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KT
2
8 tháng 5 2021
a) $2020.2020-2022.2018$
$ = 2020^2-(2020+2).(2020-2)$
$ = 2020^2 - (2020^2-2^2)$
$ = 4$
b) \(\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)...\left(\dfrac{1}{400}-1\right)\)
\(=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right)...\left(\dfrac{1}{20^2}-1\right)\)
\(=\dfrac{\left(-1\right)\cdot3\cdot\left(-2\right)\cdot4\cdot\left(-3\right)\cdot5\cdot\cdot\cdot\left(-19\right)\cdot21}{2^2\cdot3^2\cdot4^2\cdot\cdot\cdot20^2}\)
\(=-\dfrac{1}{20}\cdot\dfrac{21}{2}=-\dfrac{21}{40}\)
8 tháng 5 2021
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
LT
0
DT
0
U
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
\(A=\frac{\text{2.3.5+2.2.3.3.5.5+2.5.3.7.2.2.2.5 }}{2.3.7+2.2.3.3.7.5+2.3.3.3.7.7+2.5.3.7.7.2.2.2}\)
\(=\frac{2.3.5.\left(1+2.3.5+2^3.5.7\right)}{2.3.7.\left(1+2.3.5+3.3.7+2^3.5.7\right)}\)
\(=\frac{2.3.5}{2.3.7.3.3.7}=\frac{5}{441}\)
7 tháng 3 2021
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: \(B=\left(\dfrac{1}{4}-1\right)\left(\dfrac{1}{9}-1\right)\left(\dfrac{1}{16}-1\right)\cdot...\cdot\left(\dfrac{1}{400}-1\right)\)
\(=\dfrac{-3}{4}\cdot\dfrac{-8}{9}\cdot\dfrac{-15}{16}\cdot...\cdot\dfrac{-399}{400}\)
\(=\dfrac{-3\cdot8\cdot15\cdot...\cdot399}{4\cdot9\cdot16\cdot...\cdot400}\)
\(=\dfrac{-3\cdot2\cdot4\cdot3\cdot5\cdot...\cdot19\cdot21}{2^2\cdot3^2\cdot4^2\cdot...\cdot20^2}\)
\(=\dfrac{-2\cdot3\cdot4\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\dfrac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot20}\)
\(=\dfrac{-1}{20}\cdot\dfrac{21}{2}\)
\(=\dfrac{-21}{40}\)
BT
0
ML
1
3 tháng 5 2018
\(A=\left(\frac{1}{4}-1\right)\left(\frac{1}{9}-1\right)\left(\frac{1}{16}-1\right)...\left(\frac{1}{400}-1\right)\)
\(-A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{400}\right)\)
\(-A=\frac{3}{4}\cdot\frac{8}{9}\cdot\frac{15}{16}\cdot...\cdot\frac{399}{400}\)
\(-A=\frac{1\cdot3}{2\cdot2}\cdot\frac{2.4}{3.3}\cdot\frac{3.5}{4.4}\cdot...\cdot\frac{19.21}{20.20}\)
\(-A=\frac{1\cdot2\cdot3\cdot...\cdot19}{2\cdot3\cdot4\cdot...\cdot20}\cdot\frac{3\cdot4\cdot5\cdot...\cdot21}{2\cdot3\cdot4\cdot...\cdot20}\)
\(-A=\frac{1}{20}\cdot\frac{21}{2}=\frac{21}{40}>\frac{20}{40}=\frac{1}{2}\)
\(-A>\frac{1}{2}\Rightarrow A< \frac{1}{2}\)
