Phân tích đa thức thành nhân tử:
g) (x+7)(x+9)-17
h) x(x+2)(x+4)(x+6)+15
K
Khách
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C
1
Những câu hỏi liên quan
30 tháng 8 2021
a: \(81x^5-x^3\)
\(=x^3\left(81x^2-1\right)\)
\(=x^3\left(9x-1\right)\left(9x+1\right)\)
b: \(9x^2y-12xy+4y\)
\(=y\left(9x^2-12x+4\right)\)
\(=y\left(3x-2\right)^2\)
c: \(\left(5-x\right)^2-16\left(x-2\right)^2\)
\(=\left(x-5\right)^2-\left(4x-8\right)^2\)
\(=\left(x-5-4x+8\right)\left(x-5+4x-8\right)\)
\(=-3\left(x-1\right)\left(5x-13\right)\)
31 tháng 8 2021
d: Ta có: \(9x^2-y^2-21x-7y\)
\(=\left(3x-y\right)\left(3x+y\right)-7\left(3x+y\right)\)
\(=\left(3x+y\right)\left(3x-y-7\right)\)
e: Ta có: \(-y^2+8y-16+9x^2\)
\(=-\left(y^2-8y+16-9x^2\right)\)
\(=-\left(y-4-3x\right)\left(y-4+3x\right)\)
f: Ta có: \(5x^2-4x-1\)
\(=5x^2-5x+x-1\)
\(=5x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(5x+1\right)\)
29 tháng 8 2021
\(g,x^4-16=\left(x^2-4\right)\left(x^2+4\right)=\left(x-2\right)\left(x+2\right)\left(x^2+4\right)\\ i,-x^2+10x-25=-\left(x-5\right)^2\\ k,x^3+3x^2+3x+1-27z^3\\ =\left(x+1\right)^3-27z^3\\ =\left(x+1-3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\\ =\left(x-3z+1\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\\ m,\left(x+y\right)^2-25\left(x+y\right)+24=\left(x+y-5\right)^2-1=\left(x+y-4\right)\left(x+y-6\right)\)
KK
29 tháng 8 2021
g. x4 - 16
<=> x4 - 42
<=> (x2)2 - 42
<=> (x2 - 4)(x2 + 4)
i. -x2 + 10x - 25
<=> -(x2 - 10x + 25)
<=> -(x2 -10x + 52)
<=> -(x - 5)2
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
NH
4
DN
6 tháng 11 2016
Ta có:
\(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
\(=\left(x^9-x^8\right)+\left(x^8-x^7\right)-\left(x^6-x^5\right)-\left(2x^5-2x^4\right)-\left(x^4-x^3\right)+\left(x^2-x\right)+\left(x-1\right) \)
\(=x^8.\left(x-1\right)+x^7.\left(x-1\right)-x^5.\left(x-1\right)-2x^4.\left(x-1\right)-x^3\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^8+x^7-x^5-2x^4-x^3+x+1\right)\)
HM
0
h: Ta có: \(x\left(x+2\right)\left(x+4\right)\left(x+6\right)+15\)
\(=\left(x^2+6x\right)\left(x^2+6x+8\right)+15\)
\(=\left(x^2+6x\right)^2+8\left(x^2+6x\right)+15\)
\(=\left(x^2+6x+3\right)\left(x^2+6x+5\right)\)
\(=\left(x+1\right)\left(x+5\right)\left(x^2+6x+3\right)\)