Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Phối hợp cả 3 phương phép để phân tích các đa thức sau thành phân tử:
a) 36 - 4a2 + 20ab - 25b2
= 36 - (4a2 - 20ab + 25b2)
= 62 - (2a - 5b)2
= (6 - 2a + 5b)(6 + 2a - 5b)
b) a3 + 3a2 + 3a + 1 - 27b3
= (a + 1)3 - (3b)3
= (a + 1 - 3b)[(a + 1)2 + 3b(a + 1) + 9b2]
= (a + 1 - 3b)(a2 + 2a + 1 + 3ab + 3b + 9b2)
c) x2 + 2xy + y2 - xz - yz
= (x + y)2 - z(x + y)
= (x + y)(x + y - z)
d) 5a3 - 10a2b + 5ab2 - 10a + 10b
= 5(a3 - 2a2b + ab2 - 2a + 2b)
= 5[a(a2 - 2ab + b2) - 2(a - b)]
= 5[a(a - b)2 - 2(a - b)]
= 5(a - b)(a2 - ab - 2)
a) \(3xy^2-12xy+12x\)
\(=3x\left(y-4y+4\right)\)
b) \(3x^3y-6x^2y-3xy^3-6axy^2-3a^2xy+3xy\)
\(=3xy\left(x^2-2x-y^2-2ay-a^2+1\right)\)
\(=3xy\left[\left(x^2-2\cdot x\cdot1+1^2\right)-\left(y^2+2\cdot y\cdot a+a^2\right)\right]\)
\(=3xy\left[\left(x-1\right)^2-\left(y+a\right)^2\right]\)
\(=3xy\left(x-1-y-a\right)\left(x-1+y+a\right)\)
c) \(36-4a^2+20ab-25b^2\)
\(=6^2-\left[\left(2a\right)^2-2\cdot2a\cdot5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
d) \(5a^3-10a^2b+5ab^2-10a+10b\)
\(=5a\left(a^2-2ab+b^2\right)-10\left(a-b\right)\)
\(=5a\left(a-b\right)^2-10\left(a-b\right)\)
\(=\left(a-b\right)\left[5a\left(a-b\right)-10\right]\)
\(=5\left(a-b\right)\left[a\left(a-b\right)-2\right]\)
\(=5\left(a-b\right)\left(a^2-ab-2\right)\)
a. 3xy2-12xy+12x
= 3x(y2-4y+4)
= 3x(y-2)2
b. 3x3y-6x2y-3xy3-6axy2-3a2xy+3xy
= 3xy( x2-2x-y2-2ay-a2+1)
= 3xy ((x2-2x+1)-(a2-2ay-y2))
=3xy((x-1)2-(a-y)2)
= 3xy((x-1+a-y)(x-1-(a-y))
=3xy(x-1+a-y)(x-1-a+y)
d. =( 5a(a2-2ab+b2))-(10(a+b))
= 5a(a-b)2-10(a-b)
=5a(a-b)(a-b)-10(a-b)
=(a-b)(5a(a-b)-10)
Hình như mik làm sai hết rồi
a) 36 - 4a2 + 20ab - 25b2 = 36 - ( 4a2 - 20ab + 25b2 ) = 62 - ( 2a - 5b )2 = ( 6 - 2a + 5b )( 6 + 2a - 5b )
b) ( xy + 4 )2 - 4( x + y )2 = ( xy + 4 )2 - 22( x + y )2 = ( xy + 4 )2 - [ 2( x + y ) ]2
= ( xy + 4 )2 - ( 2x + 2y )2 = ( xy + 4 - 2x - 2y )( xy + 4 + 2x + 2y )
= [ x( y - 2 ) - 2( y - 2 ) ][ x( y + 2 ) + 2( y + 2 ) ]
= ( y - 2 )( x - 2 )( y + 2 )( x + 2 )
c) x2 + y2 - x2y2 + xy - x - y
= ( x2 - x2y2 ) + ( y2 - y ) + ( xy - x )
= x2( 1 - y2 ) + y( y - 1 ) + x( y - 1 )
= x2( 1 - y )( 1 + y ) - y( 1 - y ) - x( 1 - y )
= ( 1 - y )[ x2( 1 + y ) - y - x ) ]
= ( 1 - y )( x2 + x2y - y - x )
= ( 1 - y )[ ( x2 - x ) + ( x2y - y ) ]
= ( 1 - y )[ x( x - 1 ) + y( x2 - 1 ) ]
= ( 1 - y )[ x( x - 1 ) + y( x - 1 )( x + 1 ) ]
= ( 1 - y )( x - 1 )[ x + y( x + 1 ) ]
= ( 1 - y )( x - 1 )( x + xy + y )
d) 3x + 3y - x2 - 2xy - y2
= 3( x + y ) - ( x2 + 2xy + y2 )
= 3( x + y ) - ( x + y )2
= ( x + y )( 3 - x - y )
e) ( 2xy + 1 )2 - ( 2x + y )2
= ( 2xy + 1 - 2x - y )( 2xy + 1 + 2x + y )
= [ ( 2xy - 2x ) - ( y - 1 ) ][ ( 2xy + 2x ) + ( y + 1 ) ]
= [ 2x( y - 1 ) - ( y - 1 ) ][ 2x( y + 1 ) + ( y + 1 ) ]
= ( y - 1 )( 2x - 1 )( y + 1 )( 2x + 1 )
a) \(36-4a^2+20ab-25b^2\)
\(=36-\left(4a^2-20ab+25b^2\right)\)
\(=36-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
b) \(\left(xy+4\right)^2-4\left(x+y\right)^2\)
\(=\left(xy+4-2x-2y\right)\left(xy+4+2x+2y\right)\)
\(=\left[x\left(y-2\right)-2\left(y-2\right)\right]\left[x\left(y+2\right)+2\left(y+2\right)\right]\)
\(=\left(x+2\right)\left(x-2\right)\left(y+2\right)\left(y-2\right)\)
c) \(x^2+y^2-x^2y^2+xy-x-y\)
\(=-\left(x^2y^2-x^2\right)+\left(y^2-y\right)+\left(xy-x\right)\)
\(=-x^2\left(y-1\right)\left(y+1\right)+y\left(y-1\right)+x\left(y-1\right)\)
\(=\left(y-1\right)\left(-x^2y-x^2+y+x\right)\)
\(=\left(1-y\right)\left[\left(x^2y-y\right)+\left(x^2-x\right)\right]\)
\(=\left(1-y\right)\left(x-1\right)\left(xy+y+x\right)\)
a/ \(=3y^2-6y-2x+1\)
b/ \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
c/ \(=\left(2-x\right)^3\)
d/ \(=xy^2+x^2y+3xy+x^2y+x^3+3x^2-3xy-3x^2-9x\)
\(=xy\left(y+x+3\right)+x^2\left(y+x+3\right)-3x\left(y+x+3\right)\)
\(=\left(xy+x^2-3x\right)\left(y+x+3\right)=x\left(y+x-3\right)\left(y+x+3\right)\)
e/ \(=xy-x^2+2x-y^2+xy-2y\)
\(=x\left(y-x+2\right)-y\left(y-x+2\right)=\left(x-y\right)\left(y-x+2\right)\)
a) =(2x+3y-1)2
b)=-(x-1)3
c)=-(x3-6x2+12x-8)=-(x-2)3
d)x3 + 2x2y + xy2 – 9x
= x(x2 + 2xy + y2 -9)
= x[(x2 + 2xy + y2) - 32]
= x[(x + y)2 - 32]
= x (x + y – 3)(x + y + 3)
e) 2x-2y-x2+2xy-y2=2(x-y)-(x-y)2=(x-y)(2-x+y)
Bài 1:
a) x3 - 3x2 + 3x - 1 + 2(x2 - x)
= (x - 1)3 + 2x(x - 1)
= (x - 1)[(x - 1)2 + 2x]
= (x - 1)(x2 - 2x + 1 + 2x)
= (x - 1)(x2 + 1)
b) 36 - 4a2 + 20ab - 25b2
= 36 - (2a - 5b)2
= (6 - 2a + 5b)(6 + 2a - 5b)
c) 5a3 - 10a2b + 5ab2 - 10a + 10b
= 5(a3 - 2a2b + ab2 - 2a + 2b)
= 5[a(a2 - 2ab + b2) - 2(a - b)]
= 5[a(a - b)2 - 2(a - b)]
= 5(a - b)(a2 - ab - 2)
a)\(36-4a^2+20ab-25b^2=6^2-\left(4a^2-20ab+25b^2\right)\)
\(=6^2-\left[\left(2a\right)^2-2.2a.5b+\left(5b\right)^2\right]\)
\(=6^2-\left(2a-5b\right)^2\)
\(=\left(6-2a+5b\right)\left(6+2a-5b\right)\)
b)\(a^3+3a^2+3a+1-27b^3=\left(a+1\right)^3-\left(3b\right)^3\)(chỗ này mình sửa 27b2 thành 27b3 vì mình nghĩ nhầm đề)
\(=\left(a+1-3b\right)\left[\left(a+1\right)^2+\left(a+1\right)3b+\left(3b\right)^2\right]\)
\(=\left(a+1-3b\right)\left(a^2+2a+1+3ab+3b+9b^2\right)\)
c)\(x^3+3x^2+3x+1-3x^2-3x=\left(x+1\right)^3-3x\left(x+1\right)\)
\(=\left(x+1\right)\left[\left(x+1\right)^2-3x\right]\)
\(=\left(x+1\right)\left(x^2+2x+1-3x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)\)
a) 36-4a2+20ab-25b2
= 6^2 - (4a^2 - 20xb + 25b^2)
= 6^2 - (2a - 5b)^2
= [6 - (2a - 5b)] [6 + (2a - 5b)]
= (6 - 2a + 5b) (6 + 2a -5b)