\(8A=2^6+2^9+2^{12}+...+2^{63}\)
=>\(7A=2^{63}-8\)
hay \(A=\dfrac{2^{63}-8}{7}\)
\(8A=2^6+2^9+2^{12}+...+2^{63}\)
=>\(7A=2^{63}-8\)
hay \(A=\dfrac{2^{63}-8}{7}\)
a)5^6:5^4+3^2-2021^0
b)100-[60-(9-2^3).34].2
1) tìm x , biết :
a) 720 : [ 118 - ( 2 x - 10 ] = 60 ;
b) 2840 + [( 999 - 9 x ) : 60 ] . 24 = 3200 ;
c) ( 3 x - 48 ) . 6 = 3 mũ 3 . 2 mũ 2 - 2 mũ 3 . 3 mũ 2
Tính
a, 1-1/2-1/3-1/6
b, 17/8-11/6+2/9
c, 21/28-18/60+3/5
Rút gọn các tổng sau:
a) A = 2 - 2\(^2\) + 2\(^3\) - 2\(^4\) + ... + 2\(^{99}\) - 2\(^{100}\)
b) B = 1 + 2\(^2\) + 2\(^4\) + ... + 2\(^{98}\) + 2\(^{100}\)
c) C = 1 - 2\(^3\) + 2\(^6\) - 2\(^9\) + ... + 2\(^{60}\) - 2\(^{63}\) + 2\(^{69}\)
d) D = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^4}\) + ... + \(\dfrac{1}{3^{100}}\)
e) E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4^2}\) - \(\dfrac{1}{4^3}\) + ... + \(\dfrac{1}{4^{98}}\) - \(\dfrac{1}{4^{99}}\) + \(\dfrac{1}{4^{100}}\)
a) A=1-2+3-4+5-6+.........+99-100+101
b) B=1+4+5+9+14+......+60+97
c) C=100+98+96+....+2-97-95-.......-1
d) D=1+2-3-4+5+6-7-8+9+10-11-12+.......+299-300+301+302
1) Tìm các số a,b biết rằng a + b = 432 và UCLN (a;b) = 36.
2) Cho A= 2 + 22 + 23 + ... + 260.Chứng tỏ A chia hết cho 6 ; 7 ; 9.
\(2\cdot x=\frac{1+2=3+...+9}{1-2+3-4+5-6+7-8+9}+\frac{25\cdot150-60\cdot5+20\cdot75}{1+2+3+...+99}\)
\(M=\frac{\text{2 . 6 . 10 + 4 . 12 . 20 + 6 . 18 . 30 + ..... + 20 . 60 . 100}}{\text{1 . 2 . 3 + 2 . 4 . 6 + 3 . 6 . 9 + ..... + 10 . 20 . 30}}\)
1.Tính:
A= 1 + 2 + 2 mũ 2 + ............+ 2 mũ 60
B = 3 + 3 mũ 3 + 3 mũ 5 + ............. + 3 mũ 81
C = 2 mũ 3 + 2 mũ 6 + 2 mũ 9 + ........... + 2 mũ 90
D = 3 mũ 100 - 3 mũ 90 + 3 mũ 98 - .................... 3
A=2 x 6 x 10 + 4 x12 x 20 +6 x 18 x 30 + ...+ 20 x 60 x 100 / 1 x 2 x 3 + 2 x 4 x 6 + 3 x 6 x 9 +...+ 10 x 20 x 30
Khó lắm đấy . Ai giải được bái luôn