a: 6x-2y=2(3x-y)

b: =(x-y)(x-2)(x+2)

Lời giải:
a. Không phân tích được nữa

b. $x^2(x-y)+4(y-x)=x^2(x-y)-4(x-y)=(x-y)(x^2-4)=(x-y)(x-2)(x+2)$
c. $x^3+2x^2y+xy^2-4x=x(x^2+2xy+y^2-4)$

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

ko phân tích dc

b: =(x-y)(x-2)(x+2)

a: \(4x^2-4x\)

\(=4x\cdot x-4x\cdot1\)

\(=4x\left(x-1\right)\)

b: \(x^2-2xy+y^2-4\)

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

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

`4x^2-4x`

`= 4x(x-1)`

__

`x^2 -2xy +y^2-4`

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

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

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

a,

\(4x^2-4x\\=4x(x-1)\)

b,

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

\(\text{#}Toru\)

a: \(=x\left(x-3\right)-4y\left(x-3\right)\)

=(x-3)(x-4y)

d: \(=\left(x-2\right)\left(x+2\right)+\left(x+2\right)^2\)

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

=2x(x+2)

\(a,=x\left(x-3\right)-4y\left(x-3\right)=\left(x-4y\right)\left(x-3\right)\\ b,=\left(x-1\right)\left(x^2+x+1\right)-4x\left(x-1\right)=\left(x-1\right)\left(x^2-3x+1\right)\\ c,=\left(x-y\right)\left(1-a\right)\\ d,=\left(x-2\right)\left(x-2+x+2\right)=2x\left(x-2\right)\\ e,=x^2\left(x+y\right)-xz\left(x+y\right)=x\left(x-z\right)\left(x+y\right)\\ f,=\left(x-y-2\right)\left(x+y\right)\)

a.

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1+3z\right)\left[\left(x+1\right)^2+3z\left(x+1\right)+9z^2\right]\)

\(=\left(x+3z+1\right)\left(x^2+2x+1+3zx+3z+9z^2\right)\)

b.

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

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

c.

\(=x^4-1+4x^2-4\)

\(=\left(x^2-1\right)\left(x^2+1\right)+4\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left(x^2+5\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

a) Ta có: \(x^3+3x^2+3x+1-27z^3\)

\(=\left(x+1\right)^3-\left(3z\right)^3\)

\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)

b) Ta có: \(x^2-2xy+y^2-zx+yz\)

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

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

c) Ta có: \(x^4+4x^2-5\)

\(=x^4+4x^2+4-9\)

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

\(=\left(x^2-1\right)\left(x^2+5\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x^2+5\right)\)

a) \(=6x^2y^2\left(6xy-7\right)\)

b) \(=3xy\left(x^3y+5x-6\right)\)

c) \(=\left(ax+ab\right)-\left(bx+x^2\right)=a\left(b+x\right)-x\left(b+x\right)=\left(a-x\right)\left(b+x\right)\)

d) \(=3\left(2x-1\right)-\left(2x-1\right)^2=\left(2x-1\right)\left(3-2x+1\right)=\left(2x-1\right)\left(4-2x\right)=2\left(2x-1\right)\left(2-x\right)\)

\(a,=6x^2y^2\left(6xy-7\right)\\ b,=3xy\left(x^3y+5x-6\right)\\ c,=x\left(a-x\right)-b\left(a-x\right)=\left(x-b\right)\left(a-x\right)\\ d,=3\left(2x-1\right)-\left(2x-1\right)^2=\left(2x-1\right)\left(3-2x+1\right)=2\left(2-x\right)\left(2x-1\right)\)

Câu 1:

Phần a đề sai nên mk sửa lại:

a, x² + 5x - 14 = x² - 2x + 7x - 14 = x(x - 2) + 7(x - 2) = (x - 2)(x + 7)

b, xz + yz - 5(x + y) = z(x + y) - 5(x + y) = (x + y)(z - 5)

Câu 2:

x² - 4x = -4

\(\Leftrightarrow\) x² - 4x + 4 = 0

\(\Leftrightarrow\) (x - 2)² = 0

\(\Leftrightarrow\) x - 2 = 0

\(\Leftrightarrow\) x = 2

Vậy x = 2

Chúc bn học tốt!

b) x²-3x+xy-3y

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

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

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

c) x²-y²-4x+4

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

=\(\left(x-2\right)^2\) \(-y^2\)

=(\(x-y-2\)) \(\left(x+y-2\right)\)

a)6x³y-9x²y²+3xy

=\(3xy\left(2x^2-3xy+1\right)\)

1/

a, \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2+2x+1\right)\left(2x^2-2x+1\right)\)

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

c, \(x^4+324=x^4+36x^2+324-36x^2=\left(x^2+18\right)^2-\left(6x\right)^2=\left(x^2+6x+18\right)\left(x^2-6x+18\right)\)

2/

a, \(x^2+\frac{1}{3}x+\frac{1}{36}=\left(x+\frac{1}{6}\right)^2=\left(\frac{35}{6}+\frac{1}{6}\right)^2=6^2=36\)

b, \(x^2-y^2+2y-1=x^2-\left(y-1\right)^2=\left(x+y-1\right)\left(x-y+1\right)=\left(100+1-1\right)\left(100-1+1\right)=100.100=10000\)

\(A=x^2+3x+2=\left(x+1\right)\left(x+2\right)\)

\(B=x^2-4x-5=\left(x-5\right)\left(x+1\right)\)

\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)

\(A=x^2+3x+2=\left(x+1\right)\left(x+2\right)\)

\(B=x^2-4x-5=\left(x-5\right)\left(x+1\right)\)

\(C=3x^2+7x+4=\left(x+1\right)\left(3x+4\right)\)

A = x² + 3x + 2

A = x² + x + 2x + 2

A = x(x + 1) + 2(x + 1)

A = (x + 2)(x + 1)