\)
\(\left(b\right)298+2998+29998+299998\)
\(=300-2+3000-2+30000-2+300000-2\)
\(=\left(300+3000+30000+300000\right)-\left(2+2+2+2\right)\)
\(=333300-8=333296\)
\(\left(c\right)9+99+999+9999+99999\)
\(=10-1+100-1+1000-1+10000-1+100000-1\)
\(=\left(10+100+1000+10000+100000\right)-\left(1+1+1+1+1\right)\)
\(=111110-5=111105\)
2)
\(\left(a\right)\left(2+4+6+...+2002+2004+2006\right)-\left(1+3+5+...+2001+2003+2005\right)\)
\(=\left(2-1\right)+\left(4-3\right)+\left(6-5\right)+...+\left(2002-2001\right)+\left(2004-2003\right)+\left(2006-2005\right)\)
\(=1+1+1+...+1+1+1\)( 1003 số 1 )
\(=1003\)
\(\left(b\right)88-87+86-85+84-83+...+6-5+4-3+2-1\)
\(=\left(88-87\right)+\left(86-85\right)+\left(84-83\right)+...+\left(6-5\right)+\left(4-3\right)+\left(2-1\right)\)
\(=1+1+1+...+1+1+1\)( 44 số 1 )
\(=44\)
\(\left(c\right)100-98+96-94+92-90+...+12-10+8-6+4-2\)
\(=\left(100-98\right)+\left(96-94\right)+\left(92-90\right)+...+\left(12-10\right)+\left(8-6\right)+\left(4-2\right)\)
\(=2+2+2+...+2+2+2\) ( 25 số 2 )
\(=50\)
3)
\(\left(a\right)360-357+354-351+348-345+...+312-309+306-303+300-297\)
\(=\left(360-357\right)+\left(354-351\right)+\left(348-345\right)+...+\left(312-309\right)+\left(306-303\right)+\)\(\left(300-297\right)\)
\(=3+3+3+3+3+3+3+3+3+3+3=33\)
\(\left(b\right)2006-1-2-3-4-...-47-48-49-50\)
\(=2006-\left(1+2+3+4+...+47+48+49+50\right)\)
\(=2006-\frac{\left(50+1\right)\left[\left(50-1\right)+1\right]}{2}\)
\(=2006-1275=731\)
\(\left(c\right)280-276+272-268+264-260+...+216-212+208-204+200-196\)
\(=\left(280-276\right)+\left(272-268\right)+\left(264-260\right)+...+\left(216-212\right)+\left(208-204\right)+\)\(\left(200-196\right)\)
\(=4+4+4+4+4+4+4+4+4+4+4=44\)
a)
`127+246+273+354`
`=(127+273)+(246+354)`
`=400+600`
`=1000`
b)
`1,58+3,04+6,96+3,42`
`=(1,58+3,42)+(3,04+6,96)`
`=5+10`
`=15`
c)
`1/2+1/3+1/5+1/6`
`=(1/2+1/3+1/6)+1/5`
`=(3/6+2/6+1/6)+1/5`
`=6/6+1/5`
`=1+1/5`
`=5/5+1/5`
`=6/5`
a, 127 + 246 + 273 + 354
= ( 127 + 273) + ( 246 + 354)
= 400 + 600
= 1000
b, 1,58 + 3,04 + 6,96 + 3,42
= ( 1,58 + 3,42) + ( 3,04 + 6,96)
= 5 + 10
= 15
c, \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{5}\) + \(\dfrac{1}{6}\)
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + \(\dfrac{1}{6}\)) + 0,2
= 1 + 0,2
= 1,2
a) \(\dfrac{122}{123}\) và \(\dfrac{10}{11}\)
\(1-\dfrac{122}{123}=\dfrac{1}{123}\)
\(1-\dfrac{10}{11}=\dfrac{1}{11}\)
Vì \(\dfrac{1}{123}< \dfrac{1}{11}\) nên ⇒ \(\dfrac{122}{123}< \dfrac{10}{11}\)
b) \(\dfrac{16}{12}\) và \(\dfrac{99}{100}\)
\(\dfrac{16}{12}>1\) và \(\dfrac{99}{100}< 1\)
⇒ \(\dfrac{16}{12}>\dfrac{99}{100}\)
c) \(\dfrac{35}{70}\) và \(\dfrac{6}{11}\)
\(\dfrac{35}{70}=\dfrac{1}{2}\) = \(\dfrac{6}{12}\)
Vì \(\dfrac{6}{12}< \dfrac{6}{11}\) nên ⇒ \(\dfrac{35}{70}< \dfrac{6}{11}\)
sai đề rồi