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a, \(x^2+2xy+y^2+x^2-2xy+y^2\)
\(=2\left(x^2+y^2\right)\)
b, \(\left(x+y\right)^2+2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2\)
\(=\left(x+y+x-y\right)^2\)\(=\left(2x\right)^2=4x^2\)
xin tiick
a) \(\left(x+y\right)^2+\left(x-y\right)^2\)
\(=x^2+y^2+2xy+x^2+y^2-2xy\)
\(=2x^2+2y^2\)
b) \(2\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2+\left(x+y\right)^2\)
\(=2\left(x^2-y^2\right)+x^2+y^2-2xy+x^2+y^2+2xy\)
\(=2\left(x^2-y^2\right)+2\left(x^2+y^2\right)\)
\(=2\left(2x^2\right)=4x^2\)
\(A=\left(x-2\right)^2+\left(x+3\right)^2-2\left(x+1\right)\left(x-1\right).\)
\(A=x^2-4x+4+x^2+6x+9-2\left(x^2-1\right)\)
\(A=x^2-4x+4+x^2+6x+9-2x^2+2\)
\(A=2x+15\)
6) c) x3 - x2 + x = 1
<=> x3 - x2 + x - 1 = 0
<=> (x3 - x2) + (x - 1) = 0
<=> x2 (x - 1) + (x - 1) = 0
<=> (x - 1) (x2 + 1) = 0
=> x - 1 = 0 hoặc x2 + 1 = 0
* x - 1 = 0 => x = 1
* x2 + 1 = 0 => x2 = -1 => x = -1
Vậy x = 1 hoặc x = -1
Bài 5:
a) Đặt \(A=\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^2-1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=\left(3^{16}-1\right)\left(3^{16}+1\right)\)
\(\Rightarrow8A=3^{32}-1\)
\(\Rightarrow A=\frac{3^{32}-1}{8}\)
b) (7x+6)2 + (5-6x)2 - (10-12x)(7x+6)
=(7x+6)2 + (5-6x)2 - 2(5-6x)(7x+6)
\(=\left(7x+6-5+6x\right)^2\)
\(=\left(13x+1\right)^2\)
a) \(\left(xy+1\right)^2-\left(x+y\right)\)
\(=\left(xy+1-x-y\right)\left(xy+1+x+y\right)\)
\(=\left[x\left(y-1\right)-\left(y-1\right)\right]\left[x\left(y+1\right)+\left(y+1\right)\right]\)
\(=\left(x-1\right)\left(y-1\right)\left(x+1\right)\left(y+1\right)\)
b) \(\left(x+y\right)^3-\left(x-y\right)^3\)
\(=\left(x+y-x+y\right)\left[\left(x+y\right)^2+\left(x-y\right)\left(x+y\right)+\left(x-y\right)^2\right]\)
\(=2y\left(x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2\right)\)
\(=2y\left(3x^2+y^2\right)\)
a) ( xy + 1 )2 - ( x + y )2
= [ ( xy + 1 ) - ( x + y ) ][ ( xy + 1 ) + ( x + y ) ]
= ( xy - x - y + 1 )( xy + x + y + 1 )
b) ( x + y )3 - ( x - y )3
C1. = x3 + 3x2y + 3xy2 + y3 - ( x3 - 3x2y + 3xy2 - y3 )
= x3 + 3x2y + 3xy2 + y3 - x3 + 3x2y - 3xy2 + y3
= 6x2y + 2y3
= 2y( 3x2 + y2 )
C2. = [ ( x + y ) - ( x - y ) ][ ( x + y )2 + ( x + y )( x - y ) + ( x - y )2 ]
= ( x + y - x + y )( x2 + 2xy + y2 + x2 - y2 + x2 - 2xy + y2 )
= 2y( 3x2 + y2 )
c) 3x4y2 + 3x3y2 + 3xy2 + 3y2
= 3( x4y2 + x3y2 + xy2 + y2 )
= 3[ ( x4y2 + x3y2 ) + ( xy2 + y2 ) ]
= 3[ x3y2( x + 1 ) + y2( x + 1 ) ]
= 3( x + 1 )( x3y2 + y2 )
= 3y2( x + 1 )( x3 + 1 )
= 3y2( x + 1 )( x + 1 )( x2 - x + 1 )
= 3y2( x + 1 )2( x2 - x + 1 )
d) 4( x2 - y2 ) - 8( x - ay ) - 4( a2 - 1 )
= 4[ ( x2 - y2 ) - 2( x - ay ) - ( a2 - 1 )
= 4( x2 - y2 - 2x + 2ay - a2 + 1 )
= 4[ ( x2 - 2x + 1 ) - ( y2 - 2ay + a2 ) ]
= 4[ ( x - 1 )2 - ( y - a )2 ]
= 4[ ( x - 1 ) - ( y - a ) ][ ( x - 1 ) + ( y + a ) ]
= 4( x - y + a - 1 )( x + y + a - 1 )
\(x\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\left(x-y\right)+xy^{16}\\ =x\left(x+y\right)\left(x-y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^8-y^8\right)\left(x^8+y^8\right)+xy^{16}\\ =x\left(x^{16}-y^{16}\right)+xy^{16}\\ =x^{17}-xy^{16}+xy^{16}\\ =x^{17}\)
\(x\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)\left(x-y\right)+xy^{16}\)
\(=x\left(x-y\right)\left(x+y\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^2-y^2\right)\left(x^2+y^2\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^4-y^4\right)\left(x^4+y^4\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^8-y^8\right)\left(x^8+y^8\right)+xy^{16}\)
\(=x\left(x^{16}-y^{16}\right)+xy^{16}\)
\(=x^{17}-xy^{16}+xy^{16}\)
\(=x^{17}\)