A = 3 x 1/1x2 -5 x 1/2x3 + 7 x 1/3x4 -.....+ 15 x 1/7x8 - 17 x 1/8x9
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\(A=3\cdot\frac{1}{1\cdot2}-5\cdot\frac{1}{2\cdot3}+7\cdot\frac{1}{3\cdot4}-\cdots+15\cdot\frac{1}{7\cdot8}-17\cdot\frac{1}{8\cdot9}\)
\(=\frac{3}{1\cdot2}-\frac{5}{2\cdot3}+\frac{7}{3\cdot4}-\cdots+\frac{15}{7\cdot8}-\frac{17}{8\cdot9}\)
\(=1+\frac12-\frac12-\frac13+\frac13+\frac14-\cdots+\frac17+\frac18-\frac18-\frac19\)
\(=1-\frac19=\frac89\)
1/1*2+1/2*3+1/3*4+...+1/9*10
=1-1/2+1/2-1/3+1/3-1/4+...+1/9-1/10
=1-1/10
=9/10
nho k cho minh voi nhe
\(\frac{1}{1.2}\)+ \(\frac{1}{2.3}\)+ \(\frac{1}{3.4}\)+ ......... + \(\frac{1}{7.8}\)+ \(\frac{1}{8.9}\)+ \(\frac{1}{9.10}\)
\(=\)\(1\)\(-\)\(\frac{1}{10}\)
\(=\)\(\frac{9}{10}\)
A = ( -4/5 + 4/3 ) + (-5/4 + 14/5) - 7/3
= 8/15 + 31/20 - 7/3
= 25/12 - 7/3
= -1/4
B = 8/3 x 2/5 x 3/8 x 10x 19/92
= 16/15 x 15/4 x 19/92
= 4x19/92
= 19/23
C = - \(\dfrac{5}{7}\) x \(\dfrac{2}{11}\) + \(\dfrac{-5}{7}\) x \(\dfrac{9}{14}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{10}{77}\) - \(\dfrac{45}{98}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{635}{1078}\) + \(\dfrac{1}{57}\)
= - \(\dfrac{36195}{61446}\) + \(\dfrac{1078}{61446}\)
= - \(\dfrac{35117}{61446}\)
Mình không thể giải thích được nhưng kết quả chắc chắn là : \(\frac{8}{9}\)
Bài 1:
Đặt \(A=\frac{2}{1x2}+\frac{2}{2x3}+\frac{2}{3x4}+...+\frac{2}{18x19}+\frac{2}{19x20}\)
\(\frac{A}{2}=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{18x19}+\frac{1}{19x20}\)
\(\frac{A}{2}=\frac{2-1}{1x2}+\frac{3-2}{2x3}+\frac{4-3}{3x4}+...+\frac{19-18}{18x19}+\frac{20-19}{19x20}\)
\(\frac{A}{2}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{18}-\frac{1}{19}+\frac{1}{19}-\frac{1}{20}=1-\frac{1}{20}=\frac{19}{20}\)
\(A=\frac{2x19}{20}=\frac{19}{10}\)
Bài 2:
Đặt \(B=\frac{1}{1x2}+\frac{1}{2x3}+\frac{1}{3x4}+...+\frac{1}{8x9}+\frac{1}{9x10}\)
Làm tương tự câu 1 có \(B=1-\frac{1}{10}=\frac{9}{10}\)
\(Bx100=\frac{9}{10}x100=90\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]:\frac{1}{2}=1\)
=> \(\left[\frac{5}{2}:\left(x+\frac{206}{100}\right)\right]=\frac{1}{2}\)
=> \(x+\frac{206}{100}=\frac{5}{2}:\frac{1}{2}=5\Rightarrow x=5-\frac{206}{100}=\frac{294}{100}=\frac{147}{50}\)
a) 3/7 + 4/9 + 4/7 + 5/9
= ( 3/7 + 4/7 ) + ( 4/9 + 5/9 )
= 7/7 + 9/9
= 1 + 1
= 2
b)1/5 + 4/10 + 9/15 + 16/20 + 25/25 + 36/30 + 49/35 + 64/40 + 81/45
= 1/5 + 2/5 + 3/5 + 4/5 + 5/5 + 6/5 + 7/5 + 8/5 + 9/5
= ( 1/5 + 9/5 ) + ( 2/5 + 8/5 ) + (7/5 + 3/5 ) + ( 4/5 + 6/5 ) + 5/5
= 2 + 2 + 2 + 2 + 1
= 2 x 4 + 1
= 8 +1
= 9
c) 1/8 + 1/12 + 3/8 + 5/12
= ( 1/8 + 3/8 ) + ( 1/12 + 5/12)
= 4/8 + 6/12
= 1/2 + 1/2
= 2/4 = 1/2
mỏi tay rồi
d; (1 - \(\dfrac{1}{2}\)) x (1 - \(\dfrac{1}{3}\)) x (1 - \(\dfrac{1}{4}\)) x ... x ( 1 - \(\dfrac{1}{100}\))
= \(\dfrac{1}{2}\) x \(\dfrac{2}{3}\) x \(\dfrac{3}{4}\) x \(\dfrac{3}{4}\) x ... x \(\dfrac{99}{100}\)
= \(\dfrac{1}{100}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.10}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=1-\frac{1}{10}\)
\(=\frac{9}{10}\)
chúc bn học tốt
\(1,27+2,77+4,27+5,77+...+31,27+32,47\)
\(=\left(1,27+32,77\right)+\left(2,77+31,27\right)+....+\left(16,27+17,77\right)\)
\(=34,04+34,04+....+34,04\)( 11 số hạng)
\(=34,04.11=374,44\)
chúc bn học tốt
\(A=3\cdot\frac{1}{1\cdot2}-5\cdot\frac{1}{2\cdot3}+7\cdot\frac{1}{3\cdot4}-\cdots+15\cdot\frac{1}{7\cdot8}-17\cdot\frac{1}{8\cdot9}\)
\(=\frac{3}{1\cdot2}-\frac{5}{2\cdot3}+\frac{7}{3\cdot4}-\cdots+\frac{15}{7\cdot8}-\frac{17}{8\cdot9}\)
\(=1+\frac12-\frac12-\frac13+\frac13+\frac14-\cdots+\frac17+\frac18-\frac18-\frac19\)
\(=1-\frac19=\frac89\)
\(\) Ta có:
\(A=\frac{3\cdot1}{1\cdot2}-\frac{5\cdot1}{2\cdot3}+\frac{7\cdot1}{3\cdot4}-\cdots+\frac{15\cdot1}{7\cdot8}-\frac{17\cdot1}{8\cdot9}\)
\(A=\frac{3}{1\cdot2}-\frac{5}{2\cdot3}+\frac{7}{3\cdot4}-\cdots+\frac{15}{7\cdot8}-\frac{17}{8\cdot9}\)
\(A=\frac{1+2}{1\cdot2}-\frac{2+3}{2\cdot3}+\frac{3+4}{3\cdot4}-\cdots+\frac{7+8}{7\cdot8}-\frac{8+9}{8\cdot9}\)
\(A=\left(\frac11+\frac12\right)-\left(\frac12+\frac13\right)+\left(\frac13+\frac14\right)-\cdots+\left(\frac17+\frac18\right)-\left(\frac18+\frac19\right)\)
\(A=\frac11+\frac12-\frac12-\frac13+\frac13+\frac14-\cdots+\frac17+\frac18-\frac18-\frac19\)
\(A=1-\frac19\)
\(A=\frac89\)
Vậy \(A=\frac89\)