A=3/3x5+3/5x7+3/7x9+...+3/97x99?
Tính tổng:
A=1/1+1/2+1/3+1/4+1/5+1/6
B=1x3+3x5+5x7+7x9+...+95x97+97x99
A=1/3x5 + 1/5x7 +1/7x9 +.....+1/97x99
Giups con với ạ
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)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
a. 3/(3x5) + 3/(5x7) + 3/(7x9) +... + 3/(99x101)
b. 5/(3x5) +5/(5x7) +5/(7x9) +...+ 5/(99x101)
917749738461936926399639748776398646491639394748947630373937366
tính nhanh
1x3 + 3x5 + 5x7 + 7x9 + ......... + 97x99
1/3x5+1/5x7+1/7x9+...+1/97x99
Đặt A = 1/3×5 + 1/5×7 + 1/7×9 + ... + 1/97×99
2A = 2/3×5 + 2/5×7 + 2/7×9 + ... + 2/97×99
2A = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99
2A = 1/3 - 1/99
2A = 32/99
A = 32/99 : 2
A = 32/99 × 1/2 = 16/99
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:(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..................... >.<
tìm x :
2/3x5 + 2/5x7 + 2/7x9 + ..... +2/97x99
2/3x5 + 2/5x7 + 2/7x9 + ......+ 2/97x99 = 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + ... + 1/97 - 1/99
= 1/3 - 1/99 = 96/3.99 = 32/99
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{97\cdot99}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
=\(\frac{1}{3}-\frac{1}{99}\)
=\(\frac{32}{99}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}\)
\(=\frac{33}{99}-\frac{1}{99}\)
\(=\frac{32}{99}\)