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Lời giải:
$N=p^{m+2}q-pq^{m+3}-p^{m+3}q^{n+4}$
$=pq(p^{m+1}-q^{m+2}-p^{m+2}q^{n+3})$
a) 3x(x + 7)2 - 11x2(x + 7) + 9(x + 7) = (x + 7)[3x(x + 7) - 11x2 + 9) = (x + 7)(3x2 + 21x - 11x2 + 9)
= (x + 7)(-8x2 + 21x + 9)(-8x2 + 24x - 3x + 9) = (x + 7)[-8x(x - 3) - 3(x - 3)] = -(x + 7)(8x + 3)(x - 3)
b) 3x(x - 9)2 - (9 - x)3 = 3x(x - 9)2 + (x - 9)3 = (x - 9)2(3x + x - 9) = (x - 9)2(4x - 9)
c) pm + 2.q - pm + 1.q3 - p2.qn + 1 + p.qn + 3 = (pm + 2.q - p2.qn + 1) - (pm + 1.q3 - p.qn + 3)
= p2.q(pm - qn) - p.q3(pm - qn) = pq(pm - qn)(p - q2)
d) x2y2z + xy2z2 + x2yz = xyz(xy + yz + x)
a) \(3x\left(x+7\right)^2-11x^2\left(x+7\right)+9\left(x+7\right)\)
\(=\left(x+7\right)\left[3x\left(x+7\right)-11x^2+9\right]=\left(x+7\right)\left(3x^2+21x-11x^2+9\right)\)
\(=\left(x+7\right)\left(-8x^2+21x+9\right)=\left(x+7\right)\left[\left(-8x^2+24x\right)-\left(3x-9\right)\right]\)
\(=\left(x+7\right)\left[-8x\left(x-3\right)-3\left(x-3\right)\right]=-\left(x+7\right)\left(x-3\right)\left(8x+3\right)\)
b) \(3x\left(x-9\right)^2-\left(9-x\right)^3=3x\left(x-9\right)^2+\left(x-9\right)^3\)
\(=\left(x-9\right)^2\left(3x+x-9\right)=\left(x-9\right)^2\left(4x-9\right)\)
c) \(p^{m+2}.q-p^{m+1}.q^3-p^2.q^{n+1}+p.q^{n+3}\)
\(=p^{m+1}.q\left(p-q^2\right)-p.q^{n+1}\left(p-q^2\right)\)\(=p.q.\left(p-q^2\right).\left(p^m.q^n\right)\)
d) \(x^2y^2z+xy^2z^2+x^2yz=xyz\left(xy+yz+x\right)\)
\(p^{m+2}.q-p^{m+1}.q^3-p^2.q^{n+1}+p.q^{n+3}\)
\(=pq\left(p^{m+1}-p^mq^2-pq^n+q^{n+2}\right)\)
\(=p^m\left(p-q^2\right)-q^n\left(p-q^2\right)\)
\(=\left(p-q^2\right)\left(p^m-q^n\right)\)
..........
Đề sai nhé .Sửu lại
\(x^2-4x^2y^2+4+4x\)
\(=\left(x^2+4x+4\right)-4x^2y^2\)
\(=\left(x+2\right)^2-\left(2xy\right)^2\)
\(=\left(x+2+2xy\right)\left(x+2-2xy\right)\)
1/ \(x^2+x-90=\left(x^2-10x\right)+\left(9x-90\right)=x\left(x-10\right)+9\left(x-10\right)=\left(x-10\right)\left(x+9\right)\)
2/ \(2x^2+4xy+2y^2=\left(2x^2+2xy\right)+\left(2xy+2y^2\right)=2x\left(x+y\right)+2y\left(x+y\right)=\left(x+y\right)\left(2x+2y\right)\)
3/ \(2y^2-14y+24=2\left(y^2-7y+12\right)=2\left[\left(y^2-4y\right)+\left(12-3y\right)\right]=2\left[y\left(y-4\right)-3\left(y-4\right)\right]\)
\(=2\left(y-4\right)\left(y-3\right)\)
4/ \(x^8+x^4+1=\left(x^8+x^7+x^6\right)-\left(x^7+x^6+x^5\right)+\left(x^5+x^4+x^3\right)-\left(x^3+x^2+x\right)+\left(x^2+x+1\right)\)
\(=x^6\left(x^2+x+1\right)-x^5\left(x^2+x+1\right)+x^3\left(x^2+x+1\right)-x\left(x^2+x+1\right)+\left(x^2+x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^6-x^5+x^3-x+1\right)\)
\(=\left(x^2+x+1\right)\left[\left(x^6-x^5+x^4\right)-\left(x^4-x^3+x^2\right)+\left(x^2-x+1\right)\right]\)
\(=\left(x^2+x+1\right)\left[x^4\left(x^2-x+1\right)\right]-x^2\left(x^2-x+1\right)+\left(x^2-x+1\right)\)
\(=\left(x^2+x+1\right)\left(x^2-x+1\right)\left(x^4-x^2+1\right)\)
a) ax2 - 2bxy + 2bx2 - axy
= ( ax2 - axy ) + ( 2bx2 - 2bxy )
= ax( x - y ) + 2bx( x - y )
= ( x - y )( ax + 2bx )
= x( x - y )( a + 2b )
b) x2 + 2x - 4y2 + 8y - 3 < đã sửa >
= ( x2 + 2x + 1 ) - ( 4y2 - 8y + 4 )
= ( x + 1 )2 - ( 2y - 2 )2
= [ ( x + 1 ) - ( 2y - 2 ) ][ ( x + 1 ) + ( 2y - 2 ) ]
= ( x + 1 - 2y + 2 )( x + 1 + 2y - 2 )
= ( x - 2y + 3 )( x + 2y - 1 )
c) x4 + 5x3 + 20x - 16
= x4 + 5x3 + 4x2 - 4x2 + 20x - 16
= ( x4 + 5x3 - 4x2 ) + ( 4x2 + 20x - 16 )
= x2( x2 + 5x - 4 ) + 4( x2 + 5x - 4 )
= ( x2 + 5x - 4 )( x2 + 4 )
a) = pq(pm+1 - pm - pq2n+3)
b) = (m-3)2 - (x - 2y)2
= ( m-3 +x -2y)(m-3 -x +2y)
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