\(D=15(4^2+1)(4^4+1)...(4^{64}+1)\\=(4^2-1)(4^2+1)(4^4+1)...(4^{64}+1)\\=(4^4-1)(4^4+1)...(4^{64}+1)\\...\\=(4^{64}-1)(4^{64}+1)\\=4^{128}-1\\Vậy:D=4^{128}-1\)
\(D=15(4^2+1)(4^4+1)...(4^{64}+1)\\=(4^2-1)(4^2+1)(4^4+1)...(4^{64}+1)\\=(4^4-1)(4^4+1)...(4^{64}+1)\\...\\=(4^{64}-1)(4^{64}+1)\\=4^{128}-1\\Vậy:D=4^{128}-1\)
15(4^2+1)(4^4+1)...(4^64+1)
Tính A=(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)-1/15 . 4^64
A=(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)−1/15 . 4^64
BT7: Tính
\(3,C=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)\left(5^4+1\right)...\left(5^{16}+1\right)\)
\(4,D=15\left(4^2+1\right)\left(4^4+1\right)...\left(4^{64}+1\right)\)
\(5,E=24\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)...\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
bài 1
(4^2+1)(4^4+1)(4^8+1)(4^16+1)(4^32+1)-1/15*4^64
Bài 2:Cho x+1/x=10. Tính S=x^5+1/x^5
Tính
S=\(-1^2+2^2-3^2+4^2-...+2016^2\)
A=\(\left(4^2+1\right)\left(4^4+1\right)\left(4^8+1\right)\left(4^{16}+1\right)\left(4^{32}+1\right)-\frac{1}{15}.4^{64}\)
Mình đang gấp giúp mình nha!
thu gọn:
a) (2+1)(2^2+1)(2^4+1)..............(2^32+1)-2^64
b) (5+3)(5^2+3^2)(5^4+3^4)...................(5^64+3^64).\(\frac{5^{128}-3^{128}}{2}\)
Tính nhanh
a) 5^4×3^4-(15^2-1)*(15^2+1)
b) 50^2-49^2+48^2-47^2+...+2^2-1^2
d) (3^2+1)*(3^4+1)*(3^8+1)
phân tích đa thức thành nhân tử
a. 4x^2-3x-4
b. x^2+2x-3
c. 64+x^4+y^4
d. (x+1)(x+2)(x+3)(x+4)-24