Tính giá trị của biểu thức sau :
A = ( 1 - 1 / 22) ( 1- 1/32 ) ( 1- 1 /42 ) ...( 1 - 1 /20182)
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Những câu hỏi liên quan
T
18 tháng 10 2023
a) \(A=2+2^2+2^3+...+2^{2017}\)
\(2A=2^2+2^3+2^4+...+2^{2018}\)
\(2A-A=\left(2^2+2^3+2^4+...+2^{2018}\right)-\left(2+2^2+2^3+...+2^{2017}\right)\)
\(A=2^{2018}-2\)
b) \(C=1+3^2+3^4+...+3^{2018}\)
\(3^2\cdot C=3^2+3^4+3^6+...+3^{2020}\)
\(9C-C=\left(3^2+3^4+3^6+...+3^{2020}\right)-\left(1+3^2+3^4+...+3^{2018}\right)\)
\(8C=3^{2020}-1\)
\(\Rightarrow C=\dfrac{3^{2020}-1}{8}\)
\(Toru\)
PT
1
CM
10 tháng 12 2018
B = 2 + 2 2 − 1 + 2 − 2 2 − 1 = ( 2 − 1 + 1 ) 2 + ( 2 − 1 − 1 ) 2 = 2 − 1 + 1 + 1 − 2 − 1 = 2
LT
7
TM
3 tháng 11 2017
\(1-\frac{1}{2}-\frac{1}{4}-\frac{1}{8}-\frac{1}{16}-\frac{1}{32}\)\(-\frac{1}{64}\)
\(=1-\frac{32}{64}-\frac{16}{64}-\frac{8}{64}-\frac{4}{64}\)\(-\frac{2}{64}-\frac{1}{64}\)
\(=1-\left(\frac{32}{64}-\frac{16}{64}-\frac{8}{64}-\frac{4}{64}-\frac{2}{64}-\frac{1}{64}\right)\)
\(=1-\frac{1}{64}\)
\(=\frac{64}{64}-\frac{1}{64}\)
\(=\frac{63}{64}\)
NM
1
15 tháng 10 2023
\(4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{16}-1\right)\cdot\left(3^{16}+1\right)\)
\(=\dfrac{1}{2}\left(3^{32}-1\right)\)
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2018^2}\right)\)
\(\Rightarrow A=\left(1-\frac{1}{4}\right)\left(1-\frac{1}{9}\right)\left(1-\frac{1}{16}\right)...\left(1-\frac{1}{4072324}\right)\)
\(\Rightarrow A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{4072323}{4072324}\)
\(\Rightarrow A=\frac{3.8.15...4072323}{4.9.16...4072324}\)
\(\Rightarrow A=\frac{3.4.2.3.5...2017.2019}{2.2.3.3.4.4...2018.2018}\)
\(\Rightarrow A=\frac{\left(2.3.4...2017\right).\left(3.4.5...2019\right)}{\left(2.3.4...2018\right).\left(2.3.4...2018\right)}\)
\(\Rightarrow A=\frac{1.2019}{2018.2}\)
\(\Rightarrow A=\frac{2019}{4036}\)
Vậy ...
P/s : Mik ko chắc đâu
~ Ủng hộ nhé
\(A=\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{2018^2}\right)\)
\(A=\frac{3}{2^2}.\frac{8}{3^2}.\frac{15}{4^2}...\frac{4072323}{2018^2}\)
\(A=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}...\frac{2017.2019}{2018^2}\)
\(A=\frac{1.2.3...2017}{2.3.4...2018}.\frac{3.4.5...2019}{2.3.4...2018}\)
\(A=\frac{1}{2018}.\frac{2019}{2}\)
\(A=\frac{2019}{4036}\)
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