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a. 6x3-x2-486x+81
= 6x3-54x2+53x2-477x-9x+81
= 6x2.(x-9)+53x.(x-9)-9.(x-9)
= (x-9).(6x2+53x-9)
= (x-9)(6x2+54x-x-9)
=(x-9)[6x.(x+9)-(x+9)]=(x-9)(x+9)(6x-1)
b. x3-5x2+3x+9
= x3+x2-6x2-6x+9x+9
=x2.(x+1)-6x.(x+1)+9.(x+1)
=(x+1)(x2-6x+9)=(x+1)(x-3)2
c. x3+3x2+6x+4
= x3+x2+2x2+2x+4x+4
= x2.(x+1)+2x.(x+1)+4.(x+1)
= (x+1)(x2+2x+4)
d.
(x - 4)(x2 + 4x + 16) - x(x2 - 6) = 2
x3 - 64 - x3 + 6x = 2
6x = 2 + 64
6x = 66
x = 66 : 6
x = 11
x3 - 27 + 3x(x - 3)
= (x - 3)(x2 + 3x + 9) + 3x(x - 3)
= (x - 3)(x2 + 3x + 9 + 3x)
= (x - 3)(x2 + 6x + 9)
= (x - 3)(x + 3)2
5x3 - 7x2 + 10x - 14
= 5x(x2 + 2) - 7(x2 + 2)
= (x2 + 2)(5x - 7)
a) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
b) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
c) \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
a ) ( 3x2 + 3x + 2)2 - ( 3x2 + 3x - 2)2
=(3x2 + 3x + 2 + 3x2 + 3x - 2) [( 3x2 + 3x + 2) - ( 3x2 + 3x - 2) ]
=(6x2+6x)*4
=24x(x+1)
b ) ( xy+1)2 - ( x+y)2
=( xy+1 + x+y ) [( xy+1) - ( x+y)]
=[x(y+1)+(y+1)] [x(y-1) - (y-1)]
=(x+1)(y+1)(x-1)(y-1)
c ) ( x + y)3 - ( x - y)3
=[( x + y)-( x - y)] [( x + y)2 - ( x + y)( x - y) + ( x - y)2 ]
=2y( x2+2xy+y2 - x2+y2+ x2-2xy +y2 )
=2y(3y2+x2)
d ) 4( x2 - y2 ) - 8(x - ay) - 4(a2 - 1)
=4(-a2+2ay-y2+x2-2x+1)
=4[-(a-y)2+(x-1)2]
=-4(y-x-a+1)(y+x-a-1)
bài 1:
a. x2 - 5=0
=>x2 = 0+5 = 5
=> x = \(\sqrt{5}\)
vậy x= \(\sqrt{5}\)
sorry biết mỗi a thôi
a) x2 - 5 = 0
x2 = 0 + 5
x2 = 5
=> x = \(\sqrt{5}\)
Vậy ...
a) x.(1-x)+(x-1)2
=x.1-x2+x2-2.x2.1+12
=x-x2+x2-2.x2+1
=(-x2+x2-2x2)+1
=-2x2+1
b)(x+1)2-3.(x+1)
=x2+2.x2.1+12-3.x+3
=x2+2.x2+1-3x+3
=(x2+2x2)+(1+3)-3x
=3x2-3x+4
c)3x.(x-1)2-(1-x)3
=3x.x2-2,x2.1+12-13-3.12.x+3.x.12=3x.x2-2x2+1-1-3x+3x=(3x-3x+3x)(x2-2x2)(1-1)=3x.(-x2)
b. 2x3-3x2+3x-1=2x3-x2-2x2+x+2x-1
= x2(2x-1)-x(2x-1)+(2x-1)
=(2x-1)(x2-x-1)
c. 3x3-14x2+4x+3= 3x3+x2-15x2-5x+9x+3
=x2(3x+1)-5x(3x-1)+3(3x+1)
=(3x+1)(x2-5x+3)
Lời giải:
a) Đa thức không phân tích được thành nhân tử
b)
\((x^2-3x)^2-3(x^2-3x)+2\)
\(=a^2-3a+2\) (đặt \(x^2-3x=a)\)
\(=a^2-a-2a+2=a(a-1)-2(a-1)=(a-1)(a-2)\)
\(=(x^2-3x-1)(x^2-3x-2)\)