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\(x\left(y^2-z^2\right)+z\left(x^2-y^2\right)+y\left(z^2-x^2\right)\)
\(=x\left(y^2-z^2\right)-\left(y^2-z^2+z^2-x^2\right)z+y\left(z^2-x^2\right)\)
\(=x\left(y^2-z^2\right)-z\left(y^2-z^2\right)-z\left(z^2-x^2\right)+y\left(z^2-x^2\right)\)
\(=\left(y^2-z^2\right)\left(x-z\right)+\left(z^2-x^2\right)\left(y-z\right)\)
\(=\left(y-z\right)\left(z-x\right)\left(-\left(y+z\right)+z+x\right)\)
= \(\left(y-z\right)\left(z-x\right)\left(x-y\right)\)
a)
\(=x^2\left(2x+3\right)+\left(2x+3\right)\)
\(=\left(x^2+1\right)\left(2x+3\right)\)
b)
\(=a\left(a-b\right)+a-b\)
\(=\left(a+1\right)\left(a-b\right)\)
c)
\(=2\left(x^2+2x+1-y^2\right)\)
\(=2\left(x+1-y\right)\left(x+1+y\right)\)
d)
\(=x^3\left(x-2\right)+10x\left(x-2\right)\)
\(=x\left(x^2+10\right)\left(x-2\right)\)
e)
\(=x\left(x^2+2x+1\right)\)
\(=x\left(x+1\right)^2\)
f)
\(=y\left(x+y\right)-\left(x+y\right)\)
\(=\left(y-1\right)\left(x+y\right)\)
a,2x3+3x2+2x+3
=(2x3+2x)+(3x2+3)
=2x(x2+1)+3(x2+1)
=(x2+1)(2x+3)
b,a2-ab+a-b
=(a2-ab)+(a-b)
=a(a-b)+(a-b)
=(a-b)(a+1)
c,2x2+4x+2-2y2
=2(x2+2x+1-y2)
=2[(x2+2x+1)-y2 ]
=2[(x+1)2-y2 ]
=2(x+1-y)(x+1+y)
d,x4-2x3+10x2-20x
=(x4-2x3)+(10x2-20x)
=x3(x-2)+10x(x-2)
=(x-2)(x3+10x)
=(x-2)[x(x2+10)]
e,x3+2x2+x
=x(x2+2x+1)
=x(x+1)2
f,xy+y2-x-y
=(xy+y2)-(x-y)
=y(x+y)-(x+y)
=(x+y)(y-1)
b, \(x^3+2x^2+2x+1=\left(x^2+x+1\right)\left(x+1\right)\)
c, \(x^3-4x^2+12x-27=\left(x^2-x+9\right)\left(x-3\right)\)
d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)
e, sai đề
a, \(\left(ab-1\right)^2+\left(a+b\right)^2=\left(a^2+1\right)\left(b^2+1\right)\)
b, \(x^3+2x^2+2x+1=\left(x+1\right)\left(x^2+x+1\right)\)
c, \(x^3-4x^2+12x-27=\left(x-3\right)\left(x^2-x+9\right)\)
d, \(x^4-2x^3+2x-1=\left(x-1\right)^3\left(x+1\right)\)
e, cho mình sửa đề xíu
\(x^4+2x^3+2x^2+2x+1=\left(x+1\right)^2\left(x^2+1\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\begin{array}{l} a){\left( {ab - 1} \right)^2} + {\left( {a + b} \right)^2}\\ = {a^2}{b^2} - 2ab + 1 + {a^2} + 2ab + {b^2}\\ = {a^2}{b^2} + 1 + {a^2} + {b^2}\\ = {a^2}\left( {{b^2} + 1} \right) + \left( {{b^2} + 1} \right)\\ = \left( {{a^2} + 1} \right)\left( {{b^2} + 1} \right)\\ c){x^3} - 4{x^2} + 12x - 27\\ = {x^3} - 27 + \left( { - 4{x^2} + 12x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9} \right) - 4x\left( {x - 3} \right)\\ = \left( {x - 3} \right)\left( {{x^2} + 3x + 9 - 4x} \right)\\ = \left( {x - 3} \right)\left( {{x^2} - x + 9} \right)\\ b){x^3} + 2{x^2} + 2x + 1\\ = {x^3} + 2{x^2} + x + x + 1\\ = x\left( {{x^2} + 2x + 1} \right) + \left( {x + 1} \right)\\ = x{\left( {x + 1} \right)^2} + \left( {x + 1} \right)\\ = \left( {x + 1} \right)\left( {x\left( {x + 1} \right) + 1} \right)\\ = \left( {x + 1} \right)\left( {{x^2} + x + 1} \right)\\ d){x^4} - 2{x^3} + 2x - 1\\ = {x^4} - 2{x^3} + {x^2} - {x^2} + 2x - 1\\ = {x^2}\left( {{x^2} - 2x + 1} \right) - \left( {{x^2} - 2x + 1} \right)\\ = \left( {{x^2} - 2x + 1} \right)\left( {{x^2} - 1} \right)\\ = {\left( {x - 1} \right)^2}\left( {x - 1} \right)\left( {x + 1} \right)\\ = {\left( {x - 1} \right)^3}\left( {x + 1} \right)\\ e){x^4} + 2{x^3} + 2{x^2} + 2x + 1\\ = {x^4} + 2{x^3} + {x^2} + {x^2} + 2x + 1\\ = {x^2}\left( {{x^2} + 2x + 1} \right) + \left( {{x^2} + 2x + 1} \right)\\ = \left( {{x^2} + 2x + 1} \right)\left( {{x^2} + 1} \right)\\ = {\left( {x + 1} \right)^2}\left( {{x^2} + 1} \right) \end{array} |
a)x4-4(x2+5)-25=x4-4x2-45=(x4-9x2)+(5x2-45)=x2(x2-9)+5(x2-9)=(x2-9)(x2+5)=(x-3)(x+3)(x2+5)
b)a2-b2-2a+1=(a2-2a+1)-b2=(a-1)2-b2=(a-b-1)(a+b-1)
c)x2-2x-4y2-4y=(x2-2x+1)-(4y2+4y+1)=(x-1)2-(2y+1)2=(x-1-2y-1)(x-1+2y+1)=(x-2y-2)(x+2y)
d)x2+4x-y2+4=(x2+4x+4)-y2=(x+2)2-y2=(x-y+2)(x+y+2)
a) ( x2 - 25 )2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) ]2 - ( x - 5 )2
= [ ( x - 5 )( x + 5 ) - ( x - 5 ) ][ ( x - 5 )( x + 5 ) + ( x - 5 ) ]
= ( x - 5 )( x + 5 - 1 )( x - 5 )( x + 5 + 1 )
= ( x - 5 )2( x + 4 )( x + 6 )
b) ( 4x2 - 25 )2 - 9( 2x - 5 )2
= ( 4x2 - 25 )2 - 32( 2x - 5 )2
= ( 4x2 - 25 )2 - ( 6x - 15 )2
= [ ( 4x2 - 25 ) - ( 6x - 15 ) ][ ( 4x2 - 25 ) + ( 6x - 15 ) ]
= ( 4x2 - 25 - 6x + 15 )( 4x2 - 25 + 6x - 15 )
= ( 4x2 - 6x - 10 )( 4x2 + 6x - 40 )
= ( 4x2 + 4x - 10x - 10 )( 4x2 + 16x - 10x - 40 )
= [ 4x( x + 1 ) - 10( x + 1 ) ][ 4x( x + 4 ) - 10( x + 4 ) ]
= ( x + 1 )( 4x - 10 )( x + 4 )( 4x - 10 )
= ( 4x - 10 )2( x + 1 )( x + 4 )
c) 4( 2x - 3 )2 - 9( 4x2 - 9 )2
= 22( 2x - 3 )2 - 32( 4x2 - 9 )2
= ( 4x - 6 )2 - ( 12x2 - 27 )2
= [ ( 4x - 6 ) - ( 12x2 - 27 ) ][ ( 4x - 6 ) + ( 12x2 - 27 ) ]
= ( 4x - 6 - 12x2 + 27 )( 4x - 6 + 12x2 - 27 )
= ( -12x2 + 4x + 21 )( 12x2 + 4x - 33 )
= ( -12x2 + 18x - 14x + 21 )( 12x2 - 18x + 22x - 33 )
= [ -12x( x - 3/2 ) - 14( x - 3/2 ) ][ 12x( x - 3/2 ) + 22( x - 3/2 ) ]
= ( x - 3/2 )( -12x - 14 )( x - 3/2 )( 12x + 22 )
= ( x - 3/2 )2( -12x - 14 )( 12x + 22 )
d) a6 - a4 + 2a3 + 2a2
= a2( a4 - a2 + 2a + 2 )
= a2( a4 - 2a3 + 2a3 + 2a2 - 4a2 + a2 + 4a - 2a + 2 )
= a2[ ( a4 - 2a3 + 2a2 ) + ( 2a3 - 4a2 + 4a ) + ( a2 - 2a + 2 ) ]
= a2[ a2( a2 - 2a + 2 ) + 2a( a2 - 2a + 2 ) + 1( a2 - 2a + 2 ) ]
= a2( a2 + 2a + 1 )( a2 - 2a + 2 )
= a2( a + 1 )2( a2 - 2a + 2 )
e) ( 3x2 + 3x + 2 )2 - ( 3x2 + 3x - 2 )2
= [ ( 3x2 + 3x + 2 ) - ( 3x2 + 3x - 2 ) ][ ( 3x2 + 3x + 2 ) + ( 3x2 + 3x - 2 ) ]
= ( 3x2 + 3x + 2 - 3x2 - 3x + 2 )( 3x2 + 3x + 2 + 3x2 + 3x - 2 )
= 4( 6x2 + 6x )
= 4.6x( x + 1 )
= 24( x + 1 )
a) ( 3x -1 )2 - 16
= (3x -1 ) 2 - 42
= ( 3x -1 -4 ).( 3x -1 +4 )
b) ( 5x-4 ) 2 - 49x2
= ( 5x-4 ) 2 - (7x)2
=( 5x -4 -7x).( 5x -4 + 7x )
=( -2x -4 ) .( 12x -4 )
còn lại giống tương tự nha pạn
~ hok tốt ~
Lời giải:
a)
\(x^6+2x^5+x^4-2x^3-2x^2+1\)
\(=(x^6+2x^5+x^4)-2(x^3+x^2)+1\)
\(=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2\)
b)
\([4abcd+(a^2+b^2)(c^2+d^2)]^2-4[cd(a^2+b^2)+ab(c^2+d^2)]^2\)
\(=[4abcd+(a^2+b^2)(c^2+d^2)]^2-[2cd(a^2+b^2)+2ab(c^2+d^2)]^2\)
\(=[4abcd+(a^2+b^2)(c^2+d^2)+2cd(a^2+b^2)+2ab(c^2+d^2)][4abcd+(a^2+b^2)(c^2+d^2)-2cd(a^2+b^2)-2ab(c^2+d^2)]\)
\(=[(a^2+b^2)(c^2+d^2+2cd)+2ab(c^2+d^2+2cd)][(a^2+b^2)(c^2+d^2-2cd)-2ab(c^2+d^2-2cd)]\)
\(=[(a^2+b^2)(c+d)^2+2ab(c+d)^2][(a^2+b^2)(c-d)^2-2ab(c-d)^2]\)
\(=(c+d)^2(a^2+b^2+2ab)(c-d)^2(a^2+b^2-2ab)\)
\(=(c+d)^2(a+b)^2(c-d)^2(a-b)^2\)