phân tích đa thức sau thành nhân tử
x^9-x^7+x^6-x^5-x^4+x^3-x^2+1
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\(x^5+x+1\)
\(=x^5+x^4+x^3-x^4-x^3-x^2+x^2+x+1\)
\(=\left(x^2+x+1\right)\left(x^3-x^2+1\right)\)
\(x^4+x^3+2x^2+x+1=\left(x^4+x^3+x^2\right)+\left(x^2+x+1\right)\\ =x^2\left(x^2+x+1\right)+\left(x^2+x+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
Dễ thấy \(x^2+1>0\); \(x^2+x+1=\left(x+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0\) nên ta không thể phân tích thêm được nữa.
Vậy \(x^4+x^3+2x^2+x+1=\left(x^2+1\right)\left(x^2+x+1\right)\).
Câu 1:
\(=x^2-\left(y-4\right)^2\)
\(=\left(x-y+4\right)\cdot\left(x+y-4\right)\)
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 )
\(=x^2\left(x+y\right)-\left(x+y\right)=\left(x^2-1\right)\left(x+y\right)=\left(x-1\right)\left(x+1\right)\left(x+y\right)\)
Ta có:
\(x^9-x^7-x^6-x^5+x^4+x^3+x^2-1\)
\(=\left(x^9-x^8\right)+\left(x^8-x^7\right)-\left(x^6-x^5\right)-\left(2x^5-2x^4\right)-\left(x^4-x^3\right)+\left(x^2-x\right)+\left(x-1\right) \)
\(=x^8.\left(x-1\right)+x^7.\left(x-1\right)-x^5.\left(x-1\right)-2x^4.\left(x-1\right)-x^3\left(x-1\right)+x\left(x-1\right)+\left(x-1\right)\)
\(=\left(x-1\right)\left(x^8+x^7-x^5-2x^4-x^3+x+1\right)\)
\(x^3\left(2+x\right)^2-\left(x+2\right)^2+1-x^3\\ =\left(x+2\right)^2\left(x^3-1\right)-\left(x^3-1\right)\\ =\left[\left(x+2\right)^2-1\right]\left(x^3-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x^2+4x+3\right)=\left(x-1\right)\left(x+1\right)\left(x+3\right)\left(x^2+x+1\right)\)
1: \(3x\left(x-1\right)+7x^2\left(x-1\right)\)
\(=\left(x-1\right)\left(7x^2+3x\right)\)
=x(7x+3)(x-1)
2: \(9\left(x+5\right)^2-\left(x-7\right)^2\)
\(=\left(3x+15\right)^2-\left(x-7\right)^2\)
=(3x+15+x-7)(3x+15-x+7)
=(4x+8)(2x+22)
\(=4\left(x+2\right)\cdot2\cdot\left(x+11\right)=8\left(x+2\right)\left(x+11\right)\)
3: \(x^2-2xy+y^2-xz+yz\)
\(=\left(x-y\right)^2-z\left(x-y\right)\)
=(x-y)(x-y-z)
4: \(\left(x^2+2x\right)^2-2x^2-4x-3\)
\(=\left(x^2+2x\right)^2-2\cdot\left(x^2+2x\right)-3\)
\(=\left(x^2+2x-3\right)\left(x^2+2x+1\right)=\left(x+3\right)\left(x-1\right)\left(x+1\right)^2\)
5: \(9x^2y^2+15x^2y-21xy^2\)
\(=3xy\cdot3xy+3xy\cdot5x-3xy\cdot7y\)
=3xy(3xy+5x-7y)
6: \(x^3-2x^2y+xy^2-9xy^4\)
\(=x\left(x^2-2xy+y^2-9y^4\right)\)
\(=x\left\lbrack\left(x-y\right)^2-\left(3y^2\right)^2\right\rbrack=x\left(x-y-3y^2\right)\left(x-y+3y^2\right)\)
7: \(x^2+x-12\)
\(=x^2+4x-3x-12\)
=x(x+4)-3(x+4)
=(x+4)(x-3)
8: (x+1)(x+3)(x+5)(x+7)+15
\(=\left(x^2+8x+7\right)\left(x^2+8x+15\right)+15\)
\(=\left(x^2+8x+7\right)^2+8\left(x^2+8x+7\right)+15\)
\(=\left(x^2+8x+7+3\right)\left(x^2+8x+7+5\right)\)
\(=\left(x^2+8x+10\right)\left(x^2+8x+12\right)=\left(x^2+8x+10\right)\left(x+2\right)\left(x+6\right)\)