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TN
7 tháng 7 2017
\(A=4\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(2A=3^{32}-1\Rightarrow A=\frac{3^{32}-1}{2}\)
31 tháng 8 2019
a) (x - 1)(x + 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x2 - 1)(x2 + 1)(x4 + 1)(x8 + 1)
= (x4 - 1)(x4 + 1)(x8 + 1)
= (x8 - 1)(x8 + 1)
= x16 - 1
b) (a2 - 2b)(a2 + 2b)(a4 + 4b2)(a8 + 16b4)
= (a4 - 4b2)(a4 + 4b2)(a8 + 16b4)
= (a8 - 16b4)(a8 + 16b4)
= a16 - 256b8
TP
1
VM
19 tháng 8 2020
Đặt \(A=12.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(\Rightarrow2A=24.\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^2-1\right).\left(5^2+1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^4-1\right).\left(5^4+1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^8-1\right).\left(5^8+1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^{16}-1\right).\left(5^{16}+1\right)\)
\(2A=\left(5^{16}\right)^2-1^2\)
\(2A=5^{32}-1\)
\(\Rightarrow A=\frac{5^{32}-1}{2}.\)
ND
1
9 tháng 7 2018
D=(x2+x+1)(x2-x+1)(x4-x2+1)(x8-x4+1)
\(=\left(\left(x^2+1\right)^2-x^2\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right).\)
\(=\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)\left(x^8-x^4+1\right).\)
\(=\left(\left(x^4+1\right)^2-x^4\right)\left(x^8-x^4+1\right).\)
\(=\left(x^8+x^4+1\right)\left(x^8-x^4+1\right)=\left(x^8+1\right)-x^8=x^{16}+x^8 +1\)
M
0
A
1 tháng 10 2020
Bài 1 :
Ta có : \(VP=\left(a+b\right)^4=\left(a+b\right)\left(a+b\right)^3\)
\(=\left(a+b\right)\left(a^3+3a^2b+3ab^2+b^3\right)=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
=> HĐT ko đc CM
Bài 2 :
a, \(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)
\(=x^3+2x^2+4x-2x^2-4x-8-x+1+7=x^3-x=x\left(x^2-1\right)\)
Sửa đề : b, \(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x+1\right)\)
\(=8\left(x^3-1\right)-8x^3+1=8x^3-8-8x^3+1=-7\)
F
1 tháng 10 2020
Xin phép chủ nahf cho mjnh sửa đề:D
\(\left(a+b\right)^4=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
a,\(\left(a+b\right)^4\)
\(=\left[\left(a+b\right)^2\right]^2\)
\(=\left(a^2+2ab+b^2\right)^2\)
\(=\left[\left(a^2+2ab\right)+b^2\right]^2\)
\(=\left(a^2+2ab\right)^2+2\left(a^2+2ab\right)b^2+b^4\)
\(=a^4+4a^3b+4a^2b^2+2a^2b^2+4ab^3+b^4\)
\(=a^4+4a^3b+6a^2b^2+4ab^3+b^4\)
Bài 2:
a,\(\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)+7\)
\(=\left(x^3-8\right)-\left(x-1\right)+7\)
b,\(8\left(x-1\right)\left(x^2+x+1\right)-\left(2x-1\right)\left(4x^2+2x-1\right)\)
\(=8\left(x^3-1\right)-\left(8x^3-1\right)\)
\(=8x^3-8-8x^3+1\)
\(=-7\)
CN
0
13 tháng 11 2017
a) \(\left(x+3\right)\left(x-1\right)-2\left(x+3\right)^2+\left(x-4\right)\left(x+4\right)\)
\(=x^2-x+3x-3-2\left(x^2+6x+9\right)+x^2-16\)
\(=2x^2+2x-19-2x^2-12x-18\)
\(=-10x-37\)
b) \(\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{\left(5^2-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{24}\)
\(=\frac{\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{24}\)
\(=\frac{\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)}{24}\)
\(=\frac{\left(5^{16}-1\right)\left(5^{16}+1\right)}{24}\)
\(=\frac{5^{32}-1}{24}\)
LH
13 tháng 11 2017
a) (x+3)(x-1)-2(x+302)+(x-4)(x+4)=x2+2x-3-2x-1800+x2-16=2x2-1819
b)...=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)/(5^2-1)=(5^4-1)(5^4+1)(5^8+1)(5^16+1)/(5^2-1)
=(5^8-1)(5^8+1)(5^16+1)/(5^2-1)=(5^16-1)(5^16+1)/(5^2-1)=(5^32-1)/(5^2-1)
Đặt A = (42 + 1)(44 + 1)(48 + 1)...(42016 + 1)
15A = (42 - 1)(42 + 1)(44 + 1)(48 + 1)...(42016 + 1)
15A = (44 - 1)(44 + 1)(48 + 1)...(42016 + 1)
15A = (48 - 1)(48 + 1)...(42016 + 1)
15A = 44032 - 1
\(A=\frac{4^{4032}-1}{15}\)