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\(A=3\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right)\cdot\left(2^2+1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^4-1\right)\cdot\left(2^4+1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^8-1\right)\cdot\left(2^8+1\right)\cdot\left(2^{16}+1\right)\)
\(A=\left(2^{16}-1\right)\cdot\left(2^{16}+1\right)\)
\(A=2^{32}-1\)
Vậy...
Mình sửa đề bài nha:
\(12\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^2-1\right)\left(5^2+1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^4-1\right)\left(5^4+1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^8-1\right)\left(5^8+1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{16}-1\right)\left(5^{16}+1\right)\)
\(=\frac{1}{2}\left(5^{32}-1\right)\)
\(=\frac{5^{32}-1}{2}\)
Chúc bạn học tốt!
\(4A=4\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5-1\right)\left(5+1\right)\left(5^2+1\right)....\left(5^{2048}+1\right)=\left(5^2-1\right)\left(5^2+1\right).....\left(5^{2048}+1\right)\left(hdt\left(a-b\right)\left(a+b\right)=a^2-b^2\right)=\left(5^4-1\right)\left(5^4+1\right)......\left(5^{2048}+1\right)=\left(5^8-1\right).....\left(5^{2048}+1\right)=.....=\left(5^{1024}+1\right)\left(5^{1024}-1\right)\left(5^{2048}+1\right)=\left(5^{2048}-1\right)\left(5^{2048}+1\right)=5^{4096}-1\)
\(A=3\left(2^3+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\left(2^2-1\right).9\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(A=\frac{9}{5}.\left(2^{32}-1\right)\)
12
= \(\frac{24}{2}\)
= \(\frac{1}{2}\left(25-1\right)\)
= \(\frac{1}{2}\left(5^2-1\right)\)
Chép đề sai kìa
a) ( x + 3 )2 - ( x - 4 )( x + 8 ) = 1
<=> x2 + 6x + 9 - ( x2 + 4x - 32 ) = 1
<=> x2 + 6x + 9 - x2 - 4x + 32 = 1
<=> 2x + 41 = 1
<=> 2x = -40
<=> x = -20
b) 3( x + 2 )2 + ( 2x - 1 )2 - 7( x + 3 )( x - 3 ) = 36
<=> 3( x2 + 4x + 4 ) + 4x2 - 4x + 1 - 7( x2 - 9 ) = 36
<=> 3x2 + 12x + 12 + 4x2 - 4x + 1 - 7x2 + 63 = 36
<=> 8x + 76 = 36
<=> 8x = -40
<=> x = -5
c) ( x - 3 )( x2 + 3x + 9 ) + x( x + 2 )( 2 - x ) = 1
<=> x3 - 27 - x( x + 2 )( x - 2 ) = 1
<=> x3 - 27 - x( x2 - 4 ) = 1
<=> x3 - 27 - x3 + 4x = 1
<=> 4x - 27 = 1
<=> 4x = 28
<=> x = 7
a) ( x - 1 )2 + ( x - 2 )2 = 2( x + 4 )2 - ( 22x + 27 )
<=> x2 - 2x + 1 + x2 - 4x + 4 = 2( x2 + 8x + 16 ) - 22x - 27
<=> 2x2 - 6x + 5 = 2x2 + 16x + 32 - 22x - 27
<=> 2x2 - 6x - 2x2 - 16x + 22x = 32 - 27 - 5
<=> 0x = 0 ( đúng ∀ x ∈ R )
Vậy phương trình có vô số nghiệm
b) ( x + 2 )2 - 2( x - 3 ) = ( x + 1 )2
<=> x2 + 4x + 4 - 2x + 6 = x2 + 2x + 1
<=> x2 + 2x - x2 - 2x = 1 - 4 - 6
<=> 0x = -9 ( vô lí )
Vậy phương trình vô nghiệm
c) ( x + 1 )3 - x2( x + 3 ) = 2
<=> x3 + 3x2 + 3x + 1 - x3 - 3x2 = 2
<=> 3x + 1 = 2
<=> 3x = 1
<=> x = 1/3
a)
\(x^2-2x+1+x^2-4x+4=2\left(x^2+8x+16\right)-22x-27\)
\(2x^2-6x+5=2x^2+16x+32-22x-27\)
\(-6x+5=-6x+5\)
\(0=0\left(llđ\forall x\right)\)
Vậy \(x=R\)
b)
\(x^2+4x+4-2x+6=x^2+2x+1\)
\(x^2+2x+10=x^2+2x+1\)
\(10=1\)
\(0=-9\left(sai\right)\)
Vậy phương trình vô nghiệm
c)
\(x^3+3x^2+3x+1-x^3-3x^2=2\)
\(3x+1=2\)
\(3x=1\)
\(x=\frac{1}{3}\)
a) ( x - 1 )3 - 4x( x + 1 )( x - 1 ) + 3( x - 1 )( x2 + x + 1 )
= x3 - 3x2 + 3x - 1 - 4x( x2 - 1 ) + 3( x3 - 13 )
= x3 - 3x2 + 3x - 1 - 4x3 + 4x + 3x3 - 3
= ( x3 - 4x3 + 3x3 ) - 3x2 + ( 3x + 4x ) + ( -1 - 3 )
= -3x2 + 7x - 4
b) ( x - 1 )( x - 2 )( 1 + x + x2 )( 4 + 2x + x2 )
= [ ( x - 1 )( 1 + x + x2 ) ][ ( x - 2 )( 4 + 2x + x2 ) ]
= [ ( x - 1 )( x2 + x + 1 ) ][ ( x - 2 )( x2 + 2x + 4 ) ]
= ( x3 - 13 )( x3 - 23 )
= ( x3 - 1 )( x3 - 8 )
= x6 - 9x3 + 8
c) ( x - 1 )3 + 3( x - 1 )( x2 + x + 1 ) - 4x( x + 1 )( x - 1 )
= x3 - 3x2 + 3x - 1 + 3( x3 - 13 ) - 4x( x2 - 1 )
= x3 - 3x2 + 3x - 1 + 3x3 - 3 - 4x3 + 4x
= ( x3 + 3x3 - 4x3 ) - 3x2 + ( 3x + 4x ) + ( -1 - 3 )
= -3x2 + 7x - 4
a,\(\left(x-1\right)^3-4x\left(x+1\right)\left(x-1\right)+3\left(x-1\right)\left(x^2+x+1\right)\)
\(=x^3-3x^2+3x-1-4x\left(x^2-1\right)+3\left(x^3-1\right)\)
\(=x^3-3x^2+3x-1-4x^3+4x+3x^3-3\)
\(=-3x^2+7x-4\)
b,\(\left(x-1\right)\left(x-2\right)\left(1+x+x^2\right)\left(4+2x+x^2\right)\)
\(=\left[\left(x-1\right)\left(x^2+x+1\right)\right]\left[\left(x-2\right)\left(x^2+2x+4\right)\right]\)
\(=\left(x^3-1\right)\left(x^3-8\right)\)
\(=x^6-9x^3+8\)
c,\(\left(x-1\right)^3+3\left(x-1\right)\left(x^2+x+1\right)-4x\left(x+1\right)\left(x-1\right)\)
\(=x^3-3x^2+3x-1+3\left(x^3-1\right)-4\left(x^2-1\right)\)
\(=x^3-3x^2+3x-1+3x^3-3-4x^3+4x\)
\(=-3x^2+7x-4\)
Sửa đề
B = 2(3+1)(32+1)(34+1)(38+1)(316+1)
= (3-1)(3+1)(32+1)(34+1)(38+1)(316+1)
= (32-1)(32+1)(34+1)(38+1)(316+1)
= (34-1)(34+1)(38+1)(316+1)
= (38-1)(38+1)(316+1)
= (316-1)(316+1)
= (332-1)