\(\frac{1}{200}-\frac{1}{200.199}-\frac{1}{199.198}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{200}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{198.199}+\frac{1}{199.200}\right)\)

\(=\frac{1}{200}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{198}-\frac{1}{199}+\frac{1}{199}-\frac{1}{200}\right)\)

\(=\frac{1}{200}-\left(1+\frac{1}{200}\right)\)

\(=\left(\frac{1}{200}-\frac{1}{200}\right)-1\)

\(=0-1\)

\(=-1\)

Câu này mk giải rùi mà

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hoặc bạn sang trang 3 của hỏi đáp toán hoc24 sẽ thấy nhé

Ta có : \(A=\dfrac{1}{199}-\dfrac{1}{199.198}-\dfrac{1}{198.197}-\dfrac{1}{197.196}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(=\dfrac{1}{199}-\left(\dfrac{1}{199.198}+\dfrac{1}{198.197}+\dfrac{1}{197.196}+...+\dfrac{1}{3.2}+\dfrac{1}{2.1}\right)\)

\(=\dfrac{1}{199}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{196.197}+\dfrac{1}{197.198}+\dfrac{1}{198.199}\right)\)

\(=\dfrac{1}{199}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...-\dfrac{1}{198}+\dfrac{1}{198}-\dfrac{1}{199}\right)\)

\(=\dfrac{1}{199}-\left(1-\dfrac{1}{199}\right)\)

\(=\dfrac{1}{199}-\dfrac{198}{199}=\dfrac{-197}{199}\)

~ Học tốt ~

\(a)\) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-\left(1-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Chúc bạn học tốt ~ 

làm hộ mk lun câu b ik

a) \(A=\frac{1}{199}-\frac{1}{199.198}-\frac{1}{198.197}-\frac{1}{197.196}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(A=\frac{1}{199}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...\frac{1}{197.198}+\frac{1}{198.199}\right)\)

\(A=\frac{1}{199}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{197}-\frac{1}{198}+\frac{1}{198}-\frac{1}{199}\right)\)

\(A=\frac{1}{199}-(1-\frac{1}{199})\)

\(A=\frac{1}{199}-1+\frac{1}{199}\)

\(A=\frac{-197}{199}\)

Vậy \(A=\frac{-197}{199}\)

b) cũng thế 

Bài 1:

a) \(\left(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}\right)-\left(-\frac{1}{511}-\frac{1}{2}+\frac{5}{19}\right)\)

= \(\frac{5}{19}-\frac{1}{511}+\frac{7}{12}+\frac{1}{511}+\frac{1}{2}-\frac{5}{19}\)

= \(\left(\frac{5}{19}-\frac{5}{19}\right)+\left(\frac{1}{511}-\frac{1}{511}\right)+\left(\frac{7}{12}+\frac{1}{2}\right)\)

= 0 + 0 + \(\frac{13}{12}\)

= \(\frac{13}{12}\).

b) \(-\left(\frac{13}{25}-\frac{4}{191}+\frac{2}{51}\right)+\left(\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\right)\)

= \(-\frac{13}{25}+\frac{4}{191}-\frac{2}{51}+\frac{4}{191}+\frac{2}{51}+\frac{3}{5}\)

= \(\left(-\frac{13}{25}+\frac{3}{5}\right)+\left(\frac{4}{191}+\frac{4}{191}\right)+\left(\frac{2}{51}-\frac{2}{51}\right)\)

= \(\frac{2}{25}+\frac{8}{191}+0\)

= \(\frac{582}{4775}\).

Mình chỉ làm câu a) và câu b) thôi nhé.

Chúc bạn học tốt!