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a) ( 5x - y )( 25x2 + 5xy + y2 ) = ( 5x )3 - y3 = 125x3 - y3
b) ( x - 3 )( x2 + 3x + 9 ) - ( 54 + x3 ) = x3 - 33 - 54 - x3 = -27 - 54 = -81
c) ( 2x + y )( 4x2 - 2xy + y2 ) - ( 2x - y )( 4x2 + 2xy + y2 ) = ( 2x )3 + y3 - [ ( 2x )3 - y3 ]= 8x3 + y3 - 8x3 + y3 = 2y3
d) ( x + y )2 + ( x - y )2 + ( x + y )( x - y ) - 3x2 = x2 + 2xy + y2 + x2 - 2xy + y2 + x2 - y2 - 3x2 = y2
e) ( x - 3 )3 - ( x - 3 )( x2 + 3x + 9 ) + 6( x + 1 )2
= x3 - 9x2 + 27x - 27 - ( x3 - 33 ) + 6( x2 + 2x + 1 )
= x3 - 9x2 + 27x - 27 - x3 + 27 + 6x2 + 12x + 6
= -3x2 + 39x + 6
= -3( x2 - 13x - 2 )
f) ( x + y )( x2 - xy + y2 ) + ( x - y )( x2 + xy + y2 ) - 2x3
= x3 + y3 + x3 - y3 - 2x3
= 0
g) x2 + 2x( y + 1 ) + y2 + 2y + 1
= x2 + 2x( y + 1 ) + ( y2 + 2y + 1 )
= x2 + 2x( y + 1 ) + ( y + 1 )2
= ( x + y + 1 )2
= [ ( x + y ) + 1 ]2
= ( x + y )2 + 2( x + y ) + 1
= x2 + 2xy + y2 + 2x + 2y + 1
1, \(x^2\left(x-3\right)-4x+12=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(x-3\right)\)
2, \(2a\left(x+y\right)-x-y=2a\left(x+y\right)-\left(x+y\right)=\left(2a-1\right)\left(x+y\right)\)
3, \(2x-4+5x^2-10x=2\left(x-2\right)+5x\left(x-2\right)=\left(2+5x\right)\left(x-2\right)\)
4, sửa đề :
\(6x^2-12x-7x+14=6x\left(x-2\right)-7\left(x-2\right)=\left(6x-7\right)\left(x-2\right)\)
5, \(xy-y^2-3x+3y=y\left(x-y\right)-3\left(x-y\right)=\left(y-3\right)\left(x-y\right)\)
a) x2(x-3)-4x+12
=x2(x-3)-4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
b) 2a(x+y)-x-y
=2a(x+y)-(x+y)
=(x+y)(2a-1)
c) 2x-4+5x2-10x
=2(x-2)+5x(x-2)
=(x-2)(2+5x)
d) 5x2-12x-7x+14
=5x2-19x+14
e) xy-y2-3x+3y
=y(x-y)-3(x-y)
=(x-y)(y-3)
#H
1, \(x^2+2x-3=x^2+3x-x-3=x\left(x-1\right)+3\left(x-1\right)=\left(x+3\right)\left(x-1\right)\)
2, \(x^2+3x-10=x^2+5x-2x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)
3, \(x^2-x-12=x^2-4x+3x-12=x\left(x+3\right)-4\left(x+3\right)=\left(x-4\right)\left(x+3\right)\)
4, \(3x^2+4x-7=3x^2+7x-3x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(3x+7\right)\left(x-1\right)\)
5, \(4x^2-9y^2-5xy=4x^2-9xy+4xy-9y^2\)
\(=4x\left(x+y\right)-9y\left(x+y\right)=\left(4x-9y\right)\left(x+y\right)\)
6, \(x^2-2x-4y^2-4y=x^2-2x+1-4y^2-4y-1=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x-1-2y-1\right)\left(x-1+2y+1\right)=\left(x-2y-2\right)\left(x+2y\right)\)
\(\frac{1}{-x^2+3x-2}=\frac{-1}{x^2-3x+2}=\frac{-1}{x^2-x-2x+2}=\frac{-1}{x\left(x-1\right)-2\left(x-1\right)}=\frac{-1}{\left(x-1\right)\left(x-2\right)}\)
( ĐKXĐ : x ≠ 1 ; x ≠ 2 )
\(\frac{1}{x^2+5x-6}=\frac{1}{x^2-x+6x-6}=\frac{1}{x\left(x-1\right)+6\left(x-1\right)}=\frac{1}{\left(x-1\right)\left(x+6\right)}\)
( ĐKXĐ : x ≠ 1 ; x ≠ -6 )
\(\frac{1}{-x^2+4x-3}=\frac{-1}{x^2-4x+3}=\frac{-1}{x^2-x-3x+3}=\frac{-1}{x\left(x-1\right)-3\left(x-1\right)}=\frac{-1}{\left(x-1\right)\left(x-3\right)}\)
( ĐKXĐ : x ≠ 1 ; x ≠ 3 )
MTC : ( x - 1 )( x - 2 )( x - 3 )( x + 6 )
=> \(\frac{-1}{\left(x-1\right)\left(x-2\right)}=\frac{-1\left(x+6\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{-x^2-3x+18}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)
=>\(\frac{1}{\left(x-1\right)\left(x+6\right)}=\frac{1\left(x-2\right)\left(x-3\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{x^2-5x+6}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)
=> \(\frac{-1}{\left(x-1\right)\left(x-3\right)}=\frac{-1\left(x-2\right)\left(x+6\right)}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}=\frac{-x^2-4x+12}{\left(x-1\right)\left(x-2\right)\left(x-3\right)\left(x+6\right)}\)