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a) xy – 3x + 2y – 6
= (xy - 3x) + (2y - 6)
= x(y - 3) + 2(y - 3)
= (y - 3)(x + 2)
b) x2y + 4xy + 4y – y3
= y(x2 + 4x + 4 - y2)
= y[(x2 + 4x + 4) - y2]
= y[(x + 2)2 - y2]
= y(x + 2 + y)(x + 2 - y)
c) x2 + y2 + xz + yz + 2xy
= (x2 + 2xy + y2) + (xz + yz)
= (x + y)2 + z(x + y)
= (x + y)(x + y + z)
d) x3 + 3x2 – 3x – 1
= (x3 - 1) + (3x2 - 3x)
= (x - 1)(x2 + x + z) + 3x(x - 1)
= (x - 1)(x2 + 4x + 1)
a )
\(xy-3x+2y-6\)
\(=\left(xy+2y\right)-3x-6\)
\(=y\left(x+2\right)-3\left(x+2\right)\)
\(=\left(y-3\right)\left(x+2\right)\)
b )
\(x^2y+4xy+4y-y^3\)
\(=y\left(x^2+4x+4-y^2\right)\)
\(=y\left[\left(x+2\right)^2-y^2\right]\)
\(=y\left(x+2-y\right)\left(x+2+y\right)\)
c )
\(x^2+y^2+xz+yz+2xy\)
\(=\left(x+y\right)^2+z\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+z\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
a) \(x^4-9x^2\)
\(=x^2\left(x^2-9\right)\)
\(=x^2\left(x-3\right)\left(^{ }x+3\right)\)
b) \(3x^2-12x+12\)
\(=3x\left(x^2-4x+4\right)\)
\(=3x\left(x-2\right)^2\)
c) \(x^2+5x+6\)
\(=x^2+3x+2x+6\)
\(=x\left(x+3\right)+2\left(x+3\right)\)
\(=\left(x+3\right)\left(x+2\right)\)
x4 - 9x2
= x4 - ( 3x )2
= ( x2 - 3x ) ( x2 + 3x )
b) 3x3 - 12x2 + 12x
= 3x3 - 6x2 - 6x2 + 12x
= 3x2( x - 2 ) - 6x ( x - 2 )
= ( 3x2 - 6x ) ( x - 2 )
= 3x ( x - 2 ) ( x - 2 )
= 3x ( x- 2 )2
c) x2 + 5x + 6
= x2 + 2x + 3x + 6
= x ( x + 2 ) + 3 ( x + 2 )
= ( x + 3 ) ( x + 2 )
\(x^4-5x^2+4\)
=\(\left(x^4-x^2\right)-\left(4x^2-4\right)\)
=\(x^2\left(x^2-1\right)-4\left(x^2-1\right)\)
=\(\left(x^2-4\right)\left(x^2-1\right)\)
\(x^2+5x-6\)
=\(x^2-x+6x-6\)
=\(x\left(x-1\right)+6\left(x-1\right)\)
=\(\left(x+6\right)\left(x-1\right)\)
2 câu cuối làm tương tự nha câu 2 nha
\(a.=x^3-2x^2+x^2-2x+x-2=x^2\left(x-2\right)+x\left(x-2\right)+\left(x-2\right)=\left(x-2\right)\left(x^2+x+2\right)\)
b.\(=2x^3+x^2-2x^2-x-2x-1=x^2\left(2x+1\right)-x\left(2x-1\right)-\left(2x-1\right)\)\(=\left(2x-1\right)\left(x^2-x-1\right)\)
c.\(3x^3-x^2+6x^2-2x-12x+4=x^2\left(3x-1\right)+2x\left(3x-1\right)-4\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2+2x-4\right)\)
d.\(3x^3-x^2-6x^2+2x+15x-5=x^2\left(3x-1\right)-2x\left(3x-1\right)+5\left(3x-1\right)\)\(=\left(3x-1\right)\left(x^2-2x+5\right)\)
t i c k cho mình nha
Lời giải:
a.
$3x^2+xy-4y^2=(3x^2-3xy)+(4xy-4y^2)=3x(x-y)+4y(x-y)=(x-y)(3x+4y)$
b.
$x^8-5x^4+4=(x^8-x^4)-(4x^4-4)$
$=x^4(x^4-1)-4(x^4-1)=(x^4-1)(x^4-4)$
$=(x^2-1)(x^2+1)(x^2-2)(x^2+2)$
$=(x-1)(x+1)(x^2+1)(x-\sqrt{2})(x+\sqrt{2})(x^2+2)$
c.
$x^3+3x^2+3x-7=(x^3+3x^2+3x+1)-8$
$=(x+1)^3-2^3=(x+1-2)[(x+1)^2+2(x+1)+4]$
$=(x-1)(x^2+4x+7)$
a) \(3x^2+xy-4y^2=3x^2-3xy+4xy-4y^2\)
\(=3x(x-y)+4y(x-y)=(3x+4y)(x-y)\)
b)\(x^8-5x^4+4=x^8-x^4-4x^4+4\)
\(=x^2(x^4-1)-4(x^4-1)=(x^2-4)(x^4-1)\)
\(=(x-2)(x+2)(x^2-1)(x^2+1)=(x-2)(x+2)(x-1)(x+1)(x^2+1)\)
c)\(x^3+3x^2+3x-7=x^3+3x^2+3x+1-8\)
\(\left(x+1\right)^3-\sqrt{2}^3=\left(x+1-\sqrt[]{2}\right)\left(\left(x+1\right)^2+2\sqrt{2}x+2\right)\)