mn giúp e ik mn

\(a\text{)}x^2y+xy^2=xy\left(x+y\right)\)

\(b\text{)}x^2-2x+1=\left(x-1\right)^2\)

\(c\text{)}x^2-5x+4=\left(x-1\right)\left(x-4\right)\)

Bài 1: 

a: \(=6x^3-10x^2\)

b: \(=6x+5\)

Bài 1: 

a: \(=6x^3-10x^2+6x\)

b: \(=-2x^3-10x^2-6x\)

Bài 4: 

a: =>3x+10-2x=0

=>x=-10

c: =>3x²-3x²+6x=36

=>6x=36

hay x=6

Bài 1:

\(a,=6x^3-10x^2+6x\\ b,=-2x^3-10x^2-6x\)

Bài 4:

\(a,\Leftrightarrow3x+10-2x=0\Leftrightarrow x=-10\\ b,\Leftrightarrow x\left(2x^2+9x-5\right)-\left(2x^3+9x^2+x+4,5\right)=3,5\\ \Leftrightarrow2x^3+9x^2-5x-2x^3-9x^2-x-4,5=3,5\\ \Leftrightarrow-6x=8\Leftrightarrow x=-\dfrac{4}{3}\\ c,\Leftrightarrow3x^2-3x^2+6x=36\Leftrightarrow x=6\)

Bài 1:

\(a,=7xy\left(2x-3y+4xy\right)\\ b,=x\left(x+y\right)-5\left(x+y\right)=\left(x-5\right)\left(x+y\right)\\ c,=\left(x-y\right)\left(10x+8\right)=2\left(5x+4\right)\left(x-y\right)\\ d,=\left(3x+1-x-1\right)\left(3x+1+x+1\right)\\ =2x\left(4x+2\right)=4x\left(2x+1\right)\\ e,=5\left[\left(x-y\right)^2-4z^2\right]=5\left(x-y-2z\right)\left(x-y+2z\right)\\ f,=x^2+8x-x-8=\left(x+8\right)\left(x-1\right)\\ g,\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left[\left(x+y\right)^2-1\right]\\ =\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\\ h,=x^2+3x+x+3=\left(x+3\right)\left(x+1\right)\)

Bài 2:

a: =x(x^2-25)

=x(x-5)(x+5)

b: =x(x-2y)+3(x-2y)

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

c: =(2x-3)(4x^2+6x+9)+2x(2x-3)

=(2x-3)(4x^2+8x+9)

\(1,\\ a,=6x^4-15x^3-12x^2\\ b,=x^2+2x+1+x^2+x-3-4x=2x^2-x-2\\ c,=2x^2-3xy+4y^2\\ 2,\\ a,=7x\left(x+2y\right)\\ b,=3\left(x+4\right)-x\left(x+4\right)=\left(3-x\right)\left(x+4\right)\\ c,=\left(x-y\right)^2-z^2=\left(x-y-z\right)\left(x-y+z\right)\\ d,=x^2-5x+3x-15=\left(x-5\right)\left(x+3\right)\\ 3,\\ a,\Leftrightarrow3x\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x+2\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-2\end{matrix}\right.\)

Câu 1

a)\(3x^2\left(2x^2-5x-4\right)=6x^4-15x^3-12x^2\)

b)\(\left(x+1\right)^2+\left(x-2\right)\left(x+3\right)-4x=x^2+2x+1+x^2+3x-2x-6-4x=2x^2-x-5\)

Bài 2

a) \(7x^2+14xy=7x\left(x+2y\right)\)

b) \(3x+12-\left(x^2+4x\right)=-x^2-x+12=\left(-x+3\right)\left(x+4\right)\)

c) \(x^2-2xy+y^2=\left(x-y\right)^2\)

d) \(x^2-2x-15=x^2+3x-5x-15=\left(x+3\right)\left(x-5\right)\)

a. 

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

b.

$x^2(x-1)+16(1-x)=x^2(x-1)-16(x-1)=(x-1)(x^2-16)=(x-1)(x-4)(x+4)$

c.

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

d.

$x^3-3x^2-3x+1=(x^3+1)-(3x^2+3x)=(x+1)(x^2-x+1)-3x(x+1)$

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

e.

$x^4+4y^4=(x^2)^2+(2y^2)^2+2.x^2.2y^2-4x^2y^2$

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

f.

$x^4-13x^2+36=(x^4-4x^2)-(9x^2-36)$

$=x^2(x^2-4)-9(x^2-4)=(x^2-9)(x^2-4)=(x-3)(x+3)(x-2)(x+2)$

g.

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

$=(x^2+x)^2-2(x^2+x)+6(x^2+x)-12$

$=(x^2+x)(x^2+x-2)+6(x^2+x-2)=(x^2+x-2)(x^2+x+6)$

$=[x(x-1)+2(x-1)](x^2+x+6)=(x-1)(x+2)(x^2+x+6)$

h.

$x^6+2x^5+x^4-2x^3-2x^2+1$

$=(x^6+2x^5+x^4)-(2x^3+2x^2)+1$

$=(x^3+x^2)^2-2(x^3+x^2)+1=(x^3+x^2-1)^2$

\(2x^2-7x+5=\left(2x^2-2x\right)-\left(5x-5\right)=2x\left(x-1\right)-5\left(x-1\right)=\left(2x-5\right)\left(x-1\right)\)

\(3x^2+5x+2=\left(3x^2+3x\right)+\left(2x+2\right)=3x\left(x+1\right)+2\left(x+1\right)=\left(3x+2\right)\left(x+1\right)\)

a: \(2x^2-7x+5=\left(x-1\right)\left(2x-5\right)\)

b: \(3x^2+5x+2=\left(x+1\right)\left(3x+2\right)\)

Câu 3: 

a: 2x-8=4

nên 2x=12

hay x=6

b: 7x-3x=2x+7

\(\Leftrightarrow4x-2x=7\)

hay \(x=\dfrac{7}{2}\)

Câu 1:

a: \(5x\left(3x-4\right)=15x^2-20x\)

b: \(\left(x+5\right)\left(x-5\right)=x^2-25\)

c) \(5x^2.\left(3x^2-7x+2\right)\)

\(=5x^2.3x^2-5x^2.7x+5x^2.2\)

\(=15x^4-35x^3+10x^2\)

d) \(\dfrac{3x^2y^2+6x^2y^3-12xy}{3xy}\)

\(=\dfrac{3xy\left(xy+2y-4\right)}{3xy}\)

\(=xy+2y-4\)

a) 5(x-2y)

b) 2x(2x-1)

c) 2x(x-3y^2-4x)

d) 2(x-y)-3x(x-y)= (x-y)(2-3x)

\(a,=5\left(x-2y\right)\\ b,=2x\left(2x-1\right)\\ c,=2x\left(x-3y^2-4x\right)\\ d,=\left(x-y\right)\left(2-3x\right)\)