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\(A=\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Leftrightarrow2A=3^{32}-1\)
\(\Leftrightarrow A=3^{31}-\frac{1}{2}\)
a: \(=\dfrac{3^8-3^6+3^6\cdot2^3}{5^3}=\dfrac{3^8-3^6\left(1-2^3\right)}{5^3}=\dfrac{11664}{125}\)
b: \(=\dfrac{7^4\cdot4-7^3}{7^3}=7\cdot4-1=27\)
c: \(=28^4-28^4+1=1\)
d: \(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)+1\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)+1\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)+1\)
\(=3^{32}\)
Ít thôi -..-
a) ( 3x + 2 )( 2x + 9 ) - ( x + 3 )( 6x + 1 ) = ( x + 1 )2 - ( x + 2 )( x - 2 )
<=> 6x2 + 31x + 18 - ( 6x2 + 19x + 3 ) = x2 + 2x + 1 - ( x2 - 4 )
<=> 6x2 + 31x + 18 - 6x2 - 19x - 3 = x2 + 2x + 1 - x2 + 4
<=> 12x + 15 = 2x + 5
<=> 12x - 2x = 5 - 15
<=> 10x = -10
<=> x = -1
b) ( 2x + 3 )( x - 4 ) + ( x - 5 )( x - 2 ) = ( 3x - 5 )( x - 4 )
<=> 2x2 - 5x - 12 + x2 - 7x + 10 = 3x2 - 17x + 20
<=> 3x2 - 12x - 2 = 3x2 - 17x + 20
<=> 3x2 - 12x - 3x2 + 17x = 20 + 2
<=> 5x = 22
<=> x = 22/5
c) ( x + 2 )3 - ( x - 2 )3 - 12x( x - 1 ) = -8
<=> x3 + 6x2 + 12x + 8 - ( x3 - 6x2 + 12x - 8 ) - 12x2 + 12x = -8
<=> x3 + 6x2 + 12x + 8 - x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
<=> 12x + 16 = -8
<=> 12x = -24
<=> x = -2
d) ( 3x - 1 )2 - 5( x + 1 ) + 6x - 3.2x + 1 - ( x - 1 )2 = 16
<=> 9x2 - 6x + 1 - 5x - 5 + 6x - 6x + 1 - ( x2 - 2x + 1 ) = 16
<=> 9x2 - 11x - 3 - x2 + 2x - 1 = 16
<=> 8x2 - 9x - 4 = 16
<=> 8x2 - 9x - 4 - 16 = 0
<=> 8x2 - 9x - 20 = 0
( Đến đây bạn có hai sự lựa chọn : 1 là vô nghiệm
2 là nghiệm vô tỉ =) )
a) (3x + 2)(2x + 9) - (x + 3)(6x + 1) = (x + 1)2 - (x + 2)(x - 2)
=> 3x(2x + 9) + 2(2x + 9) - x(6x + 1) - 3(6x + 1) = x2 + 2x + 1 - x(x - 2) - 2(x - 2)
=> 6x2 + 27x + 4x + 18 - 6x2 - x - 18x - 3 = x2 + 2x + 1 - x2 + 2x - 2x + 4
=> (6x2 - 6x2) + (27x + 4x - x - 18x) + (18 - 3) = (x2 - x2) + (2x + 2x - 2x) + (1 + 4)
=> 12x + 15 = 2x + 5
=> 12x + 15 - 2x - 5 = 0
=> 10x + 10 = 0
=> 10x = -10 => x = -1
b) (2x + 3)(x - 4) + (x - 5)(x - 2) = (3x - 5)(x - 4)
=> 2x(x - 4) + 3(x - 4) + x(x - 2) - 5(x - 2) = 3x(x - 4) - 5(x - 4)
=> 2x2 - 8x + 3x - 12 + x2 - 2x - 5x + 10 = 3x2 - 12x - 5x + 20
=> (2x2 + x2) + (-8x + 3x - 2x - 5x) + (-12 + 10) = 3x2 - 17x + 20
=> 3x2 - 12x - 2 = 3x2 - 17x + 20
=> 3x2 - 12x - 2 - 3x2 + 17x - 20 = 0
=> (3x2 - 3x2) + (-12x + 17x) + (-2 - 20) = 0
=> 5x - 22 = 0
=> 5x = 22 => x = 22/5
c) (x + 2)3 - (x - 2)3 - 12x(x - 1) = -8
=> x3 + 6x2 + 12x + 8 - (x3 - 6x2 + 12x - 8) - 12x2 + 12x = -8
=> x3 + 6x2 + 12x + 8 -x3 + 6x2 - 12x + 8 - 12x2 + 12x = -8
=> (x3 - x3) + (6x2 + 6x2 - 12x2) + (12x - 12x + 12x) + (8 + 8) = -8
=> 12x + 16 = -8
=> 12x = -24
=> x = -2
Còn bài cuối làm nốt
\(8\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(=3^{32}-1\)
A=\(\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{16}+1\right)\)
\(=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)=\left(3^{16}-1\right)\left(3^{16}+1\right)\left(3^{32}+1\right)\)
\(=\left(3^{32}-1\right)\left(3^{32}+1\right)=3^{64}-1\)
đặt A=(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
=>2A=2.(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
=>2A=(3-1)(3+1)(32+1)(34+1)(38+1)(316+1)(332+1)
=(32-1)(32+1)(34+1)(38+1)(316+1)(332+1)
=(34-1)(34+1)(38+1)(316+1)(332+1)
=(38-1)(38+1)(316+1)(332+1)
=(316-1)(316+1)(332+1)
=(332-1)(332+1)
=364-1
=>2A=\(\frac{3^{64}-1}{2}\)
bạn ơi: A=\(\frac{3^{64}-1}{2}\) chứ ko phải 2A=\(\frac{3^{64}-1}{2}\)
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) (x - 1) . (x5 + x4 + x3 + x2 + x + 1) = (x . x5 + x . x4 + x . x3 + x . x2 + x . x + x . 1) - (1 . x5 + 1 . x4 + 1 . x3 + 1 . x2 + 1 . x + 1 . 1)
= (x6 + x5 + x4 + x3 + x2 + x ) - (x5 + x4 + x3 + x2 + x + 1)
= x6 + x5 + x4 + x3 + x2 + x - x5 - x4 - x3 - x2 - x - 1
= x6 + (x5 - x5) + (x4 - x4) + (x3 - x3) + (x2 - x2) + (x - x) - 1
= x6 - 1
b) (x + 1) . (x6 - x5 + x4 - x3 + x2 - x + 1) = (x . x6 - x . x5 + x . x4 - x . x3 + x . x2 - x . x + x . 1) + (1 . x6 - 1 . x5 + 1 . x4 - 1 . x3 + 1 . x2 - 1 . x + 1 . 1)
= (x7 - x6 + x5 - x4 + x3 - x2 + x ) + (x6 - x5 + x4 - x3 + x2 - x + 1)
= x7 - x6 + x5 - x4 + x3 - x2 + x + x6 - x5 + x4 - x3 + x2 - x + 1
= x7+(-x6 + x6) + (x5 - x5) + (-x4 + x4) + (x3 - x3) + (-x2 + x2) + (x - x) + 1
= x7 + 1
Đặt A = ( 3 + 1 )( 32 + 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )
=> 2A = 2.( 3 + 1 )( 32 + 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )
= ( 3 - 1 )( 3 + 1 )( 32 + 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )
= ( 32 - 1 )( 32 + 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )
= ( 34 - 1 )( 34 + 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )
= ( 38 - 1 )( 38 + 1 )( 316 + 1 )( 332 + 1 )
= ( 316 - 1 )( 316 + 1 )( 332 + 1 )
= ( 332 - 1 )( 332 + 1 )
= 364 - 1
2A = 364 - 1 => A = \(\frac{3^{64}-1}{2}\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3-1\right)\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}.\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)=\dfrac{1}{2}\left(3^{32}-1\right)=\dfrac{3^{32}}{2}-\dfrac{1}{2}\)
\(\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(=\dfrac{\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2}\)
\(=\dfrac{3^{32}-1}{2}\)