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Trả lời:
1, x3 - x2 - 4
= x3 - x2 - 4 + 2x - 2x
= x3 - 2x2 + x2 - 4 + 2x - 2x
= ( x3 + x2 + 2x ) - ( 2x2 + 2x + 4 )
= x ( x2 + x + 2 ) - 2 ( x2 + x + 2 )
= ( x - 2 )( x2 + x + 2 )
2, x3 - 2x - 4
= x3 - 2x - 4 + 2x2 - 2x2
= x3 - 4x + 2x - 4 + 2x2 - 2x2
= ( x3 + 2x2 + 2x ) - ( 2x2 + 4x + 4 )
= x ( x2 + 2x + 2 ) - 2 ( x2 + 2x + 2 )
= ( x - 2 )( x2 + 2x + 2 )
\(a^4+4b^4=a^4+4a^2b^2+4b^4-4a^2b^2=\left(a^2+2b^2\right)^2-\left(2ab\right)^2\)
\(=\left(a^2+2b^2-2ab\right)\left(a^2+2b^2+2ab\right)\)
\(x^4+4=x^4+4x^2+4-4x^2=\left(x^2+2\right)^2-\left(2x\right)^2\)
\(=\left(x^2+2-2x\right)\left(x^2+2+2x\right)\)
1, \(x^2+2x-3=x^2+3x-x-3=x\left(x-1\right)+3\left(x-1\right)=\left(x+3\right)\left(x-1\right)\)
2, \(x^2+3x-10=x^2+5x-2x-10=x\left(x-2\right)+5\left(x-2\right)=\left(x+5\right)\left(x-2\right)\)
3, \(x^2-x-12=x^2-4x+3x-12=x\left(x+3\right)-4\left(x+3\right)=\left(x-4\right)\left(x+3\right)\)
4, \(3x^2+4x-7=3x^2+7x-3x-7=3x\left(x-1\right)+7\left(x-1\right)=\left(3x+7\right)\left(x-1\right)\)
5, \(4x^2-9y^2-5xy=4x^2-9xy+4xy-9y^2\)
\(=4x\left(x+y\right)-9y\left(x+y\right)=\left(4x-9y\right)\left(x+y\right)\)
6, \(x^2-2x-4y^2-4y=x^2-2x+1-4y^2-4y-1=\left(x-1\right)^2-\left(2y+1\right)^2\)
\(=\left(x-1-2y-1\right)\left(x-1+2y+1\right)=\left(x-2y-2\right)\left(x+2y\right)\)
a, \(4x^4+81=\left(2x^2\right)^2+9^2=\left(2x^2\right)^2+36x^2+9^2-36x^2\)
\(=\left(2x^2+9\right)^2-\left(6x\right)^2=\left(2x^2-6x+9\right)\left(2x^2+6x+9\right)\)
b, \(64x^4+y^4=\left(8x^2\right)^2+\left(y^2\right)^2=\left(8x^2\right)^2+16x^2y^2+\left(y^2\right)^2-16x^2y^2\)
\(=\left(8x^2+y^2\right)^2-\left(4xy\right)^2=\left(8x^2-4xy+y^2\right)\left(8x^2+4xy+y^2\right)\)
x2 - x - y2 - y
=x2 - y2 - x - y
=(x - y)(x + y) - (x + y)
=(x + y)(x - y - 1)
x4y4 + 4
= x4y4 + 4x2y2 + 4 - 4x2y2
= (x2y2 + 2)2 - (2xy)2
= (x2y2 - 2xy + 2)(x2y2 + 2xy + 2)
x4y4 + 64
= x4y4 + 16x2y2 + 64 - 16x2y2
= (x2y2 + 8)2 - (4xy)2
= (x2y2 - 4xy + 8)(x2y2 + 4xy + 8)
x5 + x + 1
= x5 - x2 + x2 + x + 1
= x2(x3 - 1) + (x2 + x + 1)
= x2(x - 1)(x2 + x + 1) + (x2 + x + 1)
= (x2 + x + 1)[x2(x - 1) + 1]
1/ \(4x^2-12xy+9y^2=\left(2x\right)^2-2.2.3xy+\left(3y\right)^2\)
\(=\left(2x-3y\right)^2\)
2/ \(x^3-y^6=x^3-\left(y^2\right)^3\)
\(=\left(x-y^2\right)\left(x^2+xy^2+y^4\right)\)
Làm tạm 2 phần đợi mik xíu
4x2 - 12xy + 9y2 = ( 2x )2 - 2.2x.3y + ( 3y )2 = ( 2x - 3y )2
x3 - y6 = x3 - ( y2 )3 = ( x - y2 )( x2 + xy2 + y4 )
x6 - 6x4 + 12x2 - 8 = ( x2 )3 - 3.(x2)2.2 + 3.x2.22 - 23 = ( x2 - 2 )3
( x2 + 4y2 - 5 )2 - 16( x2y2 + 2xy + 1 ) = ( x2 + 4y2 - 5 )2 - 42( xy + 1 )2
= ( x2 + 4y2 - 5 )2 - ( 4xy + 4 )2
= [ ( x2 + 4y2 - 5 ) - ( 4xy + 4 ) ][ ( x2 + 4y2 - 5 ) + ( 4xy + 4 ) ]
= ( x2 + 4y2 - 5 - 4xy - 4 )( x2 + 4y2 - 5 + 4xy + 4 )
= [ ( x2 - 4xy + 4y2 ) - 9 ][ ( x2 + 4xy + 4y2 ) - 1 ]
= [ ( x - 2y )2 - 32 ][ ( x + 2y )2 - 12 ]
= ( x - 2y - 3 )( x - 2y + 3 )( x + 2y - 1 )( x + 2y + 1 )
( a + b )3 - ( a3 + b3 ) = a3 + 3a2b + 3ab2 + b3 - a3 - b3
= 3a2b + 3ab2
= 3ab( a + b )
a, \(a^4+64=\left(a^2\right)^2+8^2=\left(a^2\right)^2+16a^2+8^2-16a^2\)
\(=\left(a^2+8\right)^2-\left(4a\right)^2=\left(a^2-4a+8\right)\left(a^2+4a+8\right)\)
b, \(a^4+4b^2=\left(a^2\right)^2+\left(2b\right)^2=\left(a^2\right)^2+4a^2b+4b^2-4a^2b\)ĐK : b >= 0
\(=\left(a+2b\right)^2-\left(2a\sqrt{b}\right)^2=\left(a+2b-2a\sqrt{b}\right)\left(a+2b+2a\sqrt{b}\right)\)
p/s : mình nghĩ phần b bạn nên để đề là \(a^4+4b^4\)sẽ hợp lí hơn
Anh nhận bú lồn hoàn toàn miễn phí nha mấy em