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a) \(a^4b^4+a^3b-abc=ab\left(a^3b^3+a^2-c\right)\)
b) \(-x^2y^2z^2-6x^3y-8x^4z^2-9x^5y^5z^5\)
\(=x^2\left(-y^2z^2-6xy-8x^2z^2-9x^3y^5z^5\right)\)
c) \(16x^2-9\left(x+y\right)^2=\left(4x\right)^2-\left(3x+3y\right)^2=\left(7x+3y\right)\left(x-3y\right)\)
d) \(\left(a-b\right)^2-1=\left(a-b+1\right)\left(a-b-1\right)\)
e) \(a^6-b^6\)
\(=\left(a^3-b^3\right)\left(a^3+b^3\right)\)
\(=\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)
Để x;y;z ra ngoài làm thừa số chung rồi quất hết phần còn lại vào ngoặc thì thành 2 nhân tử thôi bạn, kiểu như phân phối ý.
a) \(12x^5y+24x^4y^2+12x^3y^3\)
\(=12x^3y\left(x^2+2xy+y^2\right)\)
\(=12x^3y\left(x+y\right)^2\)
b) \(x^2-2xy-4+y^2\)
\(=\left(x-y\right)^2-2^2\)
\(=\left(x-y-2\right)\left(x-y+2\right)\)
g) \(12xy-12xz+3x^2y-3x^2z\)
\(=12x\left(y-z\right)+3x^2\left(y-z\right)\)
\(=3x\left(4+x\right)\left(y-z\right)\)
e) \(16x^2-9\left(x^2+2xy+y^2\right)\)
\(=\left(4x\right)^2-\left[3\left(x+y\right)\right]^2\)
\(=\left(4x-3\left(x+y\right)\right)\left(4x+3\left(x+y\right)\right)\)
\(=\left(x+y\right)\left(7x+y\right)\)
d) làm tương tự như phần g chỉ khác là phải nhóm( nhóm xen kẽ), phần f cũng vậy
a) \(6x^4-9x^3\)
\(=3x^3\left(2x-3\right)\)
b) \(5y^{10}+15y^6\)
\(=5y^6\left(y^4+5\right)\)
c) \(9x^2y^2+15^2y-21xy^2\)
\(=9x^2y^2+225y-21xy^2\)
\(=3y\left(3x^2y+75-7xy\right)\)
d) \(x^2y^2z+xy^2z^2+x^2yz^2\)
\(=xyz\left(xy+yz+xz\right)\)
a) \(8a^2xy-18b^2xy=2xy\left(4a^2-9b^2\right)=2xy\left(2a-3b\right)\left(2a+3b\right)\)
b) \(32a^2b^2-4=4\left(8a^2b^2-1\right)\)
c) \(x^2-49z^2-4xy+4y^2=\left(x^2-4xy+4y^2\right)-49z^2\)
\(=\left(x-2y\right)^2-\left(7z\right)^2=\left(x-2y+7z\right)\left(x-2y-7z\right)\)
d) \(3x^2+6x+3-3y^2=3\left(x^2+2x+1-y^2\right)=3.\left[\left(x+1\right)^2-y^2\right]\)
\(=3\left(x-y+1\right)\left(x+y+1\right)\)
e) \(12x^2y-12y^3+36xy+27y=3y\left(4x^2-4y^2+12x+9\right)\)
\(=3y\left[\left(4x^2+12x+9\right)-4y^2\right]=3y\left[\left(2x+3\right)^2-\left(2y\right)^2\right]\)
\(=3y\left(2x-2y+3\right)\left(2x+2y+3\right)\)
a) 8a2xy - 18b2xy
= 2xy( 4a2 - 9b2 )
= 2xy( [ ( 2a )2 - ( 3b )2 ]
= 2xy( 2a - 3b )( 2a + 3b )
b) 32a2b2 - 4
= 4( 8a2b2 - 1 )
c) x2 - 49z2 - 4xy + 4y2
= ( x2 - 4xy + 4y2 ) - 49z2
= ( x - 2y )2 - ( 7z )2
= ( x - 2y - 7z )( x - 2y + 7z )
d) 3x2 + 6x + 3 - 3y2
= 3( x2 + 2x + 1 - y2 )
= 3[ ( x2 + 2x + 1 ) - y2 ]
= 3[ ( x + 1 )2 - y2 ]
= 3( x - y + 1 )( x + y + 1 )
e) 12x2y - 12y3 + 36xy + 27y
= 3y( 4x2 - 4y2 + 12x + 9 )
= 3y[ ( 4x2 + 12x + 9 ) - 4y2 ]
= 3y[ ( 2x + 3 )2 - ( 2y )2 ]
= 3y( 2x - 2y + 3 )( 2x + 2y + 3 )
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)