Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
3. ( 22 + 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1
= ( 22 - 1 ).( 22 + 1 ).( 24 + 1 ).( 28 + 1 )....( 264 + 1 ) + 1
= ( 24 - 1 ).( 24 + 1 ).( 28 + 1 )......( 264 + 1 ) + 1
= ( 28 + 1 ).....( 264 + 1 ) + 1
= ( 264 - 1 ).( 264 + 1 ) + 1
= 2128 - 1 + 1
= 2128
8.( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 32 - 1 ).( 32 + 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 34 - 1 ).( 34 + 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 38 - 1 ).( 38 + 1 )....( 3128 + 1 ) + 1
= ( 316 - 1 )......( 3128 + 1 ) + 1
= ( 3128 - 1 ).( 3128 + 1 ) + 1
= 3256 - 1 + 1
= 3256
\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\) =\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\) =\(a^2\) b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\) =\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\) =\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\) =25 c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\) =\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\) =\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\) =... =\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\) \)
d)Tương tự
\(a,\left(a-b+c\right)^2-\left(b-c\right)^2+2ab-2ac\)
=\(a^2+b^2+c^2-2ab-2bc+2ac-b^2+2bc-c^2+2ab-2ac\)
=\(a^2\)
b)\(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
=\(\left(3x+1\right)^2-2\left(3x+3-2\right)\left(3x+3+2\right)+\left(3x+5\right)^2\)
=\(\left(3x+1\right)^2-2\left(\left(3x+3\right)^2-4\right)+\left(3x+5\right)^2\)
=\(9x^2+6x+1-18x^2-36x-9+8+9x^2+30x+25\)
=25
c)\(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)...\left(2^{64}+1\right)\)
=\(\left(2-1\right)\left(2+1\right)\left(2^2+1\right)....\left(2^{64}+1\right)\)
=\(\left(2^2-1\right)\left(2^2+1\right)...\left(2^{64}+1\right)\)
=...
=\(\left(2^{64}-1\right)\left(2^{64}+1\right)=2^{128}-1\)
d)Tương tự
a: \(=\left[a-\left(b-c\right)\right]^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2a\left(b-c\right)+\left(b-c\right)^2-\left(b-c\right)^2+2ab-2ac\)
\(=a^2-2ab+2ac+2ab-2ac=a^2\)
b: \(\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(3x+5\right)^2\)
\(=\left(3x+1-3x-5\right)^2\)
\(=\left(-4\right)^2=16\)
c: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\cdot...\cdot\left(2^{64}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\cdot\left(2^{32}+1\right)\left(2^{64}+1\right)\)
\(=2^{128}-1\)
d: \(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\dfrac{\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)}{2}\)
\(=\dfrac{3^{64}-1}{2}\)
Bài 2:
a.
$(6x+1)^2+(6x-1)^2-2(6x+1)(6x-1)$
$=[(6x+1)-(6x-1)]^2=2^2=4$
b.
$3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^2-1)(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^4-1)(2^4+1)(2^8+1)(2^{16}+1)$
$=(2^8-1)(2^8+1)(2^{16}+1)$
$=(2^{16}-1)(2^{16}+1)=2^{32}-1$
c.
$2C=(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^{16}+1)$
$=(5^4-1)(5^4+1)(5^8+1)(5^{16}+1)$
$=(5^8-1)(5^8+1)(5^{16}+1)$
$=(5^{16}-1)(5^{16}+1)=5^{32}-1$
$\Rightarrow C=\frac{5^{32}-1}{2}$
(a+b+c)3=(a+b)3+3(a+b)2c+3(a+b)c2+c3
=a3+b3+3ab.(a+b)+3(a+b)2c+3(a+b)c2+c3
=a3+b3+c3+3(a+b)(ab+ac+bc+c2)
=a3+b3+c3+3(a+b)[a.(b+c)+c.(b+c)]
=a3+b3+c3+3(a+b)(b+c)(c+a)
=>dpcm
P=12(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=>2P=24(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(52-1)(52+1)(54+1)(58+1)(516+1)
=(54-1)(54+1)(58+1)(516+1)
=(58-1)(58+1)(516+1)
=(516-1)(516+1)
=532-1
==>P=(532-1)/2
3: =(5^2-1)(5^2+1)(5^4+1)(5^8+1)(5^16+1)
=(5^4-1)(5^4+1)(5^8+1)(5^16+1)
=(5^8-1)(5^8+1)(5^16+1)
=(5^16-1)(5^16+1)
=5^32-1
4:
D=(4^4-1)(4^4+1)(4^8+1)*....*(4^64+1)
=(4^8-1)(4^8+1)*...*(4^64+1)
=...
=4^128-1
5: =(5^2-1)(5^2+1)(5^4+1)*...*(5^128+1)+(5^256-1)
=(5^4-1)(5^4+1)*...*(5^128+1)+5^256-1
=5^256-1+5^256-1
=2*5^256-2
Bài 1:
A = 8.(32 + 1)(34 + 1)(38 + 1)(316 + 1)
A = (32 - 1)(32 + 1)(34+ 1)(38 +1)(316 + 1)
A = (34 - 1)(34 + 1)(38+ 1)(316 + 1)
A = (38 - 1)(38 + 1)(316 + 1)
A = (316 - 1)(316 +1)
A = (316)2 - 12
A = 332 - 1
1: \(A=8\left(3^2+1\right)\left(3^4+1\right)\cdot...\cdot\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\cdot\left(3^4+1\right)\left(3^8+1\right)\cdot\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\cdot\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)=3^{32}-1\)
2: \(B=\left(1-3\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=-\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=-\left(3^4-1\right)\left(3^4+1\right)\cdot\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=-\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)=-\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=-\left(3^{32}-1\right)=1-3^{32}\)
3: \(C=24\left(5^2+1\right)\left(5^4+1\right)\cdot...\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^2-1\right)\left(5^2+1\right)\cdot\left(5^4+1\right)\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^4-1\right)\cdot\left(5^4+1\right)\cdot...\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^8-1\right)\left(5^8+1\right)\cdot...\cdot\left(5^{128}+1\right)+\left(5^{256}-1\right)\)
\(=\left(5^{16}-1\right)\left(5^{16}+1\right)\cdot...\cdot\left(5^{128}+1\right)+5^{256}-1\)
\(=\left(5^{32}-1\right)\left(5^{32}+1\right)\left(5^{64}+1\right)\left(5^{128}+1\right)+5^{256}-1\)
\(=\left(5^{64}-1\right)\left(5^{64}+1\right)\left(5^{128}+1\right)+5^{256}-1\)
\(=\left(5^{128}-1\right)\left(5^{128}+1\right)+5^{256}-1=2\left(5^{256}-1\right)\)