phân tích đa thức thành nhân tử
5x mũ 2 +6xy +y mũ 2
mn nào con thức thì gúp mik vs , đêm khuya khoắt trung thu phải ngồi làm bài nộp a
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
m, \(x^2+4x+4-4y^2=\left(x+2\right)^2-\left(2y\right)^2=\left(x+2-2y\right)\left(x+2+2y\right)\)
n, \(x^2+6xy+9y^2-4z^2=\left(x+3y\right)^2-\left(2z\right)^2=\left(x+3y-2z\right)\left(x+3y+2z\right)\)
k) = x( 2x - 1 ) - 3y( 2x - 1 ) = ( 2x - 1 )( x - 3y )
l) = x( x - y ) + 5( x - y ) = ( x - y )( x + 5 )
m) = ( a2 - 4a + 4 )( a2 + 4a + 4 ) = ( a - 2 )2( a + 2 )2
n) = y2( x2 - 1 ) - ( x2 - 1 ) = ( x - 1 )( x + 1 )( y - 1 )( y + 1 )
q) = 3[ ( x - y )2 - 4z2 ] = 3( x - y - 2z )( x - y + 2z )
1, \(7x^2y^5-14x^3y^4-21y^3=7y^3\left(x^2y^2-2x^3y-3\right)\)
2, \(-12x^2y+6xy^2-15xy=3xy\left(-4x+2y-5\right)\)
3, \(-14x^2y-21xy^2+28x^2y^2=-7xy\left(2x+3y-4xy\right)\)
\(a,x^2y-8x+xy-8=xy\left(x+1\right)-8\left(x+1\right)=\left(xy-8\right)\left(x+1\right)\\ b,=\left(x+3y\right)^2-9=\left(x+3y-3\right)\left(x+3y+3\right)\)
\(A=3x^2\left(2x^2-7x-2\right)-6x^2\left(x^2-4x-1\right)-3x^3+15\\ A=6x^4-21x^3-6x^2-6x^4+24x^3+6x^2-3x^3+15\\ A=15\left(đpcm\right)\)
\(Sửa:\left(6x^3-7x^2+2x\right):\left(2x+1\right)\\ =\left(6x^3+3x^2-10x^2-5x\right):\left(2x+1\right)\\ =\left[3x^2\left(2x+1\right)-5x\left(2x+1\right)\right]:\left(2x+1\right)\\ =3x^2-5x\)
3x2 + 9x - 30
= 3 ( x2 + 3x - 10 )
= 3 ( x2 + 5x - 2x - 10 )
= 3 [ x ( x + 5 ) - 2 ( x + 5 ) ]
= 3 ( x + 5 ) ( x - 2 )
3x2 + 9x - 30
= 3 ( x2 + 3x - 10 )
= 3 [ ( x2 - 2x ) + ( 5x - 10 ) ]
= 3 [ x ( x - 2 ) + 5 ( x - 2 ) ]
= 3 ( x + 5 ) ( x - 2 )
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 )
7(x - 3) - x(3 - x)
= (x - 3)(7 + x)
chỉ bt có v mà k bt có đúng k
1 ) 7 ( x - 3 ) - x ( 3 - x )
= 7 ( x - 3 ) + x ( x - 3 )
= ( x - 3 ) ( 7 + x )
2 ) 4x2 - 6x + 3 - 2x
= 4x2 - 2x - 6x + 3
= 2x ( 2x - 1 ) - 3 ( 2x - 1 )
= ( 2x - 1 ) ( 2x - 3 )
3 ) ( 4 - x ) - 4x + x2
= ( 4 - x ) - x ( 4 - x )
= ( 4 - x ) ( 1 - x )
4 ) x2 - 2xy + y2
= ( x - y )2
5x2 + 6xy + y2
= 5x2 + 5xy + xy + y2
= ( 5x2 + 5xy ) + ( xy + y2 )
= 5x( x + y ) + y( x + y )
= ( x + y )( 5x + y )
\(5x^2+6xy+y^2\)
\(=5x^2+5xy+xy+y^2\)
\(=5x\left(x+y\right)+y\left(x+y\right)\)
\(=\left(x+y\right)\left(5x+y\right)\)