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\(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right)\cdot\frac{1-3-5-7-....-49}{89}\)
\(\text{Đặt }:\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+...+\frac{1}{44.49}\right)\)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
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\right)\)\(.\frac{1-\left(3+5+...+49\right)}{89}\)
=\(\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right).\frac{\left(1-\frac{\left(49+3\right).24}{2}\right)}{89}\)
=\(\frac{1}{5}.\frac{45}{196}.\frac{1-\left(\frac{52.24}{2}\right)}{89}\)
=\(\frac{9}{196}.\left(1-\frac{624}{89}\right)=\frac{9}{196}.\left(\frac{-623}{89}\right)\)
=\(\frac{-9}{28}\)
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
Đặt \(A=\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right).\dfrac{1-3-5-7-...-49}{89}\)
\(=\dfrac{9}{196}.\dfrac{1-3-5-7-...-49}{89}\)
Đặt \(B=1-3-5-7-..-49\)
\(=1-\left(3+5+7+...+49\right)\)
\(=1-\left\{\left(49+3\right).\left[\left(49-3\right):2+1\right]:2\right\}\)
\(=1-624\)
\(=-623\)
\(\Rightarrow\dfrac{9}{196}.\left(\dfrac{-623}{89}\right)=-\dfrac{9}{28}\)
Vậy: \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right).\dfrac{1-3-5-7-...-49}{89}=-\dfrac{9}{28}\)
Xét \(\left(\dfrac{1}{4.9}+\dfrac{1}{9.14}+\dfrac{1}{14.19}+...+\dfrac{1}{44.49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{5}{4.9}+\dfrac{5}{9.14}+\dfrac{5}{14.19}+...+\dfrac{5}{44.49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{14}+\dfrac{1}{14}-\dfrac{1}{19}+...+\dfrac{1}{44}-\dfrac{1}{49}\right)\)
=\(\dfrac{1}{5}\left(\dfrac{1}{4}-\dfrac{1}{49}\right)\)
=\(\dfrac{1}{5}.\dfrac{45}{196}\)
=\(\dfrac{9}{196}\)
Xét \(\dfrac{1-3-5-7-..-49}{89}\)
=\(\dfrac{1-\left(3+5+7+...+49\right)}{89}\)
CT tính sl số hạng (số cuối - số đầu ):2+1
số lượng số hạn của dãy 3+5+7+...+49 là (49-3):2+1=24
Áp dụng CT tính tổng số hạng dãy số cách đều Tổng = [ (số đầu + số cuối) x Số lượng số hạng ] : 2
=> tổng = [(3+49).24]:2=624
=>\(\dfrac{1-624}{89}\)
=\(\dfrac{-623}{89}\)
=-7
từ đó ta có \(\dfrac{9}{196}.\left(-7\right)=\dfrac{-9}{28}\)
\(=\left[\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}\right)+\frac{1}{5}\left(\frac{1}{9}-\frac{1}{14}\right)+\frac{1}{5}\left(\frac{1}{14}-\frac{1}{19}\right)+...+\frac{1}{5}\left(\frac{1}{44}-\frac{1}{49}\right)\right]\cdot\frac{1-\left(3+5+...+49\right)}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-...+\frac{1}{44}-\frac{1}{49}\right)\cdot\frac{1-\left(52+52+...+52\right)\left\{12\text{ số 52}\right\}}{89}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\cdot\frac{1-624}{89}\)
\(=\frac{9}{196}\cdot-7=\frac{9}{28}\)
\(Xét A = 1/(4.9)+1/(9.14)+1/(14.19)+...+1/(44.49)
-> 5A = 5/(4.9) + 5/(9.14) + 5/(14.19) + ... + 5/(44.49)
= 1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 + ... + 1/44 - 1/49
= 1/4 - 1/49 = 45/196 -> A = 9 / 196
Xét B = (−1−3−5−7−...−49)/89
= (1 + 3 + 5 + ... + 49) / -89
= 625 / -89
biểu thức đầu bài có giá trị: A.B = 9/196 * 625/-89 = - 5625 / 17444\)Xét A = 1/(4.9)+1/(9.14)+1/(14.19)+...+1/(44.49)
-> 5A = 5/(4.9) + 5/(9.14) + 5/(14.19) + ... + 5/(44.49)
= 1/4 - 1/9 + 1/9 - 1/14 + 1/14 - 1/19 + ... + 1/44 - 1/49
= 1/4 - 1/49 = 45/196 -> A = 9 / 196
Xét B = (−1−3−5−7−...−49)/89
= (1 + 3 + 5 + ... + 49) / -89
= 625 / -89
biểu thức đầu bài có giá trị: A.B = 9/196 * 625/-89 = - 5625 / 17444
tick nha
Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}++\frac{1}{14.19}+......+\frac{1}{44.49}\)
\(A=\frac{1}{5}.\left(\frac{5}{4.9}+\frac{5}{9.14}+\frac{5}{14.19}+.....+\frac{5}{44.49}\right)\)
\(A=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+.....+\frac{1}{44}-\frac{1}{49}\right)\)
\(A=\frac{1}{5}.\left(\frac{1}{4}-\frac{1}{49}\right)=\frac{1}{5}.\frac{45}{196}=\frac{9}{196}\)
Đặt \(B=\frac{1-3-5-7-.......47-49}{89}\)
\(B=\frac{1-\left(3+5+7+......+47+49\right)}{89}\)
Từ 3 -> 49 có: (49-3):2+1=24(số hạng)
=>\(3+5+7+....+47+49=\frac{\left(49+3\right).24}{2}=624\)
=>\(B=\frac{1-624}{89}=\frac{-623}{89}=-7\)
Vậy \(\left(\frac{1}{4.9}+\frac{1}{9.14}+\frac{1}{14.19}+....+\frac{1}{44.49}\right).\frac{1-3-5-,,,,,-49}{89}=A.B=\frac{9}{196}.\left(-7\right)=-\frac{9}{28}\)
b) \(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+\frac{1}{14}-\frac{1}{19}+...+\frac{1}{44}-\frac{1}{49}\right)\frac{2-\left(1+3+5+7+..+49\right)}{12}\)
\(=\frac{1}{5}\left(\frac{1}{4}-\frac{1}{49}\right)\frac{2-\left(12.50+25\right)}{89}=-\frac{5.9.7.89}{5.4.7.7.89}=\frac{-9}{28}\)
Đặt \(A=\frac{1}{4.9}+\frac{1}{9.14}+...+\frac{1}{44.49}\)
\(5A=\frac{1}{4}-\frac{1}{9}+\frac{1}{9}-\frac{1}{14}+...+\frac{1}{44}-\frac{1}{49}\)
\(5A=\frac{1}{4}-\frac{1}{49}\)
\(A=\frac{45}{196}:5=\frac{9}{196}\)
Đặt \(B=\frac{1-3-...-49}{89}\)
\(B=\frac{\left(1-3\right)-\left(5-7\right)-...-\left(47-49\right)}{89}\)
\(B=\frac{\left(-2\right)-\left(-2\right)-...-\left(-2\right)}{89}\)
\(B=\frac{-2+2+...+2}{89}\)
\(B=\frac{\left(-2\right)+2\times24}{89}\)
\(B=\frac{46}{89}\)
\(P=A.B=\frac{9}{196}.\frac{46}{89}\)
\(P=\frac{207}{8722}\)
