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$a)$ \(x^{12}:\left(-x\right)^6\)
\(=x^{12}:x^6\)
\(=x^{12-6}\)
\(=x^6\)
$b) $ \(\left(-x\right)^7:\left(-x\right)^5\)
\(=\left(-x\right)^{7-5}\)
\(=\left(-x\right)^2\)
\(=x^2\)
$c)$ \(5x^2y^4:10x^2y\)
\(=\dfrac{1}{2}y^3\)
$e)$ \(\left(-xy\right)^{14}:\left(-xy\right)^7\)
\(=\left(-xy\right)^{14-7}\)
\(=\left(-xy\right)^7\)
Các câu còn lại tương tự nha bạn!
a, 5x2 - 45x = 5x(x - 9)
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)
f, 3xy2 - 12xy + 12x
= 3x(y2 - 4y + 4)
= 3x(y - 2)2
g, 2x2 - 8x + 8
= 2(x2 - 4x + 4)
= 2(x - 2)2
h, 5x3 + 10x2y + 5xy2
= 5x( x2 + 2xy + y2 )
= 5x(x + y)2
k, x2 + 4x - 2xy - 4y + y2
= (x2 - 2xy + y2) + (4x - 4y)
= (x - y)2 + 4(x - y)
= (x - y)(x - y + 4)
i, x3 + ax2 - 4a - 4x
= (x3 - 4x) + (ax2 - 4a)
= x(x2 - 4) + a(x2 - 4)
= (x + a)(x2 - 4)
= (x + a)(x + 2)(x - 2)
Chúc bạn học tốt !
a,3x2-6x+9x2
=>12x2-6x
=>6x(2x-1)
b,10x(x-y)-6y(y-x)
=>10x(x-y)-6y(-(x-y))
=>10x(x-y)+6y(x-y)
=>2(x-y)(5x+3y)
c,3x2+5y-3xy-5x
=>3x(x-y)-5(x-y)
=>(x-y)(3x-5)
d,3y2-3z2+3x2+6xy
=>3(y2-z2+x2+2xy)
=>3[(y+x)2-z2]
=>3(y+x-z)(y+x+z)
e,16x3+54y3
=>2(8x3+27y3)
=>2(2x+3y)(4x2-6xy+9y2)
g,x2-25-2xy+y2
=>(x-y)2-25
=>(x-y-5)(x-y+5)
h,x5-3x4+3x3-x2
=>x2(x3-3x2+3x-1)
=>x2(x-1)3
Nhớ tick cho mk nhé
a) \(\left(2x^3-2x^2y+3xy^2\right):\left(\dfrac{-1}{2}x\right)\)
\(=3x^2y^3:\left(\dfrac{-1}{2}x\right)\)
\(=-6xy^3\)
b) \(\left(3x^2y^2-6x^2y+12xy\right):3xy\)
\(=9xy^2:3xy\)
\(=3y\)
mik ko chắc chắn lắm
A = \(\dfrac{-1}{2}\)x(-4x2 + 4xy -6y2): \(\dfrac{-1}{2}x\)
= -4x2 + 4xy - 6y2
B = 3xy(xy - 2x + 4) : 3xy
= xy - 2x + 4
mik ko chép lại đầu bài đâu, vì mik lười đánh máy lắm
\(P=\left(x-y\right)^2+\left(x+y\right)^2-2\left(x+y\right)\left(x-y\right)-4x^2=\left(x-y-x-y\right)^2-\left(2x\right)^2=\left(-2y\right)^2-\left(2x\right)^2\)
\(=\left(2y-2x\right)\left(2y+2x\right)=2\left(y-x\right)2\left(y+x\right)=4\left(x+y\right)\left(y-x\right)\)
\(x^3-x^2y+3x-3y=x^2\left(x-y\right)+3\left(x-y\right)=\left(x-y\right)\left(x^2+3\right)\)
\(x^3-2x^2-4xy^2+x=x\left(x^2-2x+1-4y^2\right)=x\left[\left(x-1\right)^2-\left(2y\right)^2\right]=x\left(x+2y-1\right)\left(x-2y-1\right)\)
\(\left(x+2\right)\left(x+3\right)\left(x+4\right)\left(x+5\right)-8=\left(x^2+7x+10\right)\left(x^2+7x+12\right)-8\)
Đặt \(x^2+7x+10=t\), ta có:
\(t\left(t+2\right)-8=t^2+2t-8=t^2-2t+4t-8=t\left(t-2\right)+4\left(t-2\right)=\left(t-2\right)\left(t+4\right)\)
\(=\left(x^2+7x+10+4\right)\left(x^2+7x+10-2\right)=\left(x^2+7x+14\right)\left(x^2+7x-8\right)\)
\(=3x\left(x^2+2xy+y^2\right)=3x\left(x+y\right)^2\)
\(=3x\left(x^2+2xy+y^2\right)\)