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\(a^4+8a^3+14a^2-8a-15\)
\(=a^4+8a^3+15a^2-a^2-8a-15\)
\(=a^2\left(a^2+8a+15\right)-\left(a^2+8a+15\right)\)
\(=\left(a^2+8a+15\right)\left(a^2-1\right)\)
\(=\left(a+3\right)\left(a+5\right)\left(a-1\right)\left(a+1\right)\)
a^4+8a^3-8a^2+14a-15 = a^4- a^3+9a^3-9a^2+a^2-a+15a-15= a^3(a-1)+9a^2(a-1)+a(a-1)+15(a-1)
= (a-1)(a^3+9a^2+a+15)
Đến đó đưa về bậc 3 bn tự giải
Ta thấy:a=1 là nghiệm đa thức
Suy ra chia đa thức cho a-1 ta dc:
(a-1)(a3+9a2+a+15)
27x6 + 125y6 = ( 3x2 )3 + ( 5y2 )3 = ( 3x2 + 5y2 )( 9x4 - 15x2y2 + 25y4 )
8a6 - 8b6 = ( 2a2 )3 - ( 2b2 )3 = ( 2a - 2b )( 4a2 + 4ab + 4b2 ) = 2( a - b )4( a2 + ab + b2 ) = 8( a - b )( a2 + ab + b2 )
x4 + 64y4 = x4 + 16x2y2 + 64y4 - 16x2y2
= ( x4 + 16x2y2 + 64y4 ) - 16x2y2
= ( x2 + 8y2 )2 - ( 4xy )2
= ( x2 + 8y2 - 4xy )( x2 + 8y2 + 4xy )
x4 + x3 + 2x2 + x + 1 = x4 + x3 + x2 + x2 + x + 1
= ( x4 + x3 + x2 ) + ( x2 + x + 1 )
= x2( x2 + x + 1 ) + ( x2 + x + 1 )
= ( x2 + x + 1 )( x2 + 1 )
\(27x^6+125y^6=\left(3x^2\right)^3+\left(5y^2\right)^3=\left(3x^2+5y^2\right)\left(9x^4-15x^2.y^2+25y^4\right)\)
\(8a^6-8b^6=8\left(a^6-b^6\right)=8\left(\left(a^3\right)^2-\left(b^3\right)^2\right)=8\left(a^3-b^3\right)\left(a^3+b^3\right)\)
\(=8\left(a-b\right)\left(a^2+ab+b^2\right)\left(a+b\right)\left(a^2-ab+b^2\right)\)
\(x^{\text{4}}+64y^4=x^4+64y^4+16x^2y^2-16x^2y^2\)
\(=\left(8y^2+x^2\right)^2-\left(4xy\right)^2=\left(8y^2+x^2+4xy\right)\left(8y^2+x^2-4xy\right)\)
\(x^4+x^3+2x^2+x+1=\left(x^4+2x^2+1\right)+\left(x^3+x\right)\)
\(=\left(x^2+1\right)^2+x\left(x^2+1\right)=\left(x^2+1\right)\left(x^2+x+1\right)\)
a, Ta có: \(x^3+2x^2y+xy^2-4x\)
\(=x\left(x^2+2xy+y^2-4\right)\)
\(=x\left[\left(x+y\right)^2-2^2\right]\)
\(=x\left(x+y+2\right)\left(x+y-2\right)\)
b, Ý này dễ lắm, cậu tự làm nha!!!
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) 4a2b3 - 6a3b2 = 2a2b2( 2b - 3a )
b) ( a - b )2 - ( b - a ) = ( a - b )2 + ( a - b ) = ( a - b )( a - b + 1 )
c) ( 8a3 - 27b3 ) - 2a( 4a2 - 9b2 ) = 8a3 - 27b3 - 8a3 + 18ab2 = 18ab2 - 27b3 = 9b2( 2a - 3b )
d) 10x2 + 10xy + 5x + 5y = 10x( x + y ) + 5( x + y ) = ( x + y )( 10x + 5 ) = 5( x + y )( 2x + 1 )
e) 5ay - 3bx + ax - 15by = 5y( a - 3b ) + x( a - 3b ) = ( a - 3b )( 5y + x )
a) \(4a^2.b^3-6a^3.b^2=2a^2.b^2\left(2b-3a\right)\)
b) \(\left(a-b\right)^2-\left(b-a\right)=\left(a-b\right)^2+\left(a-b\right)\)
\(=\left(a-b\right).\left(a-b+1\right)\)
c) \(8a^3-27b^3-2a.\left(4a^2-9b^2\right)=8a^3-27b^3-8a^3+18ab^2\)
\(=-27b^3+18ab^2=18ab^2-27b^3=9b^2.\left(2a-3b\right)\)
d) \(10x^2+10xy+5x+5y=5.\left(2x^2+2xy+x+y\right)\)
\(=5.\left[\left(2x^2+2xy\right)+\left(x+y\right)\right]=5.\left[2x\left(x+y\right)+\left(x+y\right)\right]\)
\(=5\left(x+y\right)\left(2y+1\right)\)
e) \(5ay-3bx+ax-15by=\left(5ay-15by\right)-\left(3bx-ax\right)\)
\(=5y\left(a-3b\right)-x\left(3b-a\right)=5y\left(a-3b\right)+x\left(a-3b\right)\)
\(=\left(a-3b\right)\left(x+5y\right)\)
Có sai dấu ko bạn
Đề : \(a^4+8a^3+14a^2-8a-15\)
\(=a^4-a^2+8a^3-8a+15a^2-15\)
\(=a^2\left(a^2-1\right)+8a\left(a^2-1\right)+15\left(a^2-1\right)\)
\(=\left(a^2-1\right)\left(a^2+8a+15\right)\)
\(=\left(a+1\right)\left(a-1\right)\left(a+3\right)\left(a+5\right)\)