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(12x^2 - 12xy + 3y^2) - 10.(2x - y) + 8
= 3(4x^2 - 4xy + y^2) - 10(2x - y) + 8
= 3(2x - y)^2 - 10(2x - y) + 8
= 3(2x - y)^2 - 10(2x - y) + 8
= 3(2x - y)^2 - 6(2x - y) - 4(2x - y) + 8
= 3(2x - y)(2x - y - 2) - 4(2x - y -2)
= (2x - y -2)[3(2x - y) - 4]
= (2x - y -2)(6x - 3y -4)
Ai k mk mk k lại
(12x^2 - 12xy + 3y^2) - 10.(2x - y) + 8
= 3(4x^2 - 4xy + y^2) - 10(2x - y) + 8
= 3(2x - y)^2 - 10(2x - y) + 8
= 3(2x - y)^2 - 10(2x - y) + 8
= 3(2x - y)^2 - 6(2x - y) - 4(2x - y) + 8
= 3(2x - y)(2x - y - 2) - 4(2x - y -2)
= (2x - y -2)[3(2x - y) - 4]
= (2x - y -2)(6x - 3y -4)
1.
$4x^2y+5x^3-x^2y^2=x^2(4y+5x-y^2)$
2.
$5x(x-1)-3y(1-x)=5x(x-1)+3y(x-1)=(x-1)(5x+3y)$
3.
$4x^2-25=(2x)^2-5^2=(2x-5)(2x+5)$
4.
$6x-9-x^2=-(x^2-6x+9)=-(x-3)^2$
5.
$x^2+4y^2+4xy=x^2+2.x.2y+(2y)^2=(x+2y)^2$
6.
$\frac{1}{64}-27x^3=(\frac{1}{4})^3-(3x)^3$
$=(\frac{1}{4}-3x)(\frac{1}{16}+\frac{3x}{4}+9x^2)$
7.
$x^3-6x^2+12x-8=x^3-3.x^2.2+3.x.2^2-2^3$
$=(x-2)^3$
8.
$x^2-x-y^2-y=(x^2-y^2)-(x+y)=(x-y)(x+y)-(x+y)$
$=(x+y)(x-y-1)$
9.
$5x-5y+ax-ay=5(x-y)+a(x-y)$
$=(x-y)(5+a)$
a. 10a3 - 10
= 10(a3 - 1)
= 10(a - b)(a2 + ab + b2)
b. -12x3y + 75xy3
= 3xy[-(4x2 - 25y2)]
= 3xy{-[(2x)2 - (5y)2]}
= 3xy{-[(2x + 5y)(2x - 5y)]}
Câu a mình nhầm 1 tí
a. 10a3 - 10
= 10(a3 - 1)
= 10(a - 1)(a2 + a + 1)