2/1x3+2/3x5+2/5x7+...+2/2009x2011
4/1x3+4/3x5+4/5x7+...+4/2009x2011
Bài 1:Tính
A=1/1x3+1/3x5+1/5x7+.......+1/2009x2011
A = 1/1x3 + 1/3x5 + 1/5x7 +.........+ 1/2009x2011
= 1/1-1 +1/3-5 + 1/5-7 + .......+ 1/2009-2011
= 1/1 - 1/3 + 1/3 - 1/5 + 1/5 - 1/7 +........+ 1/2009 -1/2011
= 1/1 - 1/2011
= 2010/2011
a)1/1x3+1/3x5+1/5x7+...+1/Xx(x+3)=99/200
b)1/1x3+1/3x5+1/5x7+...+1/Xx(x+2)
a) \(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{x\times\left(x+3\right)}=\dfrac{99}{200}\)
Ta có: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
Công thức: \(\dfrac{1}{a\times b}=\) 1/ khoảng cách giữa a và b \(\times\left(\dfrac{1}{a}-\dfrac{1}{b}\right)\)
* Bạn làm theo công thức và vẫn dụng câu b nhé.
2/1x3+2/3x5+2/5x7+.....+2/41x43
2/1x3 + 2/3x5 + 2/5x7 + 2/7x9 + 2/9x11
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+\dfrac{2}{9\times11}\)
\(=2\times\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{11}\right)\)
\(=1-\dfrac{1}{11}\)
\(=\dfrac{11}{11}-\dfrac{1}{11}\)
\(=\dfrac{10}{11}\)
2/1x3+2/3x5+2/5x7+....2/99x101
Ta có : \(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+.....+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+......+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}\)
\(=\frac{100}{101}\)
Đặt : \(A=\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(A-\frac{2}{1\cdot3}=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\)
\(2A-\frac{2}{1\cdot3}=\frac{2}{3}-\frac{2}{5}+\frac{2}{5}-\frac{2}{7}+\frac{2}{7}-...+\frac{2}{99}-\frac{2}{101}\)
\(2A-\frac{2}{3}=\frac{2}{3}-\frac{2}{101}\)
\(2A-\frac{2}{3}=\frac{196}{303}\)
\(A-\frac{2}{3}=\frac{98}{303}\)
\(A=\frac{98}{303}+\frac{2}{3}=\frac{100}{101}\)
\(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{99\times101}\)
\(=\frac{3-1}{1\times3}+\frac{5-3}{3\times5}+\frac{7-5}{5\times7}+...+\frac{101-99}{99\times101}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{1}-\frac{1}{101}\)
\(=\frac{100}{101}\)
2/1x3+2/3x5+2/5x7+...+2/99x101
A = 2/1x3 + 2/3x5 + 2/5x7 + ... + 2/99x101
A = 2/1 - 2/101 = 200/101
Kết quả là 200/101 bạn nhé
2/2 + 1x3 / 3x5 + 2/2 + ······ + 5x7 / 97x99 + 2 / 99x101
= 1-1 / 3 + 1 / 3-1 / 5 + 1 / 5-1 / 7 + ... ... + 1 / 97-1 / 99 + 1 / 99-1 / 101
= 1-1 / 101
= 100/101
2/1x3+2/3x5+2/5x7+.......+2/41x43
Đặt A =2/1x3+2\3x5+................+2/41x43
A =1/1-1/3+1/3-1/5+...................+1/41-1/43
A=1-1/43
A=42/43
A=2/1X3+2/3X5+2/5X7+. . .+2/41X43
A=1/1-1/3+1/3-1/5+1/5-1/7+. . .+1/41-1/43
A=1/1-1/43
A=42/43
Tick mk nhé
2/1x3 + 2/3x5 + 2/5x7 + ...+ 2/59x61
\(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+....+\frac{2}{59x61}\)
\(=\frac{2}{2}x\left(\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+.....+\frac{2}{59x61}\right)\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+....+\frac{1}{59}-\frac{1}{61}\)
\(=1-\frac{1}{61}=\frac{60}{61}\)