a) (x - y)(x + y + 3).                    b) (x + y - 2xy)(2 + y + 2xy).

c) x 2 (x + l)( x 3  -  x 2  + 2).              d) (x – 1 - y)[ ( x   -   1 ) 2   +   ( x   -   1 ) y   +   y 2 ].

\(a.2x^3+6x=2x\left(x^2+3\right)\)

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

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

\(b.5x\left(x-2\right)-3x^2\left(x-2\right)\)

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

\(c.3x\left(x-5y\right)-2y\left(5y-x\right)\)

\(=3x\left(x-5y\right)+2\left(x-5y\right)\)

\(=\left(x-5y\right)\left(3x+2\right)\)

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

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

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

e. Cái bài này ghi lại đàng hoàng xíu nha t k hỉu

\(f.3x^2\left(y^2-2x\right)-15x\left(2x-y^2\right)\)

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

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

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

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) x³+4x-5 = x³-x²+x²+4x-5=(x³-x²)+(x²-x)+(5x-5)=x²(x-1)+x(x-1)+5(x-1)=(x²+x+5)(x-1)

b) x³-3x²+4=x³-2x²-x²+4=(x³-2x²)-(x²-4)=x²(x-2)-(x-2)(x+2)=(x²-x+2)(x-2)

c) x³+2x²+3x+2=x³+x²+x²+x+2x+2=(x³+x²)+(x²+x)+(2x+2)=x²(x+1)+x(x+1)+2(x+1)=(x²+x+2)(x+1)

d) bạn xem lại đề đúng ko

e) (x²+3x)²-2(x²+3x)-8=x⁴+6x³+9x²-2x²-6x-8=x⁴+6x³+7x²-6x-8=x⁴-x³+7x³-7x²+14x²-14x+8x-8=(x⁴-x³)+(7x³-7x²)+(14x²-14x)+(8x-8)=x³(x-1)+7x²(x-1)+14x(x-1)+8(x-1)=(x³+7x²+14x+8)(x-1)=(x³+x²+6x²+6x+8x+8)(x-1)=\(\left[\left(x^3+x^2\right)+\left(6x^2+6x\right)+\left(8x+8\right)\right]\left(x-1\right)\)\(=\left[x^2\left(x+1\right)+6x\left(x+1\right)+8\left(x+1\right)\right]\left(x-1\right)\)\(=\left(x^2+6x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left(x^2+2x+4x+8\right)\left(x+1\right)\left(x-1\right)\)\(=\left[\left(x^2+2x\right)+\left(4x+8\right)\right]\left(x+1\right)\left(x-1\right)\)\(=\left[x\left(x+2\right)+4\left(x+2\right)\right]\left(x+1\right)\left(x-1\right)\)=\(\left(x-1\right)\left(x+1\right)\left(x+2\right)\left(x+4\right)\)

f) (x²+4x+10)²-7(x²+4x+11)+7=(x²+4x+10)²-\(\left[7\left(x^2+4x+11\right)-7\right]\)\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)\)\(=\left(x^2+4x+10\right)\left(x^2+4x+3\right)\)

a) Ta có: \(x^3+4x-5\)

\(=x^3-x+5x-5\)

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

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

b) Ta có: \(x^3-3x^2+4\)

\(=x^3+x^2-4x^2+4\)

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

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

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

c) Ta có: \(x^3+2x^2+3x+2\)

\(=x^3+x^2+x^2+x+2x+2\)

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

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

d) Ta có: \(x^2+2xy+y^2+2x+2y-3\)

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

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

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

\(=\left(x+y+3\right)\left(x+y-1\right)\)

e) Ta có: \(\left(x^2+3x\right)^2-2\left(x^2+3x\right)-8\)

\(=\left(x^2+3x\right)^2-4\left(x^2+3x\right)+2\left(x^2+3x\right)-8\)

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

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

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

f) Ta có: \(\left(x^2+4x+10\right)^2-7\left(x^2+4x+11\right)+7\)

\(=\left(x^2+4x+10\right)^2-7\left(x^2+4x+10\right)-7+7\)

\(=\left(x^2+4x+10\right)\left(x^2+4x+10-7\right)\)

\(=\left(x^2+4x+3\right)\left(x^2+4x+10\right)\)

\(=\left(x+1\right)\left(x+3\right)\left(x^2+4x+10\right)\)

a: Ta có: \(x^2-36y^2-x+6y\)

\(=\left(x-6y\right)\left(x+6y\right)-\left(x-6y\right)\)

\(=\left(x-6y\right)\left(x+6y-1\right)\)

b: Ta có: \(16x-8x^2+x^3\)

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

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

c: Ta có: \(2x^2-4xy+2y^2-18\)

\(=2\left(x^2-2xy+y^2-9\right)\)

\(=2\cdot\left[\left(x-y\right)^2-9\right]\)

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

d: Ta có: \(3x^2-7x-10\)

\(=3x^2+3x-10x-10\)

\(=3x\left(x+1\right)-10\left(x+1\right)\)

\(=\left(x+1\right)\left(3x-10\right)\)

e: Ta có: \(x^4-x^2-30\)

\(=x^4-6x^2+5x^2-30\)

\(=x^2\left(x^2-6\right)+5\left(x^2-6\right)\)

\(=\left(x^2-6\right)\left(x^2+5\right)\)

f: Ta có: \(x^2-xy-2y^2\)

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

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

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

g: Ta có: \(x^4-13x^2y^2+4y^4\)

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

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

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

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

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

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

h: Ta có: \(\left(x^2-2x\right)^2-2\left(x^2-2x\right)-3\)

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

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

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

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

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:

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

b, (\(x^2\) - \(xy\) ) + (\(x-y\))

= (\(x-y\)).\(x\) + (\(x-y\))

= (\(x-y\)).(\(x\) + 1)

c, \(x^2\) - 2\(x\) - 15

= (\(x^2\) - 2\(x\) + 1) - 16

= (\(x\) - 1)² - 4²

= (\(x-1-4\)).(\(x-1+4\))

= (\(x-5\)).(\(x+3\))

d, 2\(x^2\) + 3\(x\) - 5

= 2\(x^2\) - 2 + 3\(x\) - 3

= 2.(\(x^2\) - 1) + 3.(\(x-1\))

= 2.(\(x-1\)).(\(x\) + 1) + 3.(\(x-1\))

= (\(x-1\)).(2\(x\) + 2  + 3)

= (\(x\) -1).(2\(x\) + 5)

a) x³-2x²-x+2

=x(x²-1)+2(-x²+1)

=x(x²-1)-2(x²-1)

=(x²-1)(x-2)

b)

x²+6x-y²+9

=x²+6x+9-y²

=(x+3)²-y²

^{=(x+3-y)(x+3+y)}

a) \(x^2-y^2-3x+3y\)

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

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

b) \(2x+2y-x^2+y^2\)

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

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

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

c) \(x^2-16+y^2+2xy\)

\(=x^2+y^2+2xy-16\)

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

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

a) \(x^2-y^2-3x+3y\)

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

b) \(2x+2y-x^2+y^2\)

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

c) \(x^2-16+y^2+2xy\)

\(=\left(x+y\right)\left(x-y\right)+2xy-16\)