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B = \((\frac{1}{4\cdot9}+\frac{1}{9\cdot14}+\frac{1}{14\cdot19}+...+\frac{1}{44\cdot49})\)
Ta có 5.B = \(\frac{5}{4\cdot9}+\frac{5}{9\cdot14}+\frac{5}{14\cdot19}+...+\frac{5}{44\cdot49}=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}=\frac{1}{4}-149=\frac{45}{196}\)
Suy ra B=\(\frac{9}{196}\)
\(1-3-5-...-49=1-(3+5+...+49)\)
\(3+5+...+49\)Khoảng cách là d = 2
Số các số hạng là : \((49-3)\)/ 2 + 1 = 24
Tổng : \((49+3)\)/ 2 x 24 = 624
Suy ra : = 1 - 624 = -623
Vậy B= \(\frac{9}{196}\).\((\frac{-623}{89})=-\frac{9}{28}\)
mk ko viết lại đề đâu
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)
=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)
=\(\frac{9}{196}.\left(1-\frac{624}{89}\right)=\frac{9}{196}.\left(\frac{-623}{89}\right)\)
=\(\frac{-9}{28}\)
\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-....-49}{89}\)
\(\text{Đặt }:\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\)là \(A\)
\(\frac{1-3-5-7-...-49}{89}\)là \(B\);ta có :
\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)
\(A=\frac{1}{5}\cdot\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}\cdot\frac{45}{196}=\frac{9}{196}\)
\(B=\frac{1-3-5-7-....-49}{89}=\frac{1-\left(3+5+7+...+49\right)}{89}\)
Tổng của \(3+5+7+...+49\)là:
\(\frac{\left(3+49\right).24}{2}=624\)
\(\Rightarrow\frac{1-624}{89}=\frac{-623}{89}=-7\)
\(\Rightarrow\left(\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-...-49}{89}=A.B=\frac{9}{196}\cdot-7=-\frac{9}{28}\)
mk ko viết lại đề đâu
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)
=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)
=\(\frac{9}{196}.\left(1-\frac{624}{89}\right)=\frac{9}{196}.\left(\frac{-623}{89}\right)\)
=\(\frac{-9}{28}\)
Đặt A = \(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\)
\(\Rightarrow\) A = \(\frac{1}{5}\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+...+\frac{1}{44.49}\right)\)
\(\Rightarrow\) A = \(\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\)
\(\Rightarrow\) A = \(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\)
\(\Rightarrow\)A = \(\frac{1}{5}.\frac{45}{196}=\frac{9}{196}\)
Đặt B = \(\frac{1-3-5-7-9-...-49}{89}\)
\(\Rightarrow\)B = \(\frac{1-\left(3+5+7+9+...+49\right)}{89}\)
\(\Rightarrow\)B = \(\frac{1-624}{89}=-7\)
Vậy M =\(\frac{9}{196}.-7=-\frac{9}{28}\)
ta có
1/5(5/36+5/126+...+5/44*49)1-3-5-7-9-...-49/89
=1/5(1/4-1/9+1/9-1/14+...+1/44-1/49)-623/89
=1/5*-7(1/4-1/49)
=-7/5*45/196
=-9/128
b) \(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+..+49\right)}{12}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}=-\frac{5.9.7.89}{5.4.7.7.89}=\frac{-9}{28}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+...+49\right)}{12}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}\)
\(=-\frac{5.9.7.89}{5.4.7.7.89}\)
\(=-\frac{9}{28}\)
\(S=\frac14+\frac{1}{54}+2=1.5\)
Gọi biểu thức cần tính là A
A= 13/4.9 - 23/9.14 +33/14.19 -43/19.24 +..+ 95/44.49+19.54/49.54
A= 4+9/4.9 - 9+14/9.14 +14+19/14.19 - 19+24/19.24 +..+ 44+51 /44.49+19/49
Ta có:
A= ( 1/4 +1/9 ) - (1/9+1/14) + ( 1/14 +1/19 ) - ( 1/19 +1/24) +...
+ ( 1/44+1/49) - ( 1/49 + 1/54 )
A = 1/4 +1/9 -1/9 -1/14+ 1/14 +1/19-1/19-1/24 +...1/44+1/49 -1/49 -1/54
A = 1/4 -1/54
A= 27/108 -2/108
A=25/108
Vậy A = 25/108
Con chim có trước hay con gà có trước
\(\frac{13}{4.9}-\frac{23}{9.14}+\frac{33}{14.19}-\frac{43}{19.24}+\cdots+\frac{93}{44.49}-\frac{103}{49.54}\)
\(=\left(\frac14+\frac19\right)-\left(\frac19+\frac{1}{14}\right)+\left(\frac{1}{14}+\frac{1}{19}\right)-\left(\frac{1}{19}+\frac{1}{24}\right)+\cdots+\left(\frac{1}{44}+\frac{1}{49}\right)-\left(\frac{1}{49}+\frac{1}{54}\right)\)
\(=\frac14+\frac19-\frac19-\frac{1}{14}+\frac{1}{14}+\frac{1}{19}-\frac{1}{19}-\frac{1}{24}+\cdots+\frac{1}{44}+\frac{1}{49}-\frac{1}{49}-\frac{1}{54}\)
\(=\frac14-\frac{1}{54}\)
\(=\frac{25}{108}\)
Ta có:
13/(4.9) - 23/(9.14) + 33/(14.19) - 43/(19.24) + ... + 95/(44.49) + 19.54/(49.54)
= 1/4 - 1/9 - (1/9 - 1/14) + (1/14 - 1/19) - ... + 1/44 - 1/49 + 1/49
= 1/4
Vậy kết quả là 1/4, vì tách các phân số thành hiệu hai phân số để các số hạng triệt tiêu nhau.
Sửa đề: \(\frac{13}{4\cdot9}-\frac{23}{9\cdot14}+\frac{33}{14\cdot19}-\frac{43}{19\cdot24}+\cdots+\frac{93}{44\cdot49}-\frac{103}{49\cdot54}\)
\(=\frac14+\frac19-\left(\frac19+\frac{1}{14}\right)+\left(\frac{1}{14}+\frac{1}{19}\right)-\cdots+\left(\frac{1}{44}+\frac{1}{49}\right)-\left(\frac{1}{49}+\frac{1}{54}\right)\)
\(=\frac14+\frac19-\frac19-\frac{1}{14}+\frac{1}{14}+\frac{1}{19}-\cdots+\frac{1}{44}+\frac{1}{49}-\frac{1}{49}-\frac{1}{54}\)
\(=\frac14-\frac{1}{54}=\frac{54-4}{54\cdot4}=\frac{50}{216}=\frac{25}{108}\)