a: \(x^2-y^2+10yz-25z^2\)

\(=x^2-\left(y^2-10yz+25z^2\right)\)

\(=x^2-\left(y-5z\right)^2\)

\(=\left(x-y+5z\right)\left(x+y-5z\right)\)

Bài a thì bạn thử xem lại đề nhé.

a) $x^3-3x^2y+4x-12y$

$=(x^3-3x^2y)+(4x-12y)$

$=x^2(x-3y)+4(x-3y)$

$=(x-3y)(x^2+4)$

b) $4x^2-y^2+4y-4$

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

$=(2x)^2-(y^2-2\cdot y\cdot2+2^2)$

$=(2x)^2-(y-2)^2$

$=[2x-(y-2)][2x+(y-2)]$

$=(2x-y+2)(2x+y-2)$

c) $9x^2-6x-y^2+2y$

$=(9x^2-y^2)-(6x-2y)$

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

$=(3x-y)(3x+y)-2(3x-y)$

$=(3x-y)(3x+y-2)$

$\text{#}Toru$

bạn ấn ở chỗ x² cho rõ hơn nhé

a: \(=\dfrac{2}{5}\left(xy-x-y^2+1\right)\)

\(=\dfrac{2}{5}\left[x\left(y-1\right)-\left(y-1\right)\left(y+1\right)\right]\)

\(=\dfrac{2}{5}\left(y-1\right)\left(x-y-1\right)\)

b: \(=x\left(x^2+2xy+y^2-9\right)\)

\(=x\left(x+y-3\right)\left(x+y+3\right)\)

a) $=\frac{2}{5}\left(xy-x-y^2+1\right)$

$=\frac{2}{5}\left[x\left(y-1\right)-\left(y-1\right)\left(y+1\right)\right]$

$=\frac{2}{5}\left(y-1\right)\left(x-y-1\right)$

b) $=x(x^2+2xy+y^2-9$

a, \(6x^3y^2.\left(2-x\right)+9x^2y^2\left(x-2\right)\)
\(=6x^3y^2.\left(2-x\right)-9x^2y^2\left(2-x\right)\)
\(=y^2.\left(2-x\right)\left(6x^3-9x^2\right)\)
\(=3x^2y^2.\left(2-x\right)\left(2x-3\right)\)

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

a: =(x-z)(y+8)

b; =x^2-2x-3x+6

=(x-2)(x-3)

c: =x^4+10x^2-x^2-10

=(x^2+10)(x^2-1)

=(x^2+10)(x-1)(x+1)

\(a,=ab\left(a+3\right)\\ b,=\left(x-1\right)^2\\ c,=x\left[\left(x-3\right)^2-y^2\right]=x\left(x-y-3\right)\left(x+y-3\right)\)

a: =(x-1)²