phân tích đa thức thành nhân tử
a) 3abc3-6a2b3c+12a3bc
b) 27-8y3
c) 4x2+4x-y2+11
d) 3a2(x-2)-6ab(2-x)
K
Khách
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Những câu hỏi liên quan
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.
$x^6+2x^5+x^4-2x^3-2x^2+1$
$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$
$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$
MH
25 tháng 10 2021
a) \(3x\left(2x-y\right)+5y\left(y-2x\right)\)
\(=3x\left(2x-y\right)-5y\left(2x-y\right)\)
\(=\left(3x-5y\right)\left(2x-y\right)\)
b) \(\left(x-5\right)^2-9\left(x+y\right)^2\)
\(=\left(x-5\right)^2-3^2\left(x+y\right)^2\)
\(=\left(x-5\right)^2-\left(3x+3y\right)^2\)
\(=\left(x-5+3x+3y\right)\left(x-5-3x-3y\right)\)
\(=\left(4x+3y-5\right)\left(-2x-3y-5\right)\)
25 tháng 10 2021
a: \(3x\left(2x-y\right)+5y\left(y-2x\right)=\left(2x-y\right)\left(3x-5y\right)\)
e: \(x^2-10x+24=\left(x-4\right)\left(x-6\right)\)
25 tháng 8 2021
a) \(x^2\left(x^2+4\right)-x^2-4=x^2\left(x^2+4\right)-\left(x^2+4\right)=\left(x^2+4\right)\left(x^2-1\right)=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
b) \(\left(x^2+x\right)^2+4x^2+4x-12=\left(x^2+x\right)^2+4\left(x^2+x\right)+4-16=\left(x^2+x+2\right)^2-4^2=\left(x^2+x+2-4\right)\left(x^2+x+2+4\right)=\left(x^2+x-2\right)\left(x^2+x+6\right)=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
c) \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24=\left(x^2+7x+10\right)^2+2\left(x^2+7x+10\right)+1-25=\left(x^2+7x+11\right)^2-5^2=\left(x^2+7x+11-5\right)\left(x^2+7x+11+5\right)=\left(x^2+7x+6\right)\left(x^2+7x+16\right)=\left(x+1\right)\left(x+6\right)\left(x^2+7x+16\right)\)
NT
25 tháng 8 2021
a. \(x^2\left(x^2+4\right)-x^2-4\)
\(=x^2\left(x^2+4\right)-\left(x^2+4\right)\)
\(=\left(x^2-1\right)\left(x^2+4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x^2+4\right)\)
b. \(\left(x^2+x\right)^2+4x^2+4x-12\)
\(=x^4+2x^3+5x^2+4x-12\)
\(=\left(x-1\right)\left(x+2\right)\left(x^2+x+6\right)\)
c. \(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-24\)
\(=\left(x+2\right)\left(x+5\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-24\) (*)
Đặt \(t=x^2+7x+10\), ta được
(*) \(=t\left(t+2\right)-24\)
\(=t^2+2t-24\)
\(=\left(t-4\right)\left(t+6\right)\)
hay \(\left(x^2+7x+6\right)\left(x^2+7x+18\right)\)
12 tháng 10 2023
a: \(x^2+4x+4=x^2+2\cdot x\cdot2+2^2=\left(x+2\right)^2\)
b: \(4x^2-4x+1=\left(2x\right)^2-2\cdot2x\cdot1+1^2=\left(2x-1\right)^2\)
c: \(2x-1-x^2\)
\(=-\left(x^2-2x+1\right)=-\left(x-1\right)^2\)
d: \(x^2+x+\dfrac{1}{4}=x^2+2\cdot x\cdot\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2=\left(x+\dfrac{1}{2}\right)^2\)
e: \(9-x^2=3^2-x^2=\left(3-x\right)\left(3+x\right)\)
g: \(\left(x+5\right)^2-4x^2=\left(x+5+2x\right)\left(x+5-2x\right)\)
\(=\left(5-x\right)\left(5+3x\right)\)
h: \(\left(x+1\right)^2-\left(2x-1\right)^2\)
\(=\left(x+1+2x-1\right)\left(x+1-2x+1\right)\)
\(=3x\left(-x+2\right)\)
i: \(=x^2y^2-4xy+4-3\)
\(=\left(xy-2\right)^2-3=\left(xy-2-\sqrt{3}\right)\left(xy-2+\sqrt{3}\right)\)
k: \(=y^2-\left(x-1\right)^2\)
\(=\left(y-x+1\right)\left(y+x-1\right)\)
l: \(=x^3+3\cdot x^2\cdot2+3\cdot x\cdot2^2+2^3=\left(x+2\right)^3\)
m: \(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot y+3\cdot2x\cdot y^2-y^3=\left(2x-y\right)^3\)
4 tháng 9 2021
\(4x^2-5xy+y^2=\left(4x^2-5xy+\dfrac{25}{16}y^2\right)-\dfrac{9}{16}y^2=\left(2x-\dfrac{5}{4}y\right)^2-\dfrac{9}{16}y^2=\left(2x-\dfrac{5}{4}y-\dfrac{3}{4}y\right)\left(2x-\dfrac{5}{4}y+\dfrac{3}{4}y\right)=\left(2x-2y\right)\left(2x-\dfrac{1}{2}y\right)=\left(x-y\right)\left(4x-y\right)\)
4 tháng 9 2021
\(4x^2-5xy+y^2\)
\(=4x^2-4xy-xy+y^2\)
\(=4x\left(x-y\right)-y\left(x-y\right)\)
\(=\left(x-y\right)\left(4x-y\right)\)
2 tháng 1 2023
Bài 2:
a: =x(x^2-25)
=x(x-5)(x+5)
b: =x(x-2y)+3(x-2y)
=(x-2y)(x+3)
c: =(2x-3)(4x^2+6x+9)+2x(2x-3)
=(2x-3)(4x^2+8x+9)
NV
Nguyễn Việt Lâm
Giáo viên
15 tháng 12 2020
a.
\(1-4x^2=\left(1-2x\right)\left(1+2x\right)\)
b.
\(8-27x^3=\left(2\right)^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
c.
\(27+27x+9x^2+x^3=x^3+3.x^2.3+3.3^2.x+3^3\)
\(=\left(x+3\right)^3\)
d.
\(2x^3+4x^2+2x=2x\left(x^2+2x+1\right)=2x\left(x+1\right)^2\)
e.
\(x^2-y^2-5x+5y=\left(x-y\right)\left(x+y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-5\right)\)
f.
\(x^2-6x+9-y^2=\left(x-3\right)^2-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
18 tháng 7 2021
e) Ta có: \(x^4-2x^3+2x-1\)
\(=\left(x^4-1\right)-2x\left(x^2-1\right)\)
\(=\left(x^2+1\right)\left(x-1\right)\left(x+1\right)-2x\left(x-1\right)\left(x+1\right)\)
\(=\left(x-1\right)\left(x+1\right)\cdot\left(x^2-2x+1\right)\)
\(=\left(x+1\right)\cdot\left(x-1\right)^3\)
h) Ta có: \(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3\left(x^2-y^2\right)-2\left(x-y\right)^2\)
\(=3\left(x-y\right)\left(x+y\right)-2\left(x-y\right)^2\)
\(=\left(x-y\right)\left(3x+3y-2x+2y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
18 tháng 7 2021
a) Ta có: \(x^2-y^2-2x-2y\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-2\right)\)
b) Ta có: \(x^2\left(x+2y\right)-x-2y\)
\(=\left(x+2y\right)\left(x^2-1\right)\)
\(=\left(x+2y\right)\left(x-1\right)\left(x+1\right)\)
DL
1
3 tháng 8 2021
a) \(4x^2-9y^2+6x-9y\)
\(=\left(2x-3y\right)\left(2x+3y\right)+3\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+3\right)\)
b) \(1-2x+2yz+x^2-y^2-z^2\)
\(=\left(x^2-2x+1\right)-\left(y^2-2yz+z^2\right)\)
\(=\left(x-1\right)^2-\left(y-z\right)^2\)
\(=\left(x-y+z-1\right)\left(x+y-z-1\right)\)
Tick hộ mình nha 😘
a: \(3abc^3-6a^2b^3c+12a^3bc\)
\(=3abc\cdot c^2-3abc\cdot2ab^2+3abc\cdot4a^2\)
\(=3abc\left(c^2-2ab^2+4a^2\right)\)
b: \(27-8y^3\)
\(=3^3-\left(2y\right)^3\)
\(=\left(3-2y\right)\left(9+6y+4y^2\right)\)
c: Sửa đề: \(4x^2+4x-y^2+1\)
\(=\left(4x^2+4x+1\right)-y^2\)
\(=\left(2x+1\right)^2-y^2\)
\(=\left(2x+1+y\right)\left(2x+1-y\right)\)
d: \(3a^2\cdot\left(x-2\right)-6ab\cdot\left(2-x\right)\)
\(=3a^2\cdot\left(x-2\right)+6ab\cdot\left(x-2\right)\)
\(=\left(x-2\right)\left(3a^2+6ab\right)\)
\(=3a\left(a+2b\right)\left(x-2\right)\)