Tính : 1+2+22+23+.....+21007
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=\(\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{1}{22}-\dfrac{23}{22}\right)+\dfrac{2}{3}\)
=\(\dfrac{17}{17}-\dfrac{22}{22}+\dfrac{2}{3}\)
=\(1-1+\dfrac{2}{3}\)
=0+\(\dfrac{2}{3}\)
=\(\dfrac{2}{3}\)
`#3107.101107`
Đặt $A = 1 + 2 + 2^2 + 2^3 + ... + 2^{50}$
$2A = 2 + 2^2 + 2^3 + ... + 2^{51}$
$2A - A = (2 + 2^2 + 2^3 + ... + 2^{51}) - (1 + 2 + 2^2 + ... + 2^{50})$
$A = 2 + 2^2 + 2^3 + ... + 2^{51] - 1 - 2 - 2^2 - ... - 2^{50}$
$A = 2^{51} - 1$
Vậy, `A =` $2^{51} - 1.$
\(S_1=1+2+2^2+2^3+..+2^{63}\\ \Rightarrow2S_1=2+2^2+2^3+2^4+...+2^{64}\\ \Rightarrow S_1-2S_1=1-2^{64}\\ \Rightarrow-S_1=1-2^{64}\\ \Rightarrow S_1=2^{64}-1.\)
a, \(\dfrac{7}{22}\) - \(\dfrac{15}{23}\) + \(\dfrac{2022}{2023}\) - \(\dfrac{8}{23}\) + \(\dfrac{15}{22}\)
= ( \(\dfrac{7}{22}\) + \(\dfrac{15}{22}\)) - ( \(\dfrac{15}{23}+\dfrac{18}{23}\)) + \(\dfrac{2022}{2023}\)
= \(\dfrac{22}{22}\) - \(\dfrac{23}{23}\) + \(\dfrac{2022}{2023}\)
= 1 - 1 + \(\dfrac{2022}{2023}\)
= \(\dfrac{2022}{2023}\)
b, - \(\dfrac{2}{11}\) + 5\(\dfrac{5}{6}\) ( 14\(\dfrac{1}{5}\) - 11\(\dfrac{1}{5}\)): 5\(\dfrac{1}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) ( \(\dfrac{71}{5}\) - \(\dfrac{56}{5}\)) : \(\dfrac{11}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{6}\) . \(\dfrac{15}{5}\) : \(\dfrac{11}{2}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{2}\) \(\times\) \(\dfrac{2}{11}\)
= - \(\dfrac{2}{11}\) + \(\dfrac{35}{11}\)
= \(\dfrac{33}{11}\)
= 3
c, 2000 + { 20 - [ 4.20220 - (32 + 5):2] }
= 2000 + { 20 - [ 4.1 - (9+5):2]}
= 2000 + { 20 - [ 4 - 14 : 2 ]}
= 2000 + { 20 - [ 4 -7]}
= 2000 + { 20 - (-3)}
= 2000 + 23
= 2023
A=1+2+22+23+...+262+263
2A=2+22+23+24+...+263+264
2A-A=2+22+23+24+...+263+264-1+2+22+23+...+262+263
A=264-1
A = 1 + 2 + 22 + 23 + ... + 262 + 263
2A = 2 + 22 + 23 + 24 + ... + 263 + 264
A = 264 - 1
bài này dễ mà
Đặt A = 1+2+2^2+2^3+...+2^1007
2A=2+2^2+2^3+2^4+...+2^1008
2A-A=(2+2^2+2^3+2^4+...+2^1008)-(1+2+2^2+2^3+....+2^1007)
A=2^1008-1
Đáp án là \(\frac{^{2^{2018}-1}}{2}\)