$4(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^8-1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^{16}-1)(3^{16}+1)$
$=\dfrac{3^{32}-1}{2}$
$4(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^2-1)(3^2+1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^4-1)(3^4+1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^8-1)(3^8+1)(3^{16}+1)$
$=\dfrac12\cdot(3^{16}-1)(3^{16}+1)$
$=\dfrac{3^{32}-1}{2}$
Tính nhanh
a) 2.4.(3^2+1)(3^4+1)...(3^16+1)
b) 2.(3+1).(3^2+1)(3^4+1)...(3^16+1)
c) 8.(3^2+1)(3^4+1)...(3^16+1)
Tính nhanh
C=50^2-49^2+48^2-47^2+...+2^2-1^2
D=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^16+1)
E=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)
Tính nhanh:
A= (3+1)(32+1)(34+1)(38+1)(316+1)
Tính nhanh 3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
Tính nhanh:
a) (2^2+4^2+6^2+......+100^2)-(1^2+3^2+5^2+........+99^2)
b) 8(3^2+1)(3^4+1)(3^8+1)(3^16+1)-3^32
Giải giúp mình đi . ghi ra tùng bước mình cảm ơn nhiều , cho mọi người 1tick.
tính nhanh
3(2^2+1)(2^4+1)(2^8+1)(2^16+1)(2^32+1)
Tính nhanh :
3^8 x 5^8 - ( 15^4-1 ) x ( 15^4 + 1 )
( 2+ 1 ) x ( 2^2 + 1 ) x ( 2^4 + 1 ) x ... x ( 2^16 +1 ) +1
Tính nhanh : 3(2^2+1)(2^4+1)(2^8+1)(2^16+1)
Bài 1 Tính Nhanh
a)3^4 . 5^4 - (15^2+1)(15^2-1)
b)x^4-12x^3+12x^2-12x+111 tại x = 11
Bài 2 Rút gọn
3(2^2+1)(2^4+1)(2^8+1)(2^16+1)