Phân tích đa thức thành nhân tử:
x5+x4+5x3-8x2+6x-6
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CM
1
24 tháng 9 2021
\(a,=\left(x-2\right)\left(9x^2y^2-6x^3y^2\right)=3x^2y^2\left(3-2x\right)\left(x-2\right)\\ b,=5x\left(x^2-y^2\right)+20x\left(x+y\right)=5x\left(x-y\right)\left(x+y\right)+20x\left(x+y\right)\\ =5\left(x+y\right)\left(x^2-xy+4x\right)\\ c,=8x^2+2x-12x-3=2x\left(4x+1\right)-3\left(4x+1\right)=\left(2x-3\right)\left(4x+1\right)\)
30 tháng 10 2021
\(=x\left(x^4-x^2-2x-1\right)\)
\(=x\left(x^4-x^3-x^2+x^3-x^2-x+x^2-x-1\right)\)
\(=x\left(x^2-x-1\right)\left(x^2+x+1\right)\)
NM
1
1 tháng 10 2021
`a)x^3-8x^2+16x`
`=x(x^2-8x+16)`
`=x(x-4)^2`
`b)x^2+4y^2+2x-4y-4xy-24`
`=(x-2y)^2+2(x-2y)-24`
`=(x-2y)^2-4(x-2y)+6(x-2y)-24`
`=(x-2y-4)(x-2y+6)`
`c)x^4+x^3-x^2-2x-2`
`=x^4-2x^2+x^3-2x+x^2-2`
`=x^2(x^2-2)+x(x^2-2)+x^2-2`
`=(x^2-2)(x^2+x+1)`
AH
Akai Haruma
Giáo viên
10 tháng 8 2021
1.
$4x^2y+5x^3-x^2y^2=x^2(4y+5x-y^2)$
2.
$5x(x-1)-3y(1-x)=5x(x-1)+3y(x-1)=(x-1)(5x+3y)$
3.
$4x^2-25=(2x)^2-5^2=(2x-5)(2x+5)$
4.
$6x-9-x^2=-(x^2-6x+9)=-(x-3)^2$
5.
$x^2+4y^2+4xy=x^2+2.x.2y+(2y)^2=(x+2y)^2$
6.
$\frac{1}{64}-27x^3=(\frac{1}{4})^3-(3x)^3$
$=(\frac{1}{4}-3x)(\frac{1}{16}+\frac{3x}{4}+9x^2)$
AH
Akai Haruma
Giáo viên
10 tháng 8 2021
7.
$x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3$
$=(x-2)^3$
8.
$x^2-x-y^2-y=(x^2-y^2)-(x+y)=(x-y)(x+y)-(x+y)$
$=(x+y)(x-y-1)$
9.
$5x-5y+ax-ay=5(x-y)+a(x-y)$
$=(x-y)(5+a)$
24 tháng 8 2021
a: \(x^4-4x^3-8x^2+8x\)
\(=x\left(x^3-4x^2-8x+8\right)\)
\(=x\left[\left(x+2\right)\left(x^2-2x+4\right)-4x\left(x+2\right)\right]\)
\(=x\left(x+2\right)\left(x^2-6x+4\right)\)
b: \(x^2-1-xy+y\)
\(=\left(x-1\right)\left(x+1\right)-y\left(x-1\right)\)
\(=\left(x-1\right)\left(x-y+1\right)\)
c: Ta có: \(\left(x-1\right)\left(x-2\right)\left(x-3\right)+\left(x-1\right)^2\cdot\left(x-2\right)\)
\(=\left(x-1\right)\cdot\left(x-2\right)\cdot\left(x-3-x-1\right)\)
\(=2\cdot\left(x-1\right)\cdot\left(x-2\right)^2\)
T
9 tháng 11 2023
\(a,5x^3y-10x^2y^2\\=5x^2y(x-2y)\\b,x^4-y^4\\=(x^2)^2-(y^2)^2\\=(x^2-y^2)(x^2+y^2)\\=(x-y)(x+y)(x^2+y^2)\)
\(c,(x+5)^2-16\\=(x+5)^2-4^2\\=(x+5-4)(x+5+4)\\=(x+1)(x+9)\\d,7x(y-3)-14(3-y)\\=7x(y-3)+14(y-3)\\=(7x+14)(y-3)\\=7(x+2)(y-3)\\Toru\)
L
14 tháng 8 2021
a) \(=\left(x+6y\right)\left(x-6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x-6y-1\right)\)
b) \(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c) \(=2\left(x-y\right)^2-18\)
\(=2\left[\left(x-y\right)^2-3^2\right]\)
\(=2\left(x-y+3\right)\left(x-y-3\right)\)
14 tháng 8 2021
a: \(x^2-36y^2-x+6y\)
\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x+6y-1\right)\)
b: \(x^3-8x^2+16x\)
\(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c: \(2x^2-4xy+2y^2-18\)
\(=2\left(x^2-2xy+y^2-9\right)\)
\(=2\left(x-y-3\right)\left(x-y+3\right)\)
d: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
PT
1
CM
17 tháng 5 2018
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
14 tháng 8 2021
a: Ta có: \(x^2-36y^2-x+6y\)
\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)
\(=\left(x-6y\right)\left(x+6y-1\right)\)
b: Ta có: \(16x-8x^2+x^3\)
\(=x\left(x^2-8x+16\right)\)
\(=x\left(x-4\right)^2\)
c: Ta có: \(2x^2-4xy+2y^2-18\)
\(=2\left(x^2-2xy+y^2-9\right)\)
\(=2\cdot\left[\left(x-y\right)^2-9\right]\)
\(=2\left(x-y-3\right)\left(x-y+3\right)\)
d: Ta có: \(3x^2-7x-10\)
\(=3x^2+3x-10x-10\)
\(=3x\left(x+1\right)-10\left(x+1\right)\)
\(=\left(x+1\right)\left(3x-10\right)\)
14 tháng 8 2021
e: Ta có: \(x^4-x^2-30\)
\(=x^4-6x^2+5x^2-30\)
\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)
\(=\left(x^2-6\right)\left(x^2+5\right)\)
f: Ta có: \(x^2-xy-2y^2\)
\(=x^2-2xy+xy-2y^2\)
\(=x\left(x-2y\right)+y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+y\right)\)
g: Ta có: \(x^4-13x^2y^2+4y^4\)
\(=x^4-4x^2y^2+4y^4-9x^2y^2\)
\(=\left(x^2-2y^2\right)^2-\left(3xy\right)^2\)
\(=\left(x^2-3xy-2y^2\right)\left(x^2-3xy+2y^2\right)\)
\(=\left(x^2-3xy-2y^2\right)\left(x^2-xy-2xy+2y^2\right)\)
\(=\left[x\left(x-y\right)-2y\left(x-y\right)\right]\left(x^2-3xy-2y^2\right)\)
\(=\left(x-y\right)\left(x-2y\right)\left(x^2-3xy-2y^2\right)\)
h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)
\(=\left(x^2-2x\right)^2-3\left(x^2-2x\right)+\left(x^2-2x\right)-3\)
\(=\left(x^2-2x\right)\left(x^2-2x-3\right)+\left(x^2-2x-3\right)\)
\(=\left(x^2-2x-3\right)\left(x^2-2x+1\right)\)
\(=\left(x-3\right)\left(x+1\right)\cdot\left(x-1\right)^2\)