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B=1/2.5+1/3.5+1/3.7+1/4.7+...+1/190
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a)7x-3.723=4.723
7x =4.723+3.723
7x =723.(4+3)
7x =723.7
7x = 724
=> x = 24
b)2.5x+1+3.521=522
2.5x+1 = 522-3.521
2.5x+1 = 521(5-3)
2.5x+1 =521.2
=>5x+1 = 521
=>x+1 = 21
x = 21 - 1
x = 20
\(A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+...+\dfrac{2}{2021\cdot2023}\)
\(A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2021}-\dfrac{1}{2023}\)
\(A=\dfrac{1}{1}-\dfrac{1}{2023}\\ A=\dfrac{2023}{2023}-\dfrac{1}{2023}\\ A=\dfrac{2022}{2023}\)
A=21.3+23.5+...+297.99�=21.3+23.5+...+297.99
A=11−13+13−15+...+197−199�=11−13+13−15+...+197−199
A=11−199�=11−199
A=9899
=
21−5
Ta có: \(A=\frac{9}{2\cdot5}-\frac{11}{3\cdot5}+\frac{13}{3\cdot7}-\frac{15}{4\cdot7}+\cdots+\frac{197}{49\cdot99}-\frac{199}{50\cdot99}\)
\(=2\left(\frac{9}{4\cdot5}-\frac{11}{5\cdot6}+\frac{13}{6\cdot7}-\frac{15}{7\cdot8}+\cdots+\frac{197}{98\cdot99}-\frac{199}{99\cdot100}\right)\)
\(=2\left(\frac14+\frac15-\frac15-\frac16+\frac16+\frac17-\cdots+\frac{1}{98}+\frac{1}{99}-\frac{1}{99}-\frac{1}{100}\right)\)
\(=2\left(\frac14-\frac{1}{100}\right)=2\cdot\frac{24}{100}=\frac{48}{100}=\frac{12}{25}\)
Câu 2:
\(D=\dfrac{3}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{100}{101}=\dfrac{150}{101}\)
Câu 3:
\(E=2\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{205}-\dfrac{1}{207}\right)\)
\(=2\cdot\left(1-\dfrac{1}{207}\right)=2\cdot\dfrac{206}{207}=\dfrac{412}{207}\)
Câu 5:
\(G=\dfrac{1}{4}\left(1-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{17}\right)\)
\(=\dfrac{1}{4}\cdot\dfrac{16}{17}=\dfrac{4}{17}\)
3 . 6 = 3 . 4 + 2 . 3 rùi đấy bạn, bn xét từng tích rùi sẽ thấy thôi.