Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
1. x2 - 16 - 4xy + 4y2
= ( x2 - 4xy + 4y2 ) - 16
= ( x - 2y )2 - 42
= ( x - 2y - 4 )( x - 2y + 4 )
2. 4x2 + 4x - 3
= ( 4x2 + 4x + 1 ) - 4
= ( 2x + 1 )2 - 2
= ( 2x + 1 - 2 )( 2x + 1 + 2 )
= ( 2x - 1 )( 2x + 3 )
3. x2 - x - 12
= x2 + 3x - 4x - 12
= x( x + 3 ) - 4( x + 3 )
= ( x + 3 )( x - 4 )
4. 3x + 3y - x2 - 2xy - y2
= ( 3x + 3y ) - ( x2 + 2xy + y2 )
= 3( x + y ) - ( x + y )2
= ( x + y )( 3 - x - y )
5. 4y4 + 16
= 4( y4 + 4 )
= 4( y4 + 4y2 + 4 - 4y2 )
= 4[ ( y4 + 4y2 + 4 ) - 4y2 ]
= 4[ ( y2 + 2 )2 - ( 2y )2 ]
= 4( y2 - 2y + 2 )( y2 + 2y + 2 )
a,\(x^2-16-4xy+4y^2\)
\(=\left(x^2-4xy+4y^2\right)-16\)
\(=\left(x-2y\right)^2-4^2\)
\(=\left(x-2y-4\right)\left(x-2y+4\right)\)
b,\(4x^2+4x-3\)
\(=4x^2-2x+6x-3\)
\(=\left(4x^2-2x\right)+\left(6x-3\right)\)
\(=2x\left(2x-1\right)+3\left(2x-1\right)\)
\(=\left(2x+3\right)\left(2x-1\right)\)
c,\(x^2-x-12\)
\(=x^2-4x+3x-12\)
\(=\left(x^2+3x\right)-\left(4x-12\right)\)
\(=x\left(x+3\right)-4\left(x+3\right)\)
\(=\left(x-4\right)\left(x+3\right)\)
1
a) x2 + 4y2 + 4xy - 16
=(x2 + 4xy + 4y2) - 16
=(x+2y)2 - 16
=(x+2y-4)(x+2y+4)
b)x2 + y2 - 2x + 4y + 5 =0
<=> x2 - 2x + 1 + y2 - 4y + 4=0
<=> (x-1)2 + (y-2)2 =0
<=> x=1 và y=2
a)
(x-y+5)2-2.(x-y+5)+1
=(x-y+5-1)2
=(x-y+4)2
b)
(x2+4y2-5)2-16.(x2.y2+2xy+1)
=(x2+4y2-5)2-[4.(xy+1)]2
=(x2+4y2-5-4xy-4)(x2+4y2-5+4xy+4)
=(x2+4y2-4xy-9)(x2+4y2+4xy-1)
=[(x-2y)2-9][(x+2y)2-1]
=(x-2y-3)(x-2y+3)(x+2y-1)(x+2y+1)
=(x2+x-3x-3)((x-2y+3)(x+2y-1)(x+1)2
=[x(x+1)-3(x+1)](x-2y+3)(x+2y-1)(x+1)2
=(x+1)(x-3)(x-2y+3)(x+2y-1)(x+1)2
a) x^2+4xy-16+4y^2
=(x^2+4xy+4y^2)-4^2
=(x+2y)^2-4^2
=(x+2y-4)(x+2y+4)
b)27-(x-1)^3
=3^3-(x-1)^3
=(4-x)(5+4x)
c)x^2-4x+3
=x^2-x-3x+3
=x(x-1)-3(x-1)
=(x-1)(x-3)
d) x^2-x-12
=x^2-4x+3x-12
=x(x-4)+3(x-4)
=(x-4)(x+3)
e) x^4+4
=(x^2)^2+2x^2.2+2^2-2x^2.2
=(x^2+2)^2-4x^2
=(x^2-2x+2)(x^2+2x+2)
tick nha
a, \(=12x^5+9x^3y^2-6x^2y^3-20x^4y-15x^2y^3-10xy^4-24x^3y^2-18xy^4+12y^5\)
(tự rút gọn cái :P)
b, \(8x^3+4x^2y-2xy^2-y^3\)
\(=4x^2\left(2x+y\right)-y^2\left(2x+y\right)=\left(2x+y\right)^2\left(2x-y\right)\)
\(4x^2y^2-4x^2-4xy-y^2=4x^2y^2-\left(2x+y\right)^2\)
\(=\left(2x+y+2xy\right)\left(2xy-2x+y\right)\)
Mấy cái còn lại nhân tung ra là được mà :))))
c) \(x^2+y^2+xz+yz+2xy\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
b) \(x^3+3x^2-3x-1\)
\(=\left(x^3-1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)\)
\(=\left(x-1\right)\left(x^2+4x+1\right)\)
1/
a) \(x^2+4y^2+4xy-16\)
\(=x^2+2.2xy+\left(2y\right)^2-4^2\)
\(=\left(x+2y\right)^2-4^2\)
\(=\left(x+2y-4\right)\left(x+2y+4\right)\)
b) ta có:
\(\left(2x+y\right)\left(y-2x\right)+4x^2\)
\(=-\left(2x-y\right)\left(2x+y\right)+4x^2\)
\(=\left(2x\right)^2-\left[\left(2x\right)^2-y^2\right]\)
\(=\left(2x\right)^2-\left(2x\right)^2+y^2\)
\(=y^2\)
Vậy giá trị của biểu thức trên không phụ thuộc vào giá trị của x
nên tại y = 10
giá trị của biểu thức trên bằng y2 = 102 = 100
a) \(x^2-25-4xy+4y^2\)
\(=\left(x^2-4xy+4y^2\right)-25\)
\(=\left(x-2y\right)^2-5^2\)
\(=\left(x-2y-5\right)\left(x-2y+5\right)\)
b) \(x^2-8x+15\)
\(=x^2-3x-5x+15\)
\(=x\left(x-3\right)-5\left(x-3\right)\)
\(=\left(x-3\right)\left(x-5\right)\)
a)\(x^2-25-4xy+4y^2\Leftrightarrow\left(x^2-4xy+4y^2\right)-25\)
\(\Leftrightarrow\left(x-2y\right)^2-5^2\)
\(\Leftrightarrow\left(x-2y-5\right)\left(x-2y+5\right)\)
b)\(x^2-8x+15\Leftrightarrow\left(x-3\right)\left(x-5\right)\)
1: =(16x^2-8x+1)-y^2
=(4x-1)^2-y^2
=(4x-1-y)(4x-1+y)
2: =(x^2-2xy+y^2)-z^2
=(x-y)^2-z^2
=(x-y-z)(x-y+z)
3: =(x^2+4xy+4y^2)-16
=(x+2y)^2-4^2
=(x+2y-4)(x+2y+4)
4: =(x^2-4xy+4y^2)-16
=(x-2y)^2-4^2
=(x-2y-4)(x-2y+4)