Bài 2: 

a: =>4x(x+5)=0

=>x=0 hoặc x=-5

b: =>(x+3)(x-3)=0

=>x=-3 hoặc x=3

a: \(x^4+3x^3+x^2+3x\)

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

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

c: \(x^2-xy-x+y\)

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

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

a: \(=2x^3-14x^2-6x\)

c: \(=-10x^5-15x^4+25x^3\)

a) 2x. (x2 – 7x -3)

= 2x³- 14x²- 6x

b) ( -2x3 + y2 -7xy). 4xy2 

= -8x⁴y²+ 4xy⁴- 28x²y³

c)(-5x3).(2x2+3x-5)

= -10x⁵-15x⁴+25x³

d) (2x2 - xy+ y2).(-3x3)

=-6x⁵+ 3x⁴y -3x³y²

e)(x2 -2x+3). (x-4) 

=x³-2x²+3x -4x²+8x-12

=x³-6x²+11x-12

f) ( 2x3 -3x -1). (5x+2)

=10x⁴-15x²-5x +4x³-6x-2

=10x⁴+4x³-15x²-11x-2

a,3x(5x^2-2x-1)

=15x^3-6x^2-3x

a) \(3x\left(5x^2-2x-1\right)\)

\(=15x^3-6x^2-3x\)

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

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

c) \(\dfrac{1}{2}x^2y\left(2x^3-\dfrac{2}{5}xy^2-1\right)\)

\(=x^5y-\dfrac{1}{5}x^3y^3-\dfrac{1}{2}x^2y\)

a) 3x(5x²-2x-1) = 15x³ - 6x² - 3x

b) (x²+2xy-3)(-xy) = -x³y - 2x²y² + 3xy

c) \(\dfrac{1}{2}\)x²y\((\)2x³ - \(\dfrac{2}{5}\)xy² - 1) = x⁵y - \(\dfrac{1}{5}\)x³y³ - \(\dfrac{1}{2}\)x²y

1.

7x(2x-1)=14x²-7x

2

a. x²+2x=x(x+2)

b.x²-xy+3x-3y

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

=(x+3)(x-y)

Câu 2:

 1. 2x/2x-5 -  5/2x-5

=2x-5/2x-5

=1

2. (6x³-7x²-x+2) : (x-1)=6x²-x-2

1: =(x+y-3x)(x+y+3x)

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

2: =(3x-1-4)(3x-1+4)

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

=3(x+1)(3x-5)

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

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

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

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

4: =(2x+1+x-1)(2x+1-x+1)

=3x(x+2)

5: =[(x+1)^2-(x-1)^2][(x+1)^2+(x-1)^2]

=(2x^2+2)*4x

=8x(x^2+1)

6: =(5x-5y)^2-(4x+4y)^2

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

=(x-9y)(9x-y)

7: =(x^2+xy+y^2+xy)(x^2+xy-y^2-xy)

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

=(x+y)^3*(x-y)

8: =(x^2+4y^2-20-4xy+16)(x^2+4y^2-20+4xy-16)

=[(x-2y)^2-4][(x+2y)^2-36]

=(x-2y-2)(x-2y+2)(x+2y-6)(x+2y+6)

1)\(y=\frac{x^2+3x+7}{x+3}=\frac{x\left(x+3\right)+7}{x+3}=x+\frac{7}{x+3}\)= > x +3 thuoc\(U_{\left(7\right)}=\left\{1;-1;7;-7\right\}\)

                                                                                                                  x thuoc \(\left\{-2;-4;3;-11\right\}\)

 2)\(y=\frac{4x+3}{2x+6}=\frac{4x+12-8}{2x+6}=\frac{2\left(2x+6\right)-8}{2x+6}=2-\frac{8}{2x+6}\)   =>2x+6 thuoc 

\(U_{\left(8\right)}=\left\{1;-1;2;-2;4;-4;8;-8\right\}\) 

=>x thuoc \(\left\{-2;-4;-1;-5;1;-7\right\}\)

4)\(y=\frac{4x+1}{3x-1}\)

\(3y=\frac{12x+3}{3x-1}=\frac{12x-4+7}{3x-1}=\frac{4\left(3x-1\right)+7}{3x-1}=4+\frac{7}{3x-1}\)

3x+1 thuoc {1;-1;7;-7}

3x thuoc {0;-2;6;-8}

x thuoc {0;2}

thanks 

a,(x+3)(x²+3x-5)

=x³+3x²-5x+3x²+9x-15

=x³+3x²+3x²-5x+9x-15

=x³+6x²+4x-15

b,(xy-1)(xy+5)

=x²y²+5xy-xy-5

=x²y²+4xy-5

c,(x²y²-1/2xy+2y)

Đa thức này đã được rút ngọn

d,(x²-xy+y²)(x+y)

=x³-x²y+xy²+x²y-xy²+y³

=x³-x²y+x²y+xy²-xy²+y³

=x³+y³

Nhớ k mk nhé , thanks !

\(a.\left(x+3\right)\cdot\left(x^2+3x-5\right)=x^3+3x^2-5x+3x^2+9x-15=x^3+6x^2+4x-15\)

\(b.\left(xy-1\right)\cdot\left(xy+5\right)=x^2y^2+5xy-xy-5=x^2y^2+4xy-5\)

\(c.\left(x^2y^2-\frac{1}{2}xy+2y\right)\)

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

(ý c hình như thiếu bạn ạ)

a,(x+3)(x²+3x-5)
=x³+3x²-5x+3x²+9x-15
=x³+6x²+4x-15
b, (xy-1)(xy+5)
=x²y²+5xy-xy-5
=x²y²+4xy-5
c, pn viết nguyên lại đề bài nhé
d, (x²-xy+y²)(x+y)
=x³+x^2y-x²y-xy²+xy²+y²
=x³+y³