tính tổng s=3x5+5x7+7x9+...+47x49
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)Tính nhanh:
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)
Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
Chúc em học tốt!
Tính nhanh
Q=\(\frac{3}{3x5}\)+\(\frac{3}{5x7}\)+\(\frac{3}{7x9}\)+....+\(\frac{3}{47x49}\)
Q= 3/3x5 + 3/5x7 + 3/7x9 +...+ 3/47x49
Q= (3/3 -3/5) + (3/5-3/7) + (3/7-3/9)+...+(3/47-3/49)
Q= 3/3 - 3/5 + 3/5 - 3/7 + 3/7 - 3/9 + ... + 3/47 - 3/49
Q=3/3 - 3/49
Q= 46/49
Tính
a. 3/(3x5) + 3/(5x7) + 3/(7x9) +... + 3/(99x101)
b. 5/(3x5) +5/(5x7) +5/(7x9) +...+ 5/(99x101)
917749738461936926399639748776398646491639394748947630373937366
Tính tổng:(giúp em nhá) B=1x3+3x5+5x7+7x9+...+95x97+97x99
B=1x3+3x5+5x7+7x9+...+95x97+97x99
= 1.(1+2)+3.(3+2)+5.(5+2)+....+95.(95+2)+97.(97+2)
= 12+1.2+32+3.2 +52+5.2+...+952+95.2+ 972+97.2
= (12+32 +52+...+952+ 972)+(1.2+3.2 +5.2+...+95.2+97.2)
= (12+32 +52+...+952+ 972)+ 2.(1+3 +5+...+95+97)
Đặt : A = 12+32 +52+...+952+ 972
C =1+3 +5+...+95+97
tính A và C (tìm câu hỏi tương tự hình như anh thấy họ làm rồi đấy) sau đó thay vào tính B
Tính tổng sau:
\(\dfrac{2}{3X5}+\dfrac{2}{5X7}+\dfrac{2}{7X9}+...+\dfrac{2}{19X21}\)
`2/[3.5]+2/[5.7]+2/[7.9]+....+2/[19.21]`
`=1/3-1/5+1/5-1/7+1/7-1/9+....+1/19-1/21`
`=1/3-1/21`
`=6/21`
Bài tập tham khảo:
Bài 1: Tính tổng A = 1/3x5 + 1/5x7 + 1/7x9 + ......+1/37x39
\(A=\dfrac{1}{3.5}+\dfrac{1}{7.9}+...+\dfrac{1}{37.39}\\ =\dfrac{1}{2}\left(\dfrac{2}{3.5}+\dfrac{2}{7.9}+...+\dfrac{2}{37.39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{39}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{3}-\dfrac{1}{39}\right)\\ =\dfrac{1}{2}.\dfrac{4}{13}\\ =\dfrac{2}{13}\)
A=13.5+17.9+...+137.39=12(23.5+27.9+...+237.39)=12(13−15+15−17+...+137−139)=12(13−139)=12.413=213
Tính tổng:(giúp em nhá)
B=1x3+3x5+5x7+7x9+...+95x97+97x99
Ta có \(6B=1\times3\times6+3\times5\times6+...+97\times99\times6\)
\(=1\times3\times\left(5+1\right)+3\times5\times\left(7-1\right)+5\times7\times\left(9-3\right)+...+97\times99\times\left(101-95\right)\)
\(=1\times3\times5+1.3+3\times5\times7-3\times5\times1+...-97\times99\times95\)
\(=97\times99\times101+3\)
\(\Rightarrow B=\frac{97\times99\times101+3}{6}=161651\)
6B=1x3x6+3x5x6+5x7x6+.....+97x99x6
6B=1x3x(5+1)+3x5x(7-1)+....+97x99x(102-95)
6B=1x3x5+1x3+3x5x7-3x5+....+97x99x101-95x97x99
6B=1x3x97x99x101
6B=969906
=>B=161651
6B = 1x3x6 + 3x5x6 + 5x7x6 +...+ 99x101x6
= 1x3(5+1)+3x5(7-1)+...+97x99(101-95)
= 1x3x5+1x3+3x5x(7-1)+...-97x99x95
=97x99x101+3 rồi tự làm..................... >.<
5/1x3+5/3x5+5/5x7+...+5/47x49
Tính tổng sau bằng cách hợp lí:
A = 2/1x3 + 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11
Ta có:
A = \(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+\frac{2}{9x11}\)
= \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}\)
= \(\frac{1}{1}-\frac{1}{11}\)
=\(\frac{10}{11}\)