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`2/(1xx3)+2/(3xx5)+2/(5xx7)+...+2/(99xx101)` đề phải ntn chứ mà nhỉ
`=1/1-1/3+1/3-1/5+1/5-1/7+...+1/99-1/101`
`=1/1-1/101`
`=101/101-1/101`
`=100/101`
(Sửa phần 3 / 3 x 5 = 2 / 3 x 5)
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{99\times101}\)
Ta có: \(=2\times\left(\dfrac{1}{1\times3}+\dfrac{1}{3\times5}+\dfrac{1}{5\times7}+...+\dfrac{1}{99\times101}\right)\)
\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\right)\)
\(=2\times\left(1-\dfrac{1}{101}\right)\)
\(=2\times\dfrac{100}{101}\)
\(=\dfrac{200}{101}\)
Giải:
\(B=\dfrac{3}{3\times5}+\dfrac{3}{5\times7}+\dfrac{3}{7\times9}+...+\dfrac{3}{48\times50}\)
\(B=\dfrac{3}{2}\times\left(\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+\dfrac{2}{7\times9}+...+\dfrac{2}{48\times50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{48}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\left(\dfrac{1}{3}-\dfrac{1}{50}\right)\)
\(B=\dfrac{3}{2}\times\dfrac{47}{150}\)
\(B=\dfrac{47}{100}\)
Chúc em học tốt!
\(I=\dfrac{1}{1\cdot3}+\dfrac{1}{3\cdot5}+...+\dfrac{1}{199\cdot201}\)
\(=\dfrac{1}{2}\left(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{199\cdot201}\right)\)
\(=\dfrac{1}{2}\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{199}-\dfrac{1}{201}\right)\)
\(=\dfrac{1}{2}\cdot\dfrac{200}{201}=\dfrac{100}{201}\)
Lời giải:
\(2\times I=\frac{2}{1\times 3}+\frac{2}{3\times 5}+\frac{2}{5\times 7}+...+\frac{2}{199\times 201}\)
\(=\frac{3-1}{1\times 3}+\frac{5-3}{3\times 5}+\frac{7-5}{5\times 7}+....+\frac{201-199}{199\times 201}\)
\(=1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{199}-\frac{1}{201}\)
\(=1-\frac{1}{201}=\frac{200}{201}\)
\(I=\frac{200}{201}:2=\frac{100}{201}\)
\(K=\dfrac{4}{1\times3}+\dfrac{4}{3\times5}+...+\dfrac{4}{299\times301}\)
\(=2\times\left(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+...+\dfrac{2}{299\times301}\right)\)
\(=2\times\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)
\(=2\times\left(1-\dfrac{1}{301}\right)=2\times\dfrac{300}{301}=\dfrac{600}{301}\)
\(K=\dfrac{4}{1\cdot3}+\dfrac{4}{3\cdot5}+...+\dfrac{4}{299\cdot301}\)
\(=2\left(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{299}-\dfrac{1}{301}\right)\)
\(=2\cdot\dfrac{300}{301}=\dfrac{600}{301}\)
C = \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\) + \(\dfrac{1}{128}\)
2\(\times\)C = 1 + \(\dfrac{1}{2}\) + \(\dfrac{1}{4}\) + \(\dfrac{1}{8}\) + \(\dfrac{1}{16}\) + \(\dfrac{1}{32}\) + \(\dfrac{1}{64}\)
2 \(\times\) C - C = 1 - \(\dfrac{1}{128}\)
C = \(\dfrac{127}{128}\)
( 5/7 + x ) : 4/9 = 2/3
5/7 + x = 2/3 x 4/9
5/7 + x = 8/27
x = 8/27 - 5/7
x = -79/189
Mình tính được bài bạn nhưng hình như bạn làm sai đề hay sao mà mình làm ra âm luôn ! Mà âm thì chưa học lớp 5 , mong bạn xem lại đề xem có sai không nhé !
\(\dfrac{2}{1\times3}+\dfrac{2}{3\times5}+\dfrac{2}{5\times7}+...+\dfrac{2}{13\times15}+\dfrac{2}{15\times17}\)
\(=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}+\dfrac{1}{15}-\dfrac{1}{17}\)
\(=1-\dfrac{1}{17}\)
\(=\dfrac{16}{17}\)
\(\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{15\cdot17}\)
\(=2-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{17}\)
\(=2-\dfrac{1}{17}\)
\(=\dfrac{33}{17}\)