Lời giải:

a.

$64x^2-24y^2=8(8x^2-3y^2)=8(\sqrt{8}x-\sqrt{3}y)(\sqrt{8}x+\sqrt{3}y)$

b.

$64x^3-27y^3=(4x)^3-(3y)^3=(4x-3y)(16x^2+12xy+9y^2)$

c.

$x^4-2x^3+x^2=(x^2-x)^2=[x(x-1)]^2=x^2(x-1)^2$

d.

$(x-y)^3+8y^3=(x-y)^3+(2y)^3=(x-y+2y)[(x-y)^2-2y(x-y)+(2y)^2]$

$=(x+y)(x^2-4xy+7y^2)$

a) \(64x^2-24y^2\)

\(=8\left(8x^2-3y^2\right)\)

b) \(64x^3-27y^3\)

\(=\left(4x\right)^3-\left(3y\right)^3\)

\(=\left(4x-3y\right)\left(16x^2+12xy+9y^2\right)\)

c) \(x^4-2x^3+x^2\)

\(=x^2\left(x^2-2x+1\right)\)

\(=x^2\left(x-1\right)^2\)

d) \(\left(x-y\right)^3+8y^3\)

\(=\left(x-y+2y\right)\left(x^2-2xy+y^2-2xy+2y^2+4y^2\right)\)

\(=\left(x+y\right)\left(x^2-4xy+7y^2\right)\)

\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)

\(x^3-6x^2y+12xy^2-8y^3\)

\(=\left(x-2y\right)^3\)

\(x^3+8y^3\)

\(=x^3+\left(2y\right)^3\)

\(=\left(x+2y\right)\left(x^2-2xy+4y^2\right)\)

Bài này khá đơn giản nhé bạn.

x³+8y³

=x³+(2y)³

=(x+2y).(x²-2xy+4y²)

Như vậy đó bạn :

Co saI đề không bạn

\(10x\left(x-y\right)-8y\left(y-x\right)\)

\(=10x\left(x-y\right)+8y\left(x-y\right)\)

\(=\left(x-y\right)\left(10x+8y\right)\)

\(=2\left(x-y\right)\left(5x+4y\right)\)

 1/x-1/y=1/6 
<=> 6(y-x) = xy 
<=>y(6-x) = 6x (1) 
<=>y= 6x/(6-x) (2) 
(1) => x<6 => x=1,2,3,4,5 
Thay vào (2) ta có các cặp số nguyên thỏa đề bài là : 
(x;y)= (2;3);(3;6)

10x (x-y) -8y(y-x)

=10x(x-y) + 8y(x-y)

=(x-y)(10x+8y)

=2(x-y)(5x+4y)

\(=x^2-\left(y-4\right)^2\)

\(=\left(x-y+4\right)\left(x+y-4\right)\)

\(=x^2-\left(y^2-8y+16\right)=x^2-\left(y-4\right)^2=\left(x-y+4\right)\left(x+y-4\right)\)

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

=x^3 -8y^3 -2(x-2y)

=(x-2y)(x^2 +2xy +4y^2)- 2(x-2y)

=(x-2y)(x^2+2x +4y^2-2)

k day nhe

Ta co:    x³ - 2x + 4y -8y³ = (x³ -8y³) -(2x -4y) = (x - 2y)(x² + 2xy +y²) -2(x-2y) = (x-2y)(x² + 2xy + y² -2)

Đường Quỳnh Giang chep lai bai tui lam