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Sửa đề: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)

Ta có: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)

\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-399}{400}\)

\(=\dfrac{-3\cdot8\cdot15\cdot...\cdot399}{4\cdot9\cdot16\cdot...\cdot400}\)

\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot19\cdot21}{2^2\cdot3^2\cdot4^2\cdot...\cdot20^2}\)

\(=\dfrac{-2\cdot3\cdot4\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot20}\)

\(=\dfrac{-1}{20}\cdot\dfrac{21}{2}\)

\(=\dfrac{-21}{40}\)

11 tháng 2 2017

Đề bài là gì hả bạn

a) $2020.2020-2022.2018$

$ = 2020^2-(2020+2).(2020-2)$

$ = 2020^2 - (2020^2-2^2)$

$ = 4$

b) \(\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{400}-1\right)\)

\(=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{20^2}-1\right)\)

\(=\dfrac{\left(-1\right)\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot\cdot\cdot\left(-19\right)\cdot21}{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot20^2}\)

\(=-\dfrac{1}{20}\cdot\dfrac{21}{2}=-\dfrac{21}{40}\)

Giải:

a) 2020.2020−2022.2018

=20202−(2020+2).(2020−2)

=20202−(20202−22)

=4

b) (1/4−1)(1/9−1)(1/16−1)...(/1400−1)

=(1/22−1)(1/32−1)(1/42−1)...(1/202-1)

=(−1)⋅3⋅(−2)⋅4⋅(−3)⋅5⋅⋅⋅(−19)⋅21/22⋅32⋅42⋅⋅⋅202

=−1/20⋅21/2

=−21/40

21 tháng 8 2018

\(B=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{400}-1\right)\)

\(-B=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{400}\right)\)

\(-B=\frac{3}{4}\cdot\frac{5}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{399}{400}\)

\(-B=\frac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(19\cdot21\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(20\cdot20\right)}\)

\(-B=\frac{\left(1\cdot2\cdot3\cdot...\cdot19\right)\left(3\cdot4\cdot5\cdot...\cdot21\right)}{\left(2\cdot3\cdot4\cdot...\cdot20\right)\left(2\cdot3\cdot4\cdot...\cdot20\right)}\)

\(-B=\frac{1\cdot21}{20\cdot2}\)

\(-B=\frac{21}{40}\)

\(B=\frac{-21}{40}\)

21 tháng 8 2018

\(A=\frac{\text{2.3.5+2.2.3.3.5.5+2.5.3.7.2.2.2.5 }}{2.3.7+2.2.3.3.7.5+2.3.3.3.7.7+2.5.3.7.7.2.2.2}\)

    \(=\frac{2.3.5.\left(1+2.3.5+2^3.5.7\right)}{2.3.7.\left(1+2.3.5+3.3.7+2^3.5.7\right)}\)

     \(=\frac{2.3.5}{2.3.7.3.3.7}=\frac{5}{441}\)

3 tháng 4 2018

\(=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{399}{400}\)

\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{19.21}{20.20}\)

\(=\frac{1.2.3.....19}{2.3.4.....20}.\frac{3.4.5.....21}{2.3.4.....20}\)

\(=\frac{1}{20}.\frac{21}{2}\)

\(=\frac{21}{40}\)

3 tháng 4 2018

\(\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{400}\right)\)

\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{399}{400}\)

\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{19.21}{20.20}\)

\(\frac{1.2.3...19}{2.3.4...20}.\frac{3.4.5...21}{2.3.4...20}\)

\(\frac{1}{20}.\frac{21}{2}=\frac{21}{40}\)