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\(\dfrac{5xy^3\left(x-y\right)}{10xy\left(x-y\right)^3}=\dfrac{y^2}{2\left(x-y\right)^2}\)
\(\dfrac{5xy^3\left(x-y\right)}{10xy\left(x-y\right)^3}=\dfrac{y^2}{2\left(x-y\right)^2}\)
a) 3x+9y
=3(x+3y)
b) 4x2-16xy
=4x(x-4y)
c) x2-6x+9
=(x-3)2
d) 4y2-36
=(2y-6)(2y+6)
e) 4y2-4
=(2y-2)(2y+2)
f) 5a2-9
=\(\left(\sqrt{5}a-3\right)\left(\sqrt{5}a+3\right)\)
g) x3-2x2+x
=x(x2-2x+1)
=x(x-1)2
h) 5xy3-10xy2+5xy
=5xy(y2-2y+1)
=5xy(y-1)2
i) 12x2y+12xy+3y
=3y(4x2+4x+1)
=3y(2x+1)2
k) 5x-5y+y2-xy
=(5x-5y)+(y2-xy)
=5(x-y)-y(x-y)
=(x-y)(5-y)
#H
a) 15x2y : ( - 5xy)
= -5xy( -3x) : ( - 5xy)
= -3x
b) ( 15x2y - 10xy2 + 5xy) : 5xy
= 5xy( 3x - 2y + 1) : 5xy
= 3x - 2y + 1
c) ( x +y)3 : ( x+y)
= (x + y)( x +y)2 : ( x+ y)
= ( x +y)2
Bài 1:
a, (\(x\) - 4).(\(x\) + 4) - (5 - \(x\)).(\(x\) + 1)
= \(x^2\) - 16 - 5\(x\) - 5 + \(x^2\) + \(x\)
= (\(x^2\) + \(x^2\)) - (5\(x\) - \(x\)) - (16 + 5)
= 2\(x^2\) - 4\(x\) - 21
b, (3\(x^2\) - 2\(xy\) + 4) + (5\(xy\) - 6\(x^2\) - 7)
= 3\(x^2\) - 2\(xy\) + 4 + 5\(xy\) - 6\(x^2\) - 7
= (3\(x^2\) - 6\(x^2\)) + (5\(xy\) - 2\(xy\)) - (7 - 4)
= - 3\(x^2\) + 3\(xy\) - 3
a) -5xy(3x2y – 5xy +y2)
=-15x3y2+25x2y2-5xy3
b) (x + 8)2 -2(x + 8) (x – 2) + (x – 2)2
=[(x+8)-(x-2)]2
=(x+8-x+2)2
=102
=100
\(a,-5xy\left(3x^2y-5xy+y^2\right)=-15x^3y^2+25x^2y^2-5xy^3\)
\(b,\left(x+8\right)^2-2\left(x+8\right)\left(x-2\right)+\left(x-2\right)^2=\left[x+8-\left(x-2\right)\right]^2=\left[x+8-x+2\right]^2=10^2=100\)