phân tích đa thức thành nhân tử
a, ab-ac-b2+2bc-c2
b, x6+8
c, 64x3-8
d, x3-2x2+4x-8
K
Khách
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T
4 tháng 9 2023
\(a,a^2-2a-4b^2-4b\)
\(=\left(a^2-4b^2\right)-\left(2a+4b\right)\)
\(=\left(a-2b\right)\left(a+2b\right)-2\left(a+2b\right)\)
\(=\left(a+2b\right)\left(a-2b-2\right)\)
\(b,x^3-2x^2+4x-8\)
\(=x^2\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x-2\right)\left(x^2+4\right)\)
\(c,x^3+36x-12x^2\)
\(=x^3-6x^2-6x^2+36x\)
\(=x^2\left(x-6\right)-6x\left(x-6\right)\)
\(=\left(x-6\right)\left(x^2-6x\right)\)
\(=x\left(x-6\right)^2\)
\(d,5a^2+3\left(a+b\right)^2-5b^2\)
\(=\left(5a^2-5b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a^2-b^2\right)+3\left(a+b\right)^2\)
\(=5\left(a-b\right)\left(a+b\right)+3\left(a+b\right)^2\)
\(=\left(a+b\right)\left[5\left(a-b\right)+3\left(a+b\right)\right]\)
\(=\left(a+b\right)\left(5a-5b+3a+3b\right)\)
\(=\left(a+b\right)\left(8a-2b\right)\)
\(=2\left(a+b\right)\left(4a-b\right)\)
\(e,x^3-3x^2+3x-1-y^3\)
\(=\left(x^3-3x^2+3x-1\right)-y^3\)
\(=\left(x-1\right)^3-y^3\)
\(=\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right)y+y^2\right]\)
\(=\left(x-y-1\right)\left(x^2-2x+1+xy-y+y^2\right)\)
\(=\left(x-y-1\right)\left(x^2+y^2-xy-y+1\right)\)
#Urushi☕
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
4 tháng 9 2023
\(c.\\ x^3+36x-12x^2\\ =x\left(x^2-12x+36\right)\\ =x.\left(x^2-2.x.6+6^2\right)\\ =x.\left(x-6\right)^2\\ ---\\ d.\\ 5a^2+3\left(a+b\right)^2-5b^2\\ =\left(5a^2-5b^2\right)+3\left(a+b\right)^2\\ =5.\left(a^2-b^2\right)+3.\left(a+b\right)\left(a+b\right)\\ =5\left(a+b\right)\left(a-b\right)+3\left(a+b\right)\left(a+b\right)\\ =\left(a+b\right)\left(5a-5b+3a+3b\right)\\ =\left(a+b\right)\left(8a-2b\right)\\ =2\left(a+b\right)\left(4a-b\right)\)
\(e.\\ x^3-3x^2+3x-1-y^3\\ =\left(x-1\right)^3-y^3\\ =\left(x-1-y\right)\left[\left(x-1\right)^2+\left(x-1\right).y+y^2\right]\\ =\left(x-y-1\right).\left[\left(x^2-2x+1\right)+y\left(x+y-1\right)\right]\)
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
a.
$x^2-y^2-2x+2y=(x^2-y^2)-(2x-2y)=(x-y)(x+y)-2(x-y)=(x-y)(x+y-2)$
b.
$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$
c.
$x^2+4x-y^2+4=(x^2+4x+4)-y^2=(x+2)^2-y^2=(x+2-y)(x+2+y)$
d.
$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$
$=(x+1)(x^2-4x+1)$
AH
Akai Haruma
Giáo viên
25 tháng 10 2021
e.
$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$
$=(x^2+2y^2)^2-(2xy)^2=(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
f.
$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$
$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$
g.
$(x^2+x)^2+4x^2+4x-12=(x^2+x)^2+4(x^2+x)-12$
$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$
$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$
$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$
h.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
NH
1
NV
Nguyễn Việt Lâm
Giáo viên
5 tháng 8 2021
a.
\(x^3-y^3+2x^2-2y^2\)
\(=\left(x-y\right)\left(x^2+xy+y^2\right)+\left(x-y\right)\left(2x+2y\right)\)
\(=\left(x-y\right)\left(x^2+xy+y^2+2x+2y\right)\)
b.
\(x^3+1-x^2-x\)
\(=\left(x+1\right)\left(x^2-x+1\right)-x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\left(x-1\right)^2\)
NM
3
NH
1
2
24 tháng 4 2022
`a) 8x^2 - 8xy - 4x + 4y`
`= 8x ( x - y ) - 4 ( x - y )`
`= ( x - y ) ( 8x - 4 )`
__________________________
`b) x^3 + 10x^2 + 25x - xy^2`
`=x ( x^2 + 10x + 25 ) - xy^2`
`= x ( x + 5 )^2 - xy^2`
`= x [ ( x + 5 )^2 - y^2 ]`
`= x ( x + 5 - y ) ( x + 5 + y )`
________________________________
`c) x^2 + x - 6`
`= x^2 + 3x - 2x - 6`
`= x ( x + 3 ) - 2 ( x + 3 )`
`= ( x + 3 ) ( x - 2 )`
_______________________________
`d) 2x^2 + 4x - 16`
`= 2x^2 - 4x + 8x - 16`
`= 2x ( x - 2 ) + 8 ( x - 2 )`
`= ( x - 2 ) ( 2x + 8 )`
LS
24 tháng 4 2022
a) x2 + xy –x – y = x(x + y) – (x + y) = (x + y)(x -1 ).
b) a2 – b2 + 8a + 16 = (a2 + 8a + 16) – b2 = (a + 4)2 – b2
= (a + 4 – b)(a + 4 + b).
tui chỉ làm dc này thui
a) \(ab-ac-b^2+2bc-c^2\)
\(=\left(ab-ac\right)-\left(b^2-2bc+c^2\right)\)
\(=a\left(b-c\right)-\left(b-c\right)^2\)
\(=\left(a-b+c\right)\left(b-c\right)\)
b) \(x^6+8=\left(x^2\right)^3+2^3\)
\(=\left(x^2+2\right)\left(x^4-2x^2+4\right)\)
c) \(64x^3-8=\left(4x\right)^3-2^3\)
\(=\left(4x-2\right)\left(16x^2+8x+4\right)\)
\(=8\left(2x-1\right)\left(4x^2+2x+1\right)\)
d) \(x^3-2x^2+4x-8\)
\(=x^2\left(x-2\right)+4\left(x-2\right)\)
\(=\left(x^2+4\right)\left(x-2\right)\)