2/2x5+2/5x8+2/8x11+...+2/95x98+2/98x101
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\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(A=\frac{1}{3}\cdot\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\cdot\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}\cdot\frac{24}{49}\)
\(A=\frac{8}{49}\)
Đề sai rồi bạn nhé, đề là như thế này:
\(A=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{92.95}+\frac{1}{95.98}\)
\(A=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{92.95}+\frac{3}{95.98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{92}-\frac{1}{95}+\frac{1}{95}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{98}\right)\)
\(A=\frac{1}{3}.\frac{24}{49}\)
\(A=\frac{24}{147}=\frac{8}{49}\)
$\frac{2}{5\times 8}+\frac{2}{8\times 11}+\frac{2}{11\times 14}+...+\frac{2}{95\times 98}$
$=\left(\frac{3}{5\times 8}+\frac{3}{8\times 11}+\frac{3}{11\times 14}+...+\frac{3}{95\times 98}\right)\times \frac{2}{3}$
$=\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{95}-\frac{1}{98}\right)\times \frac{2}{3}$
$=\left(\frac{1}{5}-\frac{1}{98}\right)\times \frac{2}{3}$
$=\frac{93}{490}\times \frac{2}{3}$
$=\frac{93\times 2}{490\times 3}$
$=\frac{31\times 1}{245\times 1}$
$=\frac{31}{245}$
\(\frac{9}{2.5}+\frac{39}{5.8}+\frac{87}{8.11}+...+\frac{9897}{98.101}\)
\(=1-\frac{1}{2.5}+1-\frac{1}{5.8}+1-\frac{1}{8.11}+...+1-\frac{1}{98.101}\)
\(=1+1+...+1-\left(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+...+\frac{1}{98.101}\right)\) \(\left(\text{33 chữ số 1}\right)\)
\(=33-\frac{1}{3}\left(\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+...+\frac{3}{98.101}\right)\)
\(=33-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+...+\frac{1}{98}-\frac{1}{101}\right)\)
\(=3-\frac{1}{3}\left(\frac{1}{2}-\frac{1}{101}\right)\)
\(=3-\frac{1}{3}-\frac{99}{202}\)
\(=\frac{1319}{606}\)
\(A=\dfrac{2}{5\times8}+\dfrac{2}{8\times11}+\dfrac{2}{11\times14}+...+\dfrac{2}{95\times98}\)
\(=2\times\dfrac{1}{3}\times\left(\dfrac{1}{5}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{11}+\dfrac{1}{11}-\dfrac{1}{14}+...+\dfrac{1}{95}-\dfrac{1}{98}\right)\)
\(=\dfrac{2}{3}\times\left(\dfrac{1}{5}-\dfrac{1}{98}\right)\)
\(=\dfrac{2}{3}\times\dfrac{93}{490}\)
\(=\dfrac{31}{245}\)
\(\dfrac{2}{5x8}\) + \(\dfrac{2}{8x11}\) + \(\dfrac{2}{11x14}\)+........+\(\dfrac{2}{95x98}\)
Đặt \(Shin=\frac{6}{2.5}+\frac{6}{5.8}+\frac{6}{8.11}+...+\frac{6}{98.101}\)
\(\Rightarrow3Shin=\frac{6}{2}-\frac{6}{5}+\frac{6}{5}-\frac{6}{8}+\frac{6}{8}-\frac{6}{11}+...+\frac{6}{98}-\frac{6}{101}\)
\(\Leftrightarrow3Shin=\frac{6}{2}-\frac{6}{101}=\frac{297}{101}\)
N/t: . là dấu nhân nha! Cái đó lớp 5 chưa biết đâu! Lên cấp 2 mới học. Trong bài làm bạn cứ ghi là dấu " x" thay cho dấu "." của mình nha!
Đặt A = 6/2.5 + 6/5.8 + ... + 6/98.101
=> A = 6.(1/2.5 + 1/5.8 + ... + 1/98.101)
=> 3A = 6.(3/2.5 + 3/5.8 + ... + 3/98.101)
=> 3A = 6.(1/2 - 1/5 + 1/5 - 1/8 + ... + 1/98 - 1/101)
=> 3A = 6.99/202
=> 3A = 297/101
=> A = 99/101
\(\frac{2}{5\times8}+\frac{2}{8\times11}+\frac{2}{11\times14}+...+\frac{2}{95\times98}\)
\(=\left(\frac{3}{5\times8}+\frac{3}{8\times11}+\frac{3}{11\times14}+...+\frac{3}{95\times98}\right)\times\frac{2}{3}\)
\(=\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+\frac{1}{11}-\frac{1}{13}+\frac{1}{13}-\frac{1}{14}+...+\frac{1}{95}-\frac{1}{98}\right)\times\frac{2}{3}\)
\(=\left(\frac{1}{5}-\frac{1}{98}\right)\times\frac{2}{3}\)
\(=\frac{93}{490}\times\frac{2}{3}\)
\(=\frac{93\times2}{490\times3}\)
\(=\frac{31\times1}{245\times1}\)
\(=\frac{31}{245}\)
2/5x8+2/8x11+2/11x14+...+2/95x98
=2(1/5x8+1/8x11+1/11x14+...+1/95x98) (khoang cach tu 5-8;8-11;11-14;...;95-98 la 3) suy ra =2/3(1/5-1/8+1/8-1/11+1/11-1/14+...+1/95-1/98)
=2/3(1/5-1/98)=2/3x93/5x98=31/245
top scorer cop tại:tính nhanh:2/2*5+2/5*8+2/8*11+2/11*14+2/14*17? | Yahoo Hỏi & Đáp
có cách làm tại:Giúp tôi giải toán - Hỏi đáp, thảo luận về toán học - Học toán với OnlineMath
b)
S2=6/2x5+6/5x8+6/8x11+...+6/29x32
=2.(3/2.5+3/5.8+...+3/29.32)
=2.(1/2-1/5+1/5-1/8+...+1/29-1/32)
=2.(1/2-1/32)
=2.15/32
=15/16
a)
Ta có:
S1=2/3x5+2/5x7+2/7x9+...+2/97x99
=1/3-1/5+1/5-1/7+...+1/97-1/99
=1/3-1/99
=32/99
\(\dfrac{2}{2\cdot5}+\dfrac{2}{5\cdot8}+...+\dfrac{2}{95\cdot98}+\dfrac{2}{98\cdot101}\)
\(=\dfrac{2}{3}\left(\dfrac{3}{2\cdot5}+\dfrac{3}{5\cdot8}+...+\dfrac{3}{95\cdot98}+\dfrac{3}{98\cdot101}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{8}+...+\dfrac{1}{95}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{101}\right)\)
\(=\dfrac{2}{3}\left(\dfrac{1}{2}-\dfrac{1}{101}\right)=\dfrac{2}{3}\cdot\dfrac{99}{202}=\dfrac{33}{101}\)