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a, \(\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{49.50}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)
b, \(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}=\frac{100}{101}\)
\(\frac{1}{5\times6}+\frac{1}{6\times7}+\frac{1}{7\times8}+...+\frac{1}{49\times50}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{5}-\frac{1}{50}=\frac{9}{50}\)
~ Hok tốt ~
\(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\)
\(=2\cdot\left(\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{99\cdot101}\right):2\)
\(=\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{99\cdot101}\right):2\)
\(=\left(\frac{3-1}{1\cdot3}+\frac{5-3}{3\cdot5}+\frac{7-5}{5\cdot7}+...+\frac{101-99}{99\cdot101}\right):2\)
\(=\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right):2\)
\(=\left(\frac{1}{1}-\frac{1}{101}\right):2\)
\(=\frac{100}{101}:2=\frac{50}{101}\).
B=2/3x5 + 2/5x7 + 2/7x9 + ...+2/99x101
B= 1/3 - 1/5 + 1/5 - 1/7 + 1/7 -1/9 + ... + 1/99 - 1/101
B= 1/3 - 1/101
B=98/303
( k mk nhé ! Cách làm câu a và b của mk đều đúng 100% đấy ! Dạng này mk học từ lâu rồi ! )
sửa đề câu a và câu b nhá , mik nghĩ đề như này :
\(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
= \(\frac{1}{1}-\frac{1}{215}\)
\(=\frac{214}{215}\)
b, đặt \(A=\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+\frac{1}{7\cdot9}+...+\frac{1}{213\cdot215}\)
\(A\cdot2=\frac{2}{1\cdot3}+\frac{2}{3.5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{213\cdot215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{213}-\frac{1}{215}\)
\(A\cdot2=\frac{1}{1}-\frac{1}{215}\)
\(A\cdot2=\frac{214}{215}\)
\(A=\frac{214}{215}:2\)
\(A=\frac{107}{215}\)
2) \(\frac{3}{1\times3}+\frac{3}{3\times5}+\frac{3}{5\times7}+...+\frac{3}{99\times101}+\frac{3}{101\times103}\)
\(=\frac{3}{2}\times\left(\frac{2}{1\times3}+\frac{2}{3\times5}+\frac{2}{5\times7}+...+\frac{2}{101\times103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{101}-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\left(1-\frac{1}{103}\right)\)
\(=\frac{3}{2}\times\frac{101}{103}\)
\(=\frac{303}{206}\)
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: \(\left(1-\dfrac{1}{3}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{3}-\dfrac{1}{5}\right)\times\dfrac{1}{2}+\left(\dfrac{1}{5}-\dfrac{1}{7}\right)\times\dfrac{1}{2}+...+\left(\dfrac{1}{x}-\dfrac{1}{x+3}\right).\dfrac{1}{2}=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(\dfrac{1}{2}\times\left(1-\dfrac{1}{x+3}\right)=\dfrac{99}{200}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{200}:\dfrac{1}{2}\)
\(1-\dfrac{1}{x+3}=\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=1-\dfrac{99}{100}\)
\(\dfrac{1}{x+1}=\dfrac{1}{100}\)
\(\Rightarrow x+1=100\)
\(x=100-1\)
\(x=99\)
Câu 2: Bằng 50/101
(Đặt biểu thức đó là A. Nhân 2 vào 2A = 2/1x3+2/3x5+2/5x7+...+2/99x101
=>2A= 1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101
=1-1/101=100/101
=>A=100/101:2=50/101)
Đặt A = 1/3.5 + 1/5.7 + 1/7.9 + ..... + 1/99.101
=> 2A = 2/3.5 + 2/5.7 + 2/7.9 + ..... + 2/99.101
=> 2A = 1/3 - 1/5 + 1/5 - 1/7 + ..... + 1/99 - 1/101
=> 2A = 1/3 - 1/101
=> 2A = 88/303
=> A = 44/303