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m = 1/3-1/5+1/5-1/7+1/7-1/9+...+1/97-1/99
m = 1/3-1/99=32/99
Sorry chị em ko làm đc câu b vì em mới học lớp 4
k em ha
a) \(M=\frac{2}{3\times5}+\frac{2}{5\times7}+\frac{2}{7\times9}+...+\frac{2}{97\times99}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(\Rightarrow M=\frac{1}{3}-\frac{1}{99}\)
\(\Rightarrow M=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
b) \(N=\frac{3}{5\times7}+\frac{3}{7\times9}+\frac{3}{9\times11}+...+\frac{3}{197\times199}\)
\(\Rightarrow N=3\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\)
\(\Rightarrow N=3\times\left[2\times\left(\frac{1}{5\times7}+\frac{1}{7\times9}+\frac{1}{9\times11}+...+\frac{1}{197\times199}\right)\right]\)
\(\Rightarrow N=3\times\left(\frac{2}{5\times7}+\frac{2}{7\times9}+\frac{2}{9\times11}+...+\frac{2}{197\times199}\right)\)
\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right)\)
\(\Rightarrow N=3\times\left(\frac{1}{5}-\frac{1}{199}\right)\)
\(\Rightarrow N=3\times\frac{194}{995}=\frac{582}{995}\)
----Chúc em học giỏi !----
\(\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{x.\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{33}{99}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy \(x=97\)
\(\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{x\cdot\left(x+2\right)}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{x}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{3}-\frac{1}{x+2}=\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{3}-\frac{32}{99}\)
\(\Rightarrow\frac{1}{x+2}=\frac{1}{99}\)
\(\Rightarrow x+2=99\)
\(\Rightarrow x=99-2\)
\(\Rightarrow x=97\)
Vậy x=97
=>A=\(\frac{7}{2}\)(\(\frac{1}{1}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{5}\)+...+\(\frac{1}{99}\)-\(\frac{1}{101}\))
=>A=\(\frac{7}{2}\)(1-\(\frac{1}{101}\))
=>A=\(\frac{350}{101}\)
7/2 ( \(\frac{1}{1}-\frac{1}{3}+\frac{1}{3}-....+\frac{1}{99}-\frac{1}{101}\))
7/2 ( 1 - 1/101 )
7/2 x 100/101
=350/101
Ta có: \(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(\Leftrightarrow M=\frac{2}{3.\left(3+2\right)}+\frac{2}{5.\left(5+2\right)}+...+\frac{2}{97\left(97+2\right)}\)
\(\Leftrightarrow M=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(\Leftrightarrow M=\frac{1}{3}-\frac{1}{99}=\frac{33}{99}-\frac{1}{99}=\frac{32}{99}\)
( Dòng thứ 2 mik làm để bạn hiểu mik đã áp dụng công thức \(\frac{a}{n\left(n+a\right)}=\frac{1}{n}-\frac{1}{n+a}\) nên bạn ghi hay ko cx được)
\(M=\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
=\(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
=\(\frac{1}{3}-\frac{1}{99}\)=\(\frac{32}{99}\)
\(C=\frac{2}{3x5}+\frac{2}{5x7}+\frac{2}{7x9}+...+\frac{2}{97x99}\)
\(C=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{97}-\frac{1}{99}\)
\(C=\frac{1}{3}-\frac{1}{97}\)
\(C=\frac{94}{291}\)
\(\frac{2}{1.3}+\frac{2}{3.5}+...+\frac{2}{99.101}\)
\(=2\left(\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{99.101}\right)\)
\(=2.\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(=1-\frac{1}{100}\Rightarrowđpcm\)
Ta có :
\(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\)
\(=1-\frac{1}{101}< 1\)\(\left(đpcm\right)\)
Đặt \(A=\frac{1}{1.2.3}+\frac{1}{3.5.7}+...+\frac{1}{45.47.49}\)
\(\Rightarrow4A=\frac{4}{1.3.5}+\frac{4}{3.5.7}+...+\frac{4}{45.47.49}\)
\(=\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+...+\frac{1}{45.47}-\frac{1}{47.49}\)
\(=\frac{1}{3}-\frac{1}{47.49}\)
\(\Rightarrow A=\frac{\frac{1}{3}-\frac{1}{47.49}}{4}=\frac{575}{6909}\)
\(\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+...+\frac{4}{99.101}\)
\(=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(=2.\left(1-\frac{1}{101}\right)\)
\(=2.\frac{100}{101}=\frac{200}{101}\)
Đặt \(A=\frac{4}{1.3}+\frac{4}{3.5}+\frac{4}{5.7}+..+\frac{4}{99.101}\)
\(A=2.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{99.101}\right)\)
\(A=2.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{101}\right)\)
\(A=2.\left(1-\frac{1}{101}\right)\)
\(A=\frac{2.100}{101}=\frac{200}{101}\)
Ủng hộ mk nha !!! ^_^
1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10
=1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6+1/7-1/7+1/8-1/8+1/9+1/9-1/10
=1/2-1/10
=5/10-1/10
=4/10=2/5
\(\frac{1}{2x3}+\frac{1}{3x4}+\frac{1}{4x5}+\frac{1}{5x6}+\frac{1}{6x7}+\frac{1}{8x9}+\frac{1}{9x10}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}\)
\(\frac{1}{2}-\frac{1}{10}\)
\(\frac{2}{5}\)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(=2.\left(\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.101}\right)\)
\(=2.\left(\frac{1}{2}.\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\right)\right)\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{101}{303}-\frac{3}{303}=\frac{98}{303}\)
Đặt A = \(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{99.101}\)
\(\Leftrightarrow A=\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{99.100}\)
\(=1-\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}-\frac{1}{7}+\frac{1}{9}+...+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=1-\frac{1}{100}=\frac{99}{100}\)