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\(x^2-\left(y-3\right)^2-4x+4\)
\(=x^2-\left(y^2-6y+9\right)-4x+4\)
\(=x^2-y^2+6y-9-4x+4\)
\(=\left(x^2-4x+4\right)-\left(y^2-6y+9\right)\)
\(=\left(x-2\right)^2-\left(y-3\right)^2\)
\(=\left[\left(x-2\right)-\left(y-3\right)\right]\left[\left(x-2\right)+\left(y-3\right)\right]\)
\(=\left(x-y+5\right)\left(x+y-5\right)\)
1.
x2 - ( y - 3 )2 - 4x + 4
= ( x2 - 4x + 4 ) - ( y - 3 )2
= ( x - 2 )2 - ( y - 3 )2
= [ ( x - 2 ) - ( y - 3 ) ][ ( x - 2 ) + ( y - 3 ) ]
= ( x - 2 - y + 3 )( x - 2 + y - 3 )
= ( x - y + 1 )( x + y - 5 )
2.
a) Ta có : 2x4 + 8x3 + 9x2 - 4x - 5
= 2x4 + 10x2 - x2 + 8x3 - 4x - 5
= ( 2x4 - x2 ) + ( 8x3 - 4x ) + ( 10x2 - 5 )
= x2( 2x2 - 1 ) + 4x( 2x2 - 1 ) + 5( 2x2 - 1 )
= ( 2x2 - 1 )( x2 + 4x + 5 )
=>(2x4 + 8x3 + 9x2 - 4x - 5) : ( 2x2 - 1 ) = x2 + 4x + 5
b) Ta có : x2 + 4x + 5 = ( x2 + 4x + 4 ) + 1 = ( x + 2 )2 + 1 ≥ 1 > 0 ∀ x
=> đpcm
a,(x-y)^2-2(x+y)+1 b, x^2-y^2+4x+4 c, 4x^2-y^2+8(y-2)
=(x-y-1)^2 =(x^2+4x+4)-y^2 =4x^2-y^2+8y-16
=(x+2)^2-y^2 =4x^2-(y^2-8y+16)
=(x+2-y)(x+2+y) =4x^2-(y-4)^2
a) (x+y)2-2(x+y)+1=(x+y-1)2
b) x2-y2+4x+4 = (x2+4x+4)-y2=(x+2)2-y2=(x+y+2)(x-y+2)
c)4x2-y2+8(y-2) = 4x2-(y2-8y+16) = (2x)2-(y-4)2=(2x+y-4)(2x-y+4)
d)x3-2x2+2x-4 = x2(x-2)+2(x-2) = (x-2)(x2+2)
e)xy-4+2x-2y=x(y+2) - 2(y+2) = (x-2)(y+2)
1.
a) \(2x\left(x-4\right)+\left(x-1\right)\left(x+2\right)=2x^2-8x+x^2+x-2=x^2-7x-2\)
b) \(\left(x-3\right)^2-\left(x-2\right)\left(x^2+2x+4\right)=x^2-6x+9-x^3+8=-x^3+x^2-6x+17\)
2.
a) \(x^2y+xy^2-3x+3y=xy\left(x+y\right)-3\left(x-y\right)=???\)
b) \(x^3+2x^2y+xy^2-16x=x\left(x^2+2xy+y^2-16\right)=x\left[\left(x+y\right)^2-16\right]=\)làm tiếp chắc dễ
3.
\(\frac{x^4?2x^3+4x^2+2x+3}{x^2+1}\) Giữa x^4 và 2x^3 (vị trí dấu ? là dấu + hay -)
4) \(A=x^2-3x+4=\left(x-\frac{3}{2}\right)^2+\frac{7}{4}\)
\(A\ge\frac{7}{4}\)
Vậy GTNN của A là 7/4
a)\(\left(4x^3-xy^2+y^3\right)\left(x^2y+2xy^2-2y^3\right)\)
\(=x^2y\left(4x^3-xy^2+y^3\right)+2xy^2\left(4x^3-xy^2+y^3\right)\)
\(-2y^3\left(4x^3-xy^2+y^3\right)\)
\(=4x^5y-x^3y^3+x^2y^4+8x^4y^2-2x^2y^4+2xy^5\)
\(-8x^3y^3+2xy^5-2y^6\)
\(=-2y^6+4x^5y+\left(2xy^5+2xy^5\right)+8x^4y^2+\left(x^2y^4-2x^2y^4\right)\)
\(-\left(x^3y^3+8x^3y^3\right)\)
\(=-2y^6+4x^5y+4xy^5+8x^4y^2-x^2y^4-9x^3y^3\)
b)
(!) \(2\left(x+y\right)^2-7\left(x+y\right)+5\)
\(=2\left(x+y\right)^2-2\left(x+y\right)-5\left(x+y\right)+5\)
\(=2\left(x+y\right)\left(x+y-1\right)-5\left(x+y-1\right)\)
\(=\left(2x+2y-5\right)\left(x+y-1\right)\)
(!!) \(\left(x+y+z\right)^2-x^2-y^2-z^2\)
\(=\left(x^2+y^2+z^2+2xy+2yz+2zx\right)-x^2-y^2-z^2\)
\(=2\left(xy+yz+zx\right)\)
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 :))))