1/3x5/7x9/11x....x37/39
7/5x11/9x15/13x....x39/37
Tính tích M x N, biết:
M = 1/3x5/7x9/11x13/15x...x37/39
N= 7/5x11/9x15/13x...x39/37
bài 4:Cho
M=1/3x5/7x9/11x13/15x...x37/39
N=7/5x11/9x15/13x...x39/37
Hãy tính MxN?
trình bày chi tiết cho mik nha mik đag cần gấp
Ta có: \(M\cdot N\)
\(=\dfrac{1}{3}\cdot\dfrac{5}{7}\cdot\dfrac{7}{9}\cdot\dfrac{13}{15}\cdot...\cdot\dfrac{37}{39}\cdot\dfrac{7}{5}\cdot\dfrac{11}{9}\cdot...\cdot\dfrac{39}{37}\)
\(=\dfrac{1}{3}\)
bài 8 cho
m=1/3x5/7x9/11x13/15x.....37/39
n=7/5x11/9x15/13x....39/37
Ta có: M⋅N
=1/3⋅5/7⋅7/9⋅13/15⋅...⋅37/39⋅7/5⋅11/9⋅...⋅39/37
=1/3
Bài 1: Cho
M=1/3x5/7x9/11x13/15x....x37/39
N=7/5x11/9x15/13x...x39/37
Tính MxN=?
Bài 2: Tính tích của 10 hỗn số đầu tiên trong dãy các hỗn số sau: \(1\frac{1}{3}x1\frac{1}{8}x1\frac{1}{15}x1\frac{1}{24}x1\frac{1}{35}x...\)
tính m,n biết
m=1/5x7/5x9/11x13/15x........x37/39
n=7/5x11/9x/15/13x..........x59/57
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tính nhanh: A=1/3x5+3/5x11+2/11x15+3/15x21
\(A=\frac{1}{3\times5}+\frac{3}{5\times11}+\frac{2}{11\times15}+\frac{3}{15\times21}\)
\(A=\frac{1}{2}\times\left(\frac{2}{3\times5}+\frac{6}{5\times11}+\frac{4}{11\times15}+\frac{6}{15\times21}\right)\)
\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{21}\right)\)
\(A=\frac{1}{2}\times\left(\frac{1}{3}-\frac{1}{21}\right)\)
\(A=\frac{1}{2}\times\frac{2}{7}\)
\(A=\frac{1}{7}\)
A = \(\frac{1}{3.5}+\frac{3}{5.11}+\frac{2}{11.15}+\frac{3}{15.21}\)
A = \(\frac{1}{2}.\left(\frac{2}{3.5}+\frac{6}{5.11}+\frac{4}{11.15}+\frac{6}{15.21}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15}+\frac{1}{15}-\frac{1}{21}\right)\)
A = \(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{21}\right)\)
A = \(\frac{1}{2}.\frac{2}{7}\)
A = \(\frac{1}{7}\)
\(B=\dfrac{3}{3x5}+\dfrac{3}{5x7}+\dfrac{3}{7x9}+....+\dfrac{3}{48x50}\)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!
1/1x3 + 1/3x5 + 1/5x7 + 1/7x9
=(1-1/3+1/3-1/5+1/5-1/7+1/7-1/9) chia 2
=(1-1/9)chia 2
=8/9 chia 2
=4/9
1/3x5 + 1/5x7 + 1/7x9 + ... + 1/2009x2011
A = \(\dfrac{1}{3\times5}\) + \(\dfrac{1}{5\times7}\) + \(\dfrac{1}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\)
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{2}{3\times5}\) + \(\dfrac{2}{5\times7}\)+ \(\dfrac{2}{7\times9}\)+...+ \(\dfrac{1}{2009\times2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{1}{5}\) - \(\dfrac{1}{7}\) + \(\dfrac{1}{7}\) - \(\dfrac{1}{9}\)+...+ \(\dfrac{1}{2009}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) ( \(\dfrac{1}{3}\) - \(\dfrac{1}{2011}\))
A = \(\dfrac{1}{2}\) \(\times\) \(\dfrac{2008}{6033}\)
A = \(\dfrac{1004}{6033}\)
\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{2}{7\times9}+..+\dfrac{1}{2009\times2011}\\ =\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\\ =\dfrac{1}{3}-\dfrac{1}{2011}\)
Đến đây bn tự tính nhé.
\(\text {#DNamNgV}\)
\(\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+\dfrac{1}{7\times9}+...+\dfrac{1}{2009\times2011}\)
\(=\dfrac{1}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{2009\times2011}\right)\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2009}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\times\left[\dfrac{1}{3}-\left(\dfrac{1}{5}-\dfrac{1}{5}\right)-\left(\dfrac{1}{7}-\dfrac{1}{7}\right)-...-\dfrac{1}{2011}\right]\)
\(=\dfrac{1}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{2011}\right)\)
\(=\dfrac{1}{2}\times\dfrac{2008}{6033}\)
\(=\dfrac{1004}{6033}\)