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A = 1.2+2.3+3.4+4.5+...+98.99+99.100
3A = 1.2.(3-0)+2.3.(4-1)+3.4.(5-2)+...+99.100.(101-98)
3A = 1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+...+99.100.101-98.99.100
3A = 99.100.101
3A = 999900
A = 333300
nhấn đúng cho mk nha!!!!!!!!!!!!
Đặt A= 1.2+2.3 +.......+99.100
3A= 1.2.3+2.3.4+3.4.3 +......+ 99.100.3
3A= 1.2. (3 - 0) + 2.3.(4 - 1) +3.4. (5 - 2)....... . 99.100. (101 - 98)
3A = (1.2.3 + 2.3.4 + 3.4.5 +...... + 99.100.101) - (0.1.2 + 1.2.3 + 2.3.4 +.......+ 98.99.100)
3A = 99.100.101 - 0.1.2
3A = 999900 - 0
3A= 999900
A= 999900 : 3
A = 333300
Học tốt nha!
đặt A = 1.2 + 2.3 + 3.4 + ... + 99.100
=> 3A = 1.2.3 + 2.3.3 + 3.4.3 + ... + 99.100.3
=> 3A = 1.2.3 + 2.3.(4 - 1) + 3.4.(5 - 2) + ... + 99.100.(101 - 98)
=> 3A = 1.2.3 + 2.3.4 - 1.2.3 + 3.4.5 - 2.3.4 + ... + 99.100.101 - 98.99.100
=> 3A = 99.100.101
=> A = 99.100.101 : 3
=> A = 333300
Tính tổng dãy sau :
Bài giải :
Đặt S = 1 . 2 + 2 . 3 + 3 . 4 + .... + 99 . 100
3S = 1 . 2 . 3 + 2 . 3 . 3 + 3 . 4 . 3 + ... + 98 . 99 . 3 + 99 . 100 . 3
= 1 . 2 . 3 + 2 . 3 ( 4 - 1 ) + 3 . 4 ( 5 - 2 ) + ... + 98 . 99 ( 100 - 97 ) + 99 . 100 ( 101 - 98 )
= 1 . 2 . 3 + 2 . 3 . 4 - 1 . 2 . 3 + 3 . 4 . 5 - 2 . 3 . 4 + ... - 97 . 98 . 99 + 99 . 100 . 101 - 98 . 99 . 100
3S = 99 . 100 .101
=> S = 99 . 100 .101 : 3
= ( 99 : 3 ) . ( 100 . 101 )
= 33 . 10 100
= 333 300
A=1.2+2.3+3.4+4.5+...+99.100
=>3A=1.2.3+2.3.3+3.4.3+4.5.3+...+99.100.3
=1.2.3+2.3.(4-1)+3.4.(5-2)+4.5.(6-3)+...+99.100.(101-98)
=1.2.3+2.3.4-1.2.3+3.4.5-2.3.4+4.5.6-3.4.5+...+99.100.101-98.99.100
=1.2.3-1.2.3+2.3.4-2.3.4+3.4.5-3.4.5+4.5.6-4.5.6+...+99.100.101
=99.100.101=999900
=>A=999900:3=333300
Vậy A=333300
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+....+\frac{2}{19.20}\)
\(=2.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{19.20}\right)\)
\(=2.\left(1-\frac{1}{20}\right)\)
\(=2.\frac{19}{20}=\frac{19}{10}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{19.20}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{19}-\frac{1}{20}\)
\(=\frac{1}{1}-\frac{1}{20}\)
\(=\frac{19}{20}\)
c; 17\(\dfrac{2}{31}\) - (\(\dfrac{15}{17}\) + 6\(\dfrac{2}{31}\))
= 17 + \(\dfrac{2}{31}\) - \(\dfrac{15}{17}\) - 6 - \(\dfrac{2}{31}\)
= (17 - 6) - \(\dfrac{15}{17}\) + (\(\dfrac{2}{31}\) - \(\dfrac{2}{31}\))
= 11 - \(\dfrac{15}{17}\)+ 0
= \(\dfrac{172}{17}\)
b; 130\(\dfrac{25}{28}\) + 120\(\dfrac{17}{35}\)
= 130 + \(\dfrac{25}{28}\) + 120 + \(\dfrac{17}{35}\)
= (130 + 120) + (\(\dfrac{25}{28}\) + \(\dfrac{17}{35}\))
= 250 + (\(\dfrac{125}{140}\) + \(\dfrac{68}{140}\))
= 250 + \(\dfrac{193}{140}\)
= 250\(\dfrac{193}{140}\)
Ta có: A = 1.2 + 2.3 + 3.4 + 4.5 +.....+ 98.99
=> 3A = 1.2.(3 - 0) + 2.3.(4 - 1) + ..... +98.99.(100 - 97)
=> 3A = 1.2.3 - 1.2.3 + 2.3.4 - 2.3.4 + ..... + 98.99.100
=> 3A = 98.99.100
=> A = 98.99.100 / 3
=> A = 323400
Bạn có thể làm như vầy nè:
Đặt 2 ra ngoài,ta có dạng S = 2 x (1/2.3 + 1/3.4 + ... + 1 x 98 x 99 + 1/99.100)
Với chú ý:1/2.3 = 1/2 - 1/3
1/3.4 = 1/3 - 1/4,........
Vậy S = 2 x ( 1/2 - 1/100) = 2 x (50/100 - 1/100) = 2.49/100 = 98/100 = 49/50
Chúc bạn học thiệt là giỏi!
\(Tac\text{ó}:\frac{2}{1.2}-\frac{2}{2.3}-\frac{2}{3.4}-...-\frac{2}{98.99}-\frac{2}{99.100}\)
=\(2.\left(\frac{1}{1.2}-\frac{1}{2.3}-\frac{1}{3.4}-...-\frac{1}{98.99}-\frac{1}{99.100}\right)\)
=\(2\left(1-\frac{1}{2}\right)\)
f,F=3. (1/2 .3 + 1/3.4 +...+ 1/99.100)
= 3. (1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 +...+ 1/99 - 1/100
= 3. (1/2 - 1/100)
= 3. 49/100
= 147/100
g, G = 5/3. (3/1.4 + 3/4.7 +...+ 3/61.64)
= 5/3 . (1 - 1/4 + 1/4 - 1/7 +...+ 1/61 - 164
= 5/3 . (1-1/64)
= 5/3 . 63/64
= 105/64
f, \(F=\frac{3}{2.3}+\frac{3}{3.4}+...+\frac{3}{99.100}\)
\(\Leftrightarrow F=3\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\right)\)
\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(\Leftrightarrow F=3\left(\frac{1}{2}-\frac{1}{100}\right)\)
\(\Leftrightarrow F=3\left(\frac{49}{100}\right)=\frac{147}{100}\)
g, \(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)
\(\Leftrightarrow G=5\left(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{61.64}\right)\)
\(\Leftrightarrow G=5.\frac{1}{3}\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+...+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Leftrightarrow G=\frac{5}{3}\left(1-\frac{1}{64}\right)\)
\(\Leftrightarrow G=\frac{5}{3}.\frac{63}{64}=\frac{105}{64}\)
\(G=\frac{5}{1.4}+\frac{5}{4.7}+...+\frac{5}{61.64}\)
\(\Rightarrow G=\frac{5}{3}.\left(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+..+\frac{3}{61.64}\right)\)
\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+..+\frac{1}{61}-\frac{1}{64}\right)\)
\(\Rightarrow G=\frac{5}{3}.\left(1-\frac{1}{64}\right)=\frac{5}{3}.\frac{63}{64}\)
\(\Rightarrow G=\frac{5.63}{3.64}=\frac{5.21.3}{3.64}=\frac{5.21}{64}=\frac{105}{64}\)
Ta có:
\(G=\dfrac{2}{1.2}+\dfrac{2}{2.3}+\dfrac{2}{3.4}+...+\dfrac{2}{4.5}+...+\dfrac{2}{99.100}\)
\(=2\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\)
\(=2\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\)
\(=2\left(1-\dfrac{1}{100}\right)=\dfrac{99}{50}\)