`@` `\text {Ans}`

`\downarrow`

`x^2 + 4x + 3`

`= x^2 + 3x + x + 3`

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

`= x(x + 3) + (x + 3)`

`= (x+1)(x+3)`

____

`2x^2 + 3x - 5`

`= 2x^2 + 5x - 2x - 5`

`= (2x^2 - 2x) + (5x - 5)`

`= 2x(x - 1) + 5(x - 1)`

`= (2x + 5)(x - 1)`

____

`16x - 5x^2 - 3`

`= 15x + x - 5x^2 - 3`

`= (15x - 5x^2) + (x - 3)`

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

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

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

\(x^2+4x+3=x^2+3x+x+3=x\left(x+3\right)+\left(x+3\right)=\left(x+1\right)\left(x+3\right)\\----\\ 2x^2+3x-5=2x^2-2x+5x-5=2x\left(x-1\right)+5\left(x-1\right)=\left(2x+5\right)\left(x-1\right)\\ ----\\ 16x-5x^2-3=-5x^2+15x+x-3=-5x\left(x+3\right)-\left(x+3\right)=-\left(5x+1\right)\left(x+3\right)\)

\(x^2+4x+3=\left(x+1\right)\left(x+3\right)\\ 2x^2+3x-5=\left(2x+5\right)\left(x-1\right)\\ 16x-5x^2-3=\left(1-5x\right)\left(x-3\right)\)

nhiều quá, các bn ngại làm, chia nhỏ ra,mk làm cho 2 câu 

a) x² +5x -6 = x² -x +x + 5x -6

= x² -x +6x -6

= x( x-1) + 6(x-1) = (x-1)(x+6)

b) 5x² +5xy -x-y = 5x(x+y) -(x+y)

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

e) \(2x^2+3x-5\)

\(=2x^2+5x-2x-5\)

\(=x\cdot\left(2x+5\right)-\left(2x+5\right)\)

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

f) \(16x-5x^2-3\)

\(=-5x^2+16x-3\)

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

\(=-5\cdot\left(x-3\right)+x-3\)

\(=\left(-5x+1\right)\left(x-3\right)\)

c)7x-\(6x^2\)-2

=3x + 4x - \(6x^2\) - 2

=(3x - \(6x^2\)) - (2 - 4x)

= 3x(1 - 2x) - 2(1 - 2x)

=(1-2x)(3x-2)

a, 5x^2 +5xy - x - y 

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

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

b, 2x^2  + 3x - 5 

= 2x^2 - 2x + 5x - 5 

= 2x( x - 1) + 5( x - 1)

= ( 2x + 5 )(x- 1 )

c; 16x - 5x^2 - 3 

c, = - ( 5x^2 - 16x + 3 )

   = - ( 5x^2 - x - 15x + 3 )

= - [ x(5x - 1 ) - 3 (5x - 1) ]

= - ( x- 3)(5x -  1 )

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

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

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

\(b,16x^2y-4xy^2-4x^3\)

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

\(=-4x\left(x-2y+y\sqrt{3}\right)\left(x-2y-y\sqrt{3}\right)\)

a.5x²-10xy+5y²-20z²

  =5(x²-2xy+y²-4z²)

  =5[ (x²-2xy+y²)-(2z)² ]

  =5[ (x-y)²-(2z)² ]

  =5(x-y-2z)(x-y+2z)

b.16x-5x²-3

  =15x+x-5x²-3

  =(15x-3)+(x-5x²)

  =3(5x-1)+x(1-5x)

  =3(5x-1)-x(5x-1)

  =(5x-1)(3-x)

c.x²-5x+5y-y²

  =(5y-5x)+(x²-y²)

  =5(y-x)+(x-y)(x+y)

  =5(y-x)-(y-x)(y+x)

  =(y-x)[5-(y+x)]

  =(y-x)(5-y-x)

d.3x²-6xy+3y²-12z²     (câu này hình như ở trên đề bạn ghi sai nha! Mình sửa lại luôn rồi đó)

=3(x²-2xy+y²-4z²)

=3[ (x²-2xy+y²)-(2z)² ]

=3[ (x-y)²-(2z)² ]

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

e.x²+4x+3

=x²+3x+x+3

=(x²+x)+(3x+3)

=x(x+1)+3(x+1)

=(x+1)(x+3)

f.(x²+1)²-4x²

=(x²+1)²-(2x)²

=(x²+1-2x)(x²+1+2x)

h.x²-4x-5

=x²-5x+x-5

=(x²+x)+(-5x-5)

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

-(x+1)(x-5)

a)x^2-(a+b)x+ab

= x^2 - ax - bx + ab

= (x^2 - ax) - (bx - ab)

= x(x-a) - b(x-a)

= (x-b)(x-a) 

b)7x^3-3xyz-21x^2+9z

= 

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

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

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

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

d) y^2+y-x^2+x

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

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

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

e)4x^2-2x-y^2-y

= [(2x)^2 - y^2] - (2x +y)

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

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

f)9x^2-25y^2-6x+10y

= 

ko biết làm

a) x³ - 2x -4=x³ - 2x² + 2x² - 4x +2x -4

=x²(x-2) + 2x(x-2)+2(x-2)

=(x-2)(x² +2x +2)

b) x² + 4x +3

=x² + 2.x.2 +2² -1

=(x+2)² - 1²

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

=(x+3)(x+1)

d, 2x²+7x+5=2x²+2x+5x+5=2x(x+1)+5(x+1)=(2x+5)(x+1)

e, 2x²+3x-5=2x²-2x+5x-5=2x(x-1)+5(x-1)=(2x+5)(x-1)

bạn viết lại đề đi, có số mũ, xuống dòng chứ thế này ai mà giải được

Giải như sau.

(1)+(2)⇔x2−2x+1+√x2−2x+5=y2+√y2+4⇔(x2−2x+5)+√x2−2x+5=y2+4+√y2+4⇔√y2+4=√x2−2x+5⇒x=3y(1)+(2)⇔x2−2x+1+x2−2x+5=y2+y2+4⇔(x2−2x+5)+x2−2x+5=y2+4+y2+4⇔y2+4=x2−2x+5⇒x=3y

⇔√y2+4=√x2−2x+5⇔y2+4=x2−2x+5, chỗ này do hàm số f(x)=t2+tf(x)=t2+t đồng biến ∀t≥0∀t≥0
Công việc còn lại là của bạn ! 

\(\left(x+6\right)\left(2x+1\right)=0\)

<=>  \(\orbr{\begin{cases}x+6=0\\2x+1=0\end{cases}}\)

<=>  \(\orbr{\begin{cases}x=-6\\x=-\frac{1}{2}\end{cases}}\)

Vậy....

hk tốt

^^