Phân tích đa thức thành nhân tử:
a) \(x^5-x^4-x^3-x^2-x-2\)
b) \(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
c) \(x^5+x+1\)
Các cậu cứu tớ với
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
A
21 tháng 9 2019
a/\(\left(x^2-x\right)^2+4\left(x^2-x\right)-12.\)
cho \(\left(x^2-x\right)=a\)
\(\Rightarrow a^2+4a-12\)
\(=a^2+6a-2a-12\)
\(=\left(a^2+6a\right)-\left(2a+12\right)\)
\(=a\left(a+6\right)-2\left(a+6\right)\)
\(=\left(a+6\right)\left(a-2\right)\)
\(=\left(x^2-x+6\right)\left(x^2-x-2\right)\)
b/ \(\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)
\(=\left(x+1\right)\left(x+4\right)\left(x+2\right)\left(x+3\right)-24\)
\(=\left(x^2+4x+x+4\right)\left(x^2+3x+2x+6\right)-24\)
\(=\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24\)
Gọi \(x^2+5x+5=a\)
\(\Rightarrow\left(x^2+5x+4\right)\left(x^2+5x+6\right)-24=\left(a-1\right)\left(a+1\right)-24\)
\(=a^2-1-24\)
\(=a^2-25\)
\(=\left(a-5\right)\left(a+5\right)\)
\(\Rightarrow\left(x^2+5x+5-5\right)\left(x^2+5x+5+5\right)\)
\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
\(\Rightarrow\left(x+1\right)\left(x+2\right)\left(x+3\right)\left(x+4\right)-24\)\(=\left(x^2+5x\right)\left(x^2+5x+10\right)\)
C
1
1 tháng 11 2020
M = x9 - x7 + x6 - x5 - x4 + x3 - x2 + 1
= ( x9 - x7 ) + ( x6 - x4 ) - ( x5 - x3 ) - ( x2 - 1 )
= x7( x2 - 1 ) + x4( x2 - 1 ) - x3( x2 - 1 ) - ( x2 - 1 )
= ( x2 - 1 )( x7 + x4 - x3 - 1 )
= ( x - 1 )( x + 1 )[ x4( x3 + 1 ) - ( x3 + 1 ) ]
= ( x - 1 )( x + 1 )( x3 + 1 )( x4 - 1 )
= ( x - 1 )( x + 1 )( x + 1 )( x2 - x + 1 )( x2 - 1 )( x2 + 1 )
= ( x + 1 )2( x - 1 )( x2 - x + 1 )( x - 1 )( x + 1 )( x2 + 1 )
= ( x + 1 )3( x - 1 )2( x2 + 1 )( x2 - x + 1 )
MD
0
T
28 tháng 8 2023
1) \(3x\left(x-1\right)+5\left(x-1\right)\)
\(=\left(x-1\right)\left(3x+5\right)\)
2) \(4x(x-2y)-8y(2y-x)\)
\(=4x\left(x-2y\right)+8y\left(x-2y\right)\)
\(=\left(4x+8y\right)\left(x-2y\right)\)
\(=4\left(x+2y\right)\left(x-2y\right)\)
3) \(a^2\left(x-1\right)+b^2\left(1-x\right)\)
\(=a^2\left(x-1\right)-b^2\left(x-1\right)\)
\(=\left(a^2-b^2\right)\left(x-1\right)\)
\(=\left(a-b\right)\left(a+b\right)\left(x-1\right)\)
4) \(3x\left(x-a\right)+4a\left(a-x\right)\)
\(=3x\left(x-a\right)-4a\left(x-a\right)\)
\(=\left(x-a\right)\left(3x-4a\right)\)
5) \(5x\left(x-y\right)^2+10y^2\left(y-x\right)^2\)
\(=5x\left(x-y\right)^2+10y^2\left(x-y\right)^2\)
\(=\left(5x+10y^2\right)\left(x-y\right)^2\)
\(=5\left(x+2y^2\right)\left(x-y\right)^2\)
6) \(3x\left(x-3\right)^2+9\left(3-x\right)^2\)
\(=3x\left(x-3\right)^2+9\left(x-3\right)^2\)
\(=\left(3x+9\right)\left(x-3\right)^2\)
\(=3\left(x+3\right)\left(x-3\right)^2\)
7) \(x\left(m-a\right)^2-y\left(a-m\right)^2\)
\(=x\left(a-m\right)^2-y\left(a-m\right)^2\)
\(=\left(x-y\right)\left(a-m\right)^2\)
8) \(6y^2\left(x-1\right)^2+9y\left(1-x\right)^2\)
\(=6y^2\left(x-1\right)^2+9y\left(x-1\right)^2\)
\(=\left(6y^2+9x\right)\left(x-1\right)^2\)
\(=3\left(2y^2+3x\right)\left(x-1\right)^2\)
#Ayumu
TN
19 tháng 6 2016
a)x7+x5+1=x7+x6-x6+2x5-x5+x4-x4+x3-x3+x2-x2+1
=x7-x6+x5-x3+x2+x6-x5+x4-x2+x+x5-x4+x3-x+1
=x2(x5-x4+x3-x+1)+x(x5-x4+x3-x+1)+1(x5-x4+x3-x+1)
=(x2+x+1)(x5-x4+x3-x+1)
b)4x4-32x2+1=4x4+12x3+2x2-12x3-36x2-6x+2x2+6x+1
=2x2(2x2+6x+1)-6x(2x2+6x+1)+1(2x2+6x+1)
=(2x2-6x+1)(2x2+6x+1)
c)x6+27=(x2+3)(x2-3x+3)(x2+3x+3)
d)3(x4+x2+1)-(x2+x+1)
=3x4-3x3+2x2+3x3-3x2+2x+3x2-3x+2
=x2(3x2-3x+2)+x(3x2-3x+2)+1(3x2-3x+2)
=(x2+x+1)(3x2-3x+2)
e)bạn tự làm nhé
HM
0
c) x\(^5\)+x+1
= x\(^5\) - x\(^2\) +x\(^2\) + x + 1
= ( x\(^5\) - x\(^2\) ) + (x\(^2\) + x + 1)
= x\(^2\)(x\(^3\) - 1) + (x\(^2\) + x +1)
= x\(^2\)(x - 1)(x\(^2\) + x +1) + (x\(^2\) + x + 1)
= (x\(^2\) + x + 1)[x\(^2\)(x - 1) +1]
Cảm ơn cậu nhá