Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Trả lời:
a, 3x2y - 6xy = 3xy ( x - 2 )
b, x2 - y2 - 9x + 9y
= ( x2 - y2 ) - ( 9x - 9y )
= ( x - y )( x + y ) - 9 ( x - y )
= ( x - y )( x + y - 9 )
c, x3 - 6x2 - y2x + 9x
= x ( x2 - 6x - y2 + 9 )
= x [ ( x2 - 6x + 9 ) - y2 ]
= x [ ( x - 3 )2 - y2 ]
= x ( x - 3 - y )( x - 3 + y )
3x2y - 6xy = 3xy( x - 2 )
x2 - y2 - 9x + 9y = ( x - y )( x + y ) - 9( x - y ) = ( x - y )( x + y - 9 )
x3 - 6x2 - y2x + 9x = x( x2 - 6x - y2 + 9 ) = x[ ( x - 3 )2 - y2 ] = x( x - y - 3 )( x + y - 3 )
Trả lời:
a) x2 + 4y2 + 4xy = x2 + 2.x.2y + (2y)2 = ( x + 2y )2
b) \(\frac{1}{64}-27x^3=\left(\frac{1}{4}\right)^3-\left(3x\right)^3=\left(\frac{1}{4}-3x\right)\left(\frac{1}{16}+\frac{3}{4}x+9x^2\right)\)
c) x3 - 6x2 + 12x - 8 = x3 - 3.x2.2 + 3.x.22 - 23 = ( x - 2 )3
d) x2 - x - y2 - y = ( x2 - y2 ) - ( x + y ) = ( x - y )( x + y ) - ( x + y ) = ( x + y )( x - y - 1 )
e) 5x - 5y + ax - ay = ( 5x - 5y ) + ( ax - ay ) = 5 ( x - y ) + a ( x - y ) = ( x - y )( 5 + a )
g) \(x^5-3x^4+3x^3-x^2=x^2\left(x^3-3x^2+3x-1\right)=x^2\left(x-1\right)^3\)
f) \(x^2-25-2xy+y^2=\left(x^2-2xy+y^2\right)-25=\left(x-y\right)^2-5^2=\left(x-y-5\right)\left(x-y+5\right)\)
e) \(16x^3+54y^3=2\left(8x^3+27y^3\right)=2\left[\left(2x\right)^3+\left(3y\right)^3\right]=2\left(2x+3y\right)\left(4x^2-6xy+9y^2\right)\)
d) \(3y^2-3z^2+3x^2+6xy=3\left(x^2+2xy+y^2-z^2\right)=3\left[\left(x+y\right)^2-z^2\right]=3\left(x+y+z\right)\left(x+y-z\right)\)
\(x^2-2x-4y^2-4y\)
\(=\left(x^2-4y^2\right)-\left(2x+4y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-2\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-2\right)\)
\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} |
\(a,36x^2-\left(3x-2\right)^2=\left(6x-3x+2\right)\left(6x+3x-2\right)\)
\(=\left(3x+2\right)\left(9x-2\right)\)
phần b,c,d lm tg tự
\(e,16x^2-24xy+9y^2=\left(4x-3y\right)^2\)
e) \(a^2+2ab+b^2-ac-bc=\left(a+b\right)^2-c\left(a+b\right)=\left(a+b-c\right)\left(a+b\right)\)
f) \(x^2-2x-4y^2-4y=x^2-2x+1-\left(4y^2+4y+1\right)=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x-2y-2\right)\left(x+2y\right)\)
g) \(x^2y-x^3-9y+9x=x^2\left(y-x\right)-9\left(y-x\right)=\left(x^2-9\right)\left(y-x\right)=\left(x-3\right)\left(x+3\right)\left(y-x\right)\)
h) \(x^2\left(x-1\right)+16\left(1-x\right)=\left(x^2-16\right)\left(x-1\right)=\left(x-4\right)\left(x+4\right)\left(x-1\right)\)