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\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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} |
\(8-27x^3\)
\(=2^3-\left(3x\right)^3\)
\(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
a) \(8-27x^3=\left(2-x\right)\left(4+6x+9x^2\right)\)
b) \(27+27x+9x^2+x^3=\left(3+x\right)^3\)
c) \(x^3+8y^3=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)
\(a,36x^2-\left(3x-2\right)^2=\left(6x-3x+2\right)\left(6x+3x-2\right)\)
\(=\left(3x+2\right)\left(9x-2\right)\)
phần b,c,d lm tg tự
\(e,16x^2-24xy+9y^2=\left(4x-3y\right)^2\)
Phân tích đa thức sau thành nhân tử
a)4x^2 + y^2 - 4xy = (2x)^2 - 4xy + y^2
= (2x - y)^2
b)27+9x^2+27x+x^3 = 3^3 + 9x^2 + 27x + x^3
= (3 + x)^3
c)8z^3+1 = (2z)^3 + 1 = (2z + 1)(4z^2 - 2z + 1)
d)(2z-3)^2-16 = (2z - 3)^2 - 4^2
= (2z - 3 - 4)(2z - 3 + 4)
= (2z - 7)(2z + 1)
e)(2x-7)^2-(x+2)^2 = (2x - 7 - x - 2)(2x - 7 + x + 2)
= (x - 9)(3x - 5)
Bài 1:
\(a,5xy\left(xy-4x-7y\right)\)
\(b,\left(x-2y\right)\left(3-6y\right)\)
\(c,\left(y+1\right)\left(x+3y+3\right)\)
\(d,10y\left(x+y\right)\left(x-y\right)\)
BÀI 1: a) 5x2y2 + 20x2y - 35xy2 = 5xy .xy + 5xy .4x - 5xy .7y
=5xy .( xy + 4x - 7y )
b) 3 .( x - 2y ) + 6y .( 2y - x ) = 3 .(x - 2y ) - 6y .( x - 2y )
= ( x - 2y ) . ( 3 - 6y )
c) x .( y + 1 ) + 3 .( y2 + y + 1 ) = x .( y + 1 ) + 3 .( y + 1 )2
= ( y + 1 ) .[ x + 3 .( y + 1 ) ]
d) 10xy .( x + y ) - 5 .( 2x + y ) . y2 = 10x2y + 10xy2 - 10xy2 - 5y3
= 10x2y - 5y3 = 5y .( 2x2 - y2 )
mk làm bài 1 r nhé><
a) \(=2xy^2\left(x^2+8x+15\right)\)
\(=2xy^2\left[\left(x^2+8x+16\right)-1\right]\)
\(=2xy^2\left[\left(x+4\right)^2-1\right]\)
\(=2xy^2\left(x+4+1\right)\left(x+4-1\right)\)
\(=2xy^2\left(x+5\right)\left(x-3\right)\)
mấy câu sau tự làm nha :*
b,=(x^2-10x+25)-4
=(x-5)^2-2^2
=(x-5-2)(x-5+2)
=(x-7)(x-3)
\(1.2x^2-2y^2-6x-6y=2\left(x^2-y^2\right)-6\left(x+y\right)\)
\(=2\left(x+y\right)\left(x-y\right)-6\left(x+y\right)=\left(x+y\right)[2\left(x-y\right)-6]=\left(x+y\right)\left(2x-2y-6\right)\)
\(2.x^4+x^3-x^2-x=\left(x^4+x^3\right)-\left(x^2+x\right)\)
\(=x^3\left(x+1\right)-x\left(x+1\right)=\left(x+1\right)\left(x^3-x\right)\)
\(3.a^3+a^2b-a^2c-abc\)( mình trả lời ở câu hỏi của bạn rồi)
\(4.x^5-x^3+x^2-1=\left(x^5-x^3\right)+\left(x^2-1\right)\)
\(=x^3\left(x^2-1\right)+\left(x^2-1\right)=\left(x^2-1\right)\left(x^3+1\right)\)
\(5.x+y\left(x-1\right)-1\) ( mình trả lời ở câu hỏi của bạn rồi)
câu 6 và 7 cũng vậy
a) 9 . ( 2x + 3 )2 - 4 . ( x + 1 )2
= 32 . ( 2x + 3 )2 - 22 . ( x + 1 )
= [ 3 . ( 2x + 3 ) ] 2 - [ 2 . ( x + 1 ) ]
= ( 6x + 9 )2 - ( 2x + 2 )2
= [ ( 6x + 9 ) - ( 2x + 2 ) ] . [ ( 6x + 9 ) + ( 2x + 2 ) ]
= ( 6x + 9 - 2x - 2 ) . ( 6x + 9 + 2x + 2 )
= ( 4x + 7 ) . ( 8x + 11 )
b) 27x3 + \(\dfrac{y^3}{8}\) = ( 3x )3 + \(\left(\dfrac{y}{2}\right)^{^{ }3}\)
= ( 3x + \(\dfrac{y}{2}\) ) . ( 9x2 - \(\dfrac{3}{2}\)xy + \(\dfrac{y^2}{4}\) )
c) ( x + y )3 - ( x - y )3
= [ ( 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 . ( x2 + 2xy + y2 + x2 - y2 + x2 - 2xy + y2 )
= 2y . ( 3x2 + y2 )
= 6y . ( x2 + y2 )
bỏ 6y . (x2 - y2 ) đi