tính giá trị các biểu thức sau:
a) S=1/2.3 + 1/ 2.4 +...+ 1/99.100
b) P=3/1.4 + 3/4.7 +...+ 3/91.94
So sánh 3/1.4 + 3/4.7+ 3/7.10+...+3/91.94+3/94.97 với 1
Ta có: \(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}\)
\(\Leftrightarrow1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{94}-\frac{1}{97}\)
\(\Leftrightarrow1-\frac{1}{97}=\frac{96}{97}\)
Do \(\frac{96}{97}< 1\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.97}< 1\)
Vậy:.............................<1
a) Tính \(\frac{1}{n}-\frac{1}{n+1}\left(n\inℕ^∗\right)\)
b)Tính \(\frac{1}{n}-\frac{1}{n+3}\left(n\inℕ^∗\right)\)
c) Tính \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{3}{91.94}\)
d) Tính \(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}\)
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
\(A.\)\(\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n}+\frac{-1}{n+1}=\frac{n+1}{n\left(n+1\right)}+\frac{-n}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}\)
\(B.\)\(\frac{1}{n}-\frac{1}{n+3}=\frac{n+3}{n\left(n+3\right)}-\frac{n}{n\left(n+3\right)}=\frac{3}{n\left(n+3\right)}\)
\(C.\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}\)
\(D.\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{91.94}=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{91}-\frac{1}{94}=\frac{1}{1}-\frac{1}{94}=\frac{93}{94}\)
Tính
A= 3/1.4 + 3/4.7 + . . . + 3/91.94
B= 1/1.3 + 1/3.5 + . . . + 1/97.99
chú ý / là phần nha
\(A=\frac{3}{1.4}+\frac{3}{4.7}+..........+\frac{3}{91.94}\)
\(\Leftrightarrow A=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{91}-\frac{1}{94}\)
\(\Leftrightarrow A=1-\frac{1}{94}=\frac{93}{94}\)
\(B=\frac{1}{1.3}+\frac{1}{3.5}+......+\frac{1}{97.99}\)
\(\Leftrightarrow2B=\frac{2}{1.3}+\frac{2}{3.5}+.......+\frac{3}{97.99}\)
\(\Leftrightarrow2B=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{97}-\frac{1}{99}\)
\(\Leftrightarrow2B=1-\frac{1}{99}=\frac{98}{99}\)
\(\Leftrightarrow B=\frac{98}{99}:2=\frac{49}{99}\)
Ta có : \(A=\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{91.94}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.....+\frac{1}{91}-\frac{1}{94}\)
\(=1-\frac{1}{94}\)
\(=\frac{93}{94}\)
so sánh 3/1.4+3/4.7+3/7.10+....+3/91.94+3/94.97 với 1
\(\frac{3}{1.4}+\frac{3}{4.7}+.....+\frac{3}{94.97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+.........+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
mà \(\frac{96}{97}< 1\)
\(\Rightarrow\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{94.07}< 1\)
vậy..................
\(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{91\cdot94}+\frac{3}{94\cdot97}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}\)
\(=1-\frac{1}{97}\)
\(=\frac{96}{97}\)
\(\Rightarrow\frac{96}{97}< 1\)
\(\Rightarrow\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)
Vậy \(\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\frac{3}{7\cdot10}+...+\frac{3}{94\cdot97}< 1\)
Bài 1 Tính giá trị biểu thức :
A = 3/1.4 + 5/4.9 + 7/9.16 + 9/16.25 + 11/25.36
B = 3/1.4 + 3/4.7 + ... + 3/100.103
C = 3/1.4 + 6/4.10 + 9/10.19 + 12/19.31 + 15/31.46 + 18/46.64
Bài 2 Chứng minh rằng :
1/1.2 + 1/3.4 + 1/5.6 + ... + 1/49.50 = 1/26 + 1/27 + 1/28 + ... + 1/50
Bài 1:
\(A=\dfrac{3}{1.4}+\dfrac{5}{4.9}+\dfrac{7}{9.16}+\dfrac{9}{16.25}+\dfrac{11}{25.36}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{16}+\dfrac{1}{16}-\dfrac{1}{25}+\dfrac{1}{25}-\dfrac{1}{36}\)
\(=1-\dfrac{1}{36}=\dfrac{35}{36}\)
\(B=\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{100.103}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{100}-\dfrac{1}{103}\)
\(=1-\dfrac{1}{103}=\dfrac{102}{103}\)
\(C=\dfrac{3}{1.4}+\dfrac{6}{4.10}+\dfrac{9}{10.19}+\dfrac{12}{19.31}+\dfrac{15}{31.46}+\dfrac{18}{46.64}\)
\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{19}+\dfrac{1}{19}-\dfrac{1}{31}+\dfrac{1}{31}-\dfrac{1}{46}+\dfrac{1}{46}-\dfrac{1}{64}\)
\(=1-\dfrac{1}{64}=\dfrac{63}{64}\)
Bài 2:
\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{49.50}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{49}-\dfrac{1}{50}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{49}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{49}+\dfrac{1}{50}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{50}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{25}\right)\)
\(=\dfrac{1}{26}+\dfrac{1}{27}+\dfrac{1}{28}+...+\dfrac{1}{50}\left(đpcm\right)\)
1. Tính tổng
a, A=1/2.3 + 1/3.4 + ... + 1/99.100
b, B= 5/1.4 + 5/4.7 + ... + 5/100.103
c, C= 1/15 +1/35 + ... + 1/2499
d, D=1/1.6 + 1/6.11 + 1/11.16 + ... +1/(5n+1).(5n+6)
mn ơi mình đang cần gấp
a: =1/2-1/3+1/3-1/4+...+1/99-1/100
=1/2-1/100=49/100
b; =5/3(1-1/4+1/4-1/7+...+1/100-1/103)
=5/3*102/103
=510/309=170/103
c: =1/2(1/3-1/5+1/5-1/7+...+1/49-1/51)
=1/2*16/51=8/51
Tính giá trị biểu thức:
B= \(1-\dfrac{3}{1.4}-\dfrac{3}{4.7}-\dfrac{3}{7.10}-...-\dfrac{3}{2020.2023}\)
\(B=1-\dfrac{3}{1\cdot4}-\dfrac{3}{4\cdot7}-...-\dfrac{3}{2020\cdot2023}\\ =1-\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{2020\cdot2023}\right)\\ =1-\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{2020}-\dfrac{1}{2023}\right)\\ =1-\left(1-\dfrac{1}{2023}\right)\\ =1-\dfrac{2022}{2023}=\dfrac{1}{2023}\)
`B=1-3/(1.4)-3/(4.7)-3/(7.10)-....-3/(2020.2023)`
`B=1-(3/(1.4)+3/(4.7)+.....+3/(2020.2023))`
`B=1-(1-1/4+1/4-1/7+.....+1/2020-1/2023)`
`B=1-(1-1/2023)`
`B=1-1+1/2023=1/2023`
Tính giá trị biểu thức A=3^2/1.4+3^2/4.7+…+3^2/196.199
A = \(\dfrac{3^2}{1.4}+\dfrac{3^2}{4.7}+...+\dfrac{3^2}{196.199}\)
A = \(\dfrac{3.3}{1.4}+\dfrac{3.3}{4.7}+...+\dfrac{3.3}{196.199}\)
A = \(3.\dfrac{3}{1.4}+3.\dfrac{3}{4.7}+...+3.\dfrac{3}{196.199}\)
A = \(3\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{196.199}\right)\)
A = \(3\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{196}-\dfrac{1}{199}\right)\)
A = \(3\left(1-\dfrac{1}{199}\right)\) = \(3.\dfrac{198}{199}\) = \(\dfrac{594}{199}\)
tính
a) P = 1 / 1.2 + 2 / 2.4 + 3 / 4.7 + ...+ 10 / 46.56
b) A= 3 / 1.2 + 3 / 2.3 + 3 / 3.4 + ....+ 3 / 99.100 chú ý : / là phần nha
c) B = 3 / 1.4 + 3 / 4.7 + 3 / 7.10 + ... + 3 / 100.103
d) C= 5 / 1.4 + 5 / 4.7 + 5 / 7.10 + ...+ 5 / 100.103
e) D= 7 / 1.5 + 7 / 5.9 + 7 / 9.13 +...+ 7 / 101.105
a)\(P=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{4}+...+\frac{1}{46}-\frac{1}{56}\)
=\(1-\frac{1}{56}=\frac{55}{56}\)
b)\(A.\frac{1}{3}=\frac{1}{3}.\left(\frac{3}{1.2}+\frac{3}{2.3}+....+\frac{3}{99.100}\right)\)
= \(\frac{1}{3}A=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{3}{99.100}\)
=> \(\frac{1}{3}A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
=> \(\frac{1}{3}A=1-\frac{1}{100}=\frac{99}{100}\)
=> \(A=\frac{99}{100}.3=\frac{297}{100}\)
c)\(B=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
d) \(\frac{3}{5}C=\frac{3}{5}.\left(\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{100.103}\right)\)
=\(\frac{3}{1.4}+\frac{3}{4.7}+...+\frac{3}{100.103}\)
=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+....+\frac{1}{100}-\frac{1}{103}\)
=\(1-\frac{1}{103}=\frac{102}{103}\)
=>\(C=\frac{102}{103}.\frac{5}{3}=\frac{170}{103}\)
e) \(\frac{4}{7}D=\frac{4}{7}.\left(\frac{7}{1.5}+\frac{7}{5.9}+...+\frac{7}{101.105}\right)\)
=\(\frac{4}{1.5}+\frac{4}{5.9}+...+\frac{4}{101.105}\)
=\(1-\frac{1}{5}+\frac{1}{5}-\frac{1}{9}+...+\frac{1}{101}-\frac{1}{105}\)
=\(1-\frac{1}{105}=\frac{104}{105}\)
=< D=\(\frac{104}{105}.\frac{7}{4}=\frac{26}{15}\)