Help me ~~~
Phân tích đa thức thành nhân tử
a)x^4-3x^3+4x^2-3x+1
b)x^4+2021x^2-2020x+2021
c)6x^4+5x^3-38x^2+5x+6
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MR
1
6 tháng 11 2021
a: \(=\left(x+1\right)\left(x^2-x+1\right)+5x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2+4x+1\right)\)
22 tháng 10 2023
2:
a: \(x^2-12x+20\)
\(=x^2-2x-10x+20\)
=x(x-2)-10(x-2)
=(x-2)(x-10)
b: \(2x^2-x-15\)
=2x^2-6x+5x-15
=2x(x-3)+5(x-3)
=(x-3)(2x+5)
c: \(x^3-x^2+x-1\)
=x^2(x-1)+(x-1)
=(x-1)(x^2+1)
d: \(2x^3-5x-6\)
\(=2x^3-4x^2+4x^2-8x+3x-6\)
\(=2x^2\left(x-2\right)+4x\left(x-2\right)+3\left(x-2\right)\)
\(=\left(x-2\right)\left(2x^2+4x+3\right)\)
e: \(4y^4+1\)
\(=4y^4+4y^2+1-4y^2\)
\(=\left(2y^2+1\right)^2-\left(2y\right)^2\)
\(=\left(2y^2+1-2y\right)\left(2y^2+1+2y\right)\)
f; \(x^7+x^5+x^3\)
\(=x^3\left(x^4+x^2+1\right)\)
\(=x^3\left(x^4+2x^2+1-x^2\right)\)
\(=x^3\left[\left(x^2+1\right)^2-x^2\right]\)
\(=x^3\left(x^2-x+1\right)\left(x^2+x+1\right)\)
g: \(\left(x^2+x\right)^2-5\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)^2-2\left(x^2+x\right)-3\left(x^2+x\right)+6\)
\(=\left(x^2+x\right)\left(x^2+x-2\right)-3\left(x^2+x-2\right)\)
\(=\left(x^2+x-2\right)\left(x^2+x-3\right)\)
\(=\left(x^2+x-3\right)\left(x+2\right)\left(x-1\right)\)
h: \(\left(x^2+2x\right)^2-2\left(x+1\right)^2-1\)
\(=\left(x^2+2x+1-1\right)^2-2\left(x+1\right)^2-1\)
\(=\left[\left(x+1\right)^2-1\right]^2-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-2\left(x+1\right)^2+1-2\left(x+1\right)^2-1\)
\(=\left(x+1\right)^4-4\left(x+1\right)^2\)
\(=\left(x+1\right)^2\left[\left(x+1\right)^2-4\right]\)
\(=\left(x+1\right)^2\left(x+1+2\right)\left(x+1-2\right)\)
\(=\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)\)
i: \(x^2+4xy+4y^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-4\left(x+2y\right)+3\)
\(=\left(x+2y\right)^2-\left(x+2y\right)-3\left(x+2y\right)+3\)
\(=\left(x+2y\right)\left(x+2y-1\right)-3\left(x+2y-1\right)\)
\(=\left(x+2y-1\right)\left(x+2y-3\right)\)
j: \(x\cdot\left(x+1\right)\left(x+2\right)\left(x+3\right)-3\)
\(=\left(x^2-3x\right)\left(x^2-3x+2\right)-3\)
\(=\left(x^2-3x\right)^2+2\left(x^2-3x\right)-3\)
\(=\left(x^2-3x+3\right)\left(x^2-3x-1\right)\)
CW
15 tháng 7 2016
a)x^2-(a+b)x+ab
= x^2 - ax - bx + ab
= (x^2 - ax) - (bx - ab)
= x(x-a) - b(x-a)
= (x-b)(x-a)
b)7x^3-3xyz-21x^2+9z
=
c)4x+4y-x^2(x+y)
= 4(x + y) - x^2(x+y)
= (4-x^2) (x+y)
= (2-x)(2+x)(x+y)
d) y^2+y-x^2+x
= (y^2 - x^2) + (x+y)
= (y-x)(y+x)+ (x+y)
= (y-x+1) (x+y)
e)4x^2-2x-y^2-y
= [(2x)^2 - y^2] - (2x +y)
= (2x-y)(2x+y) - (2x+y)
= (2x -y -1)(2x+y)
f)9x^2-25y^2-6x+10y
=
10 tháng 10 2021
a) \(4\left(x+1\right)^3-x-1=4\left(x+1\right)^3-\left(x+1\right)=\left(x+1\right)\left[4\left(x+1\right)^2-1\right]=\left(x+1\right)\left[2\left(x+1\right)-1\right]\left[2\left(x+1\right)+1\right]=\left(x+1\right)\left(2x+1\right)\left(2x+3\right)\)
b) \(5x\left(x-3\right)+\left(3-x\right)^2-\left(x-3\right)=5x\left(x-3\right)+\left(x-3\right)^2-\left(x-3\right)=\left(x-3\right)\left(5x+x-3-1\right)=\left(x-3\right)\left(6x-4\right)=2\left(x-3\right)\left(3x-2\right)\)
c) \(9x^2y^3-3x^4y^2-6x^3y^2+16xy^4=xy^2\left(9xy-3x^3-6x^2+16y^2\right)\)
26 tháng 6 2023
2: =(2x+1)^2-y^2
=(2x+1+y)(2x+1-y)
3: =x^2(x^2+2x+1)
=x^2(x+1)^2
4: =x^2+6x-x-6
=(x+6)(x-1)
5: =-6x^2+3x+4x-2
=-3x(2x-1)+2(2x-1)
=(2x-1)(-3x+2)
6: =5x(x+y)-(x+y)
=(x+y)(5x-1)
7: =2x^2+5x-2x-5
=(2x+5)(x-1)
8: =(x^2-1)*(x^2-4)
=(x-1)(x+1)(x-2)(x+2)
9: =x^2(x-5)-9(x-5)
=(x-5)(x-3)(x+3)
LN
24 tháng 7 2017
1) x3 + 5x2 + 3x - 9
= x3 + 2x2 + 3x2 + 6x - 3x - 9
= ( x3 + 2x2 ) + (3x2 + 6x ) - ( 3x + 9 )
= x2 ( x+ 2 ) + 3x ( x + 2) - 3( x +2 )
= ( x + 2 ) ( x2 + 3x -3 )
2) x3 + 5x2 + 8x + 4
= ( x3 + x2 ) + ( 4x2 + 4x ) + ( 4x + 4 )
= x2 ( x + 1 ) + 4x ( x + 1 ) + 4 ( x + 1 )
= ( x + 1) ( x2 + 4x + 4 )
= (x + 1 ) ( x + 2 )2
3) x3 - 9x2 + 6x + 16
= x3 - 8x2 - x2 + 8x - 2x + 16
= ( x3 - 8x2 ) - ( x2 - 8x ) - ( 2x - 16 )
= x2 ( x - 8 ) - x ( x - 8 ) - 2 ( x - 8 )
= ( x - 8 ) ( x2 - x - 2 )
4) x3 - 4x2 + x + 6
= x3 - 3x2 - x2 + 3x - 2x + 6
= ( x3 - 3x2 ) - ( x2 - 3x ) - ( 2x - 6)
= x2 ( x - 3 ) - x ( x- 3 ) - 2 ( x - 3)
= ( x - 3 ) ( x2 - x - 2 )
a) \(=x^4-x^3-2x^3+2x^2+2x^2-2x-x+1\)
\(=x^3\left(x-1\right)-2x^2\left(x-1\right)+2x\left(x-1\right)-\left(x-1\right)\)
\(=\left(x^3-2x^2+2x-1\right)\left(x-1\right)\)
\(=\left(x^3-x^2-x^2+x+x-1\right)\left(x-1\right)\)
\(=\left(x^2-x+1\right)\left(x-1\right)^2\)
c)
\(=6x^4-12x^3+17x^3-34x^2-4x^2+8x-3x+6\)
\(=6x^3\left(x-2\right)+17x^2\left(x-2\right)-4x\left(x-2\right)-3\left(x-2\right)\)
\(=\left(6x^3+17x^2-4x-3\right)\left(x-2\right)\)
\(=\left(6x^3+18x^2-x^2-3x-x-3\right)\left(x-2\right)\)
\(=\left(6x^2-x-1\right)\left(x+3\right)\left(x-2\right)\)
\(=\left(2x-1\right)\left(3x+1\right)\left(x+3\right)\left(x-2\right)\)
b)
\(=x^4+1011x^2+1011+\left(1010x^2-2020x+1010\right)\)
\(=x^4+1011x^2+1011+1010\left(x^2-2x+1\right)\)
\(=x^4+1011x^2+1011+1010\left(x-1\right)^2\)
CÓ: \(x^4+1010\left(x-1\right)^2+1011x^2\ge0\forall x\)
=> \(x^4+1010\left(x-1\right)^2+1011x^2+1011\ge1011>0\forall x\)
=> ĐA THỨC b > 0 => Ko ph được thành nhân tử.