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

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

^^

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

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

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

=x^2(x+1)^2

4: =x^2+6x-x-6

=(x+6)(x-1)

5: =-6x^2+3x+4x-2

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

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

6: =5x(x+y)-(x+y)

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

7: =2x^2+5x-2x-5

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

8: =(x^2-1)*(x^2-4)

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

9: =x^2(x-5)-9(x-5)

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

a) x⁴ + 2x³ + x²

= x² ( x² + 2x + 1 )

= x² ( x + 1 )² 

b) 5x² - 10xy + 5y² - 20z² 

= 5 [(x² - 2xy + y² ) - 4z² ] 

= 5 [( x - y )² - ( 2z )² ]

= 5 ( x - y - 2z ) ( x - y + 2z )

 c) x³ - x + 3x²y + 3xy²+ y³- y

= ( x³ + 3x²y + 3xy² + y³ ) - ( x + y )

= (x + y )³ - ( x + y)

= ( x + y ) [( x + y )² - 1 ]

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

1)

a) => 16x² - 8x + 1 - 8(2x² + 3x - 4x - 6) = 15

=> 16x² - 8x + 1 - 8(2x² - x - 6) = 15

=> 16x² - 8x + 1 - 16x² + 8x + 48 = 15

=> 49 = 15 (?) (vô lí)

=> Không tìm được x thoả mãn

b) (5x - 2)(x - 2) - 4(x - 3) = x² + 3

=> 5x² - 10x - 2x + 4 - 4x + 12 = x² + 3

=> 5x² - 16x + 16 = x² + 3

=> 4x² - 16x + 16 = 3

=> (2x)² - 2.2x.4 + 4² = 3

=> (2x - 4)² = 3

=> \(\left[{}\begin{matrix}2x-4=\sqrt{3}\\2x-4=-\sqrt{3}\end{matrix}\right.\)           \(\Rightarrow\left[{}\begin{matrix}x=\dfrac{4+\sqrt{3}}{2}\\x=\dfrac{4-\sqrt{3}}{2}\end{matrix}\right.\)

Mong bạn xem lại đề bài!

2) 

a) 5x² - 10xy + 5y² - 20z²

= 5(x² - 2xy + y² - 4z²)

= 5[(x - y)² - (2z)²]

= 5(x - y - 2z)(x - y + 2z)

b) a³ - ay - a²x + xy

= a(a² - y) - x(a² - y)

= (a - x)(a² - y)

c) 3x² - 6xy + 3y² - 12z²

= 3(x² - 2xy + y² - 4z²)

= 3[(x - y)² - (2z)²]

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

d) x² - 2xy + tx - 2ty

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

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

a,5x^2 - 10xy + 5y^2 - 20z^2
=5(x^2 -2xy +y^2-4z^2 )
=5[(x-y)^2-(2z)^2 ]
=5 .(x-y-2z)(x-y+2z)

b,.= (5x^2+5xy)-(x+y)
=5x(x+y)-(x+y)
=(x+y)(5x-1)

d,x² - 4x + 3 = x² - x - 3x + 3
= x(x - 1) - 3(x - 1) = (x -1)(x - 3)

e,x² - x - 6 = x² +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)

f,x² - x - 6 = x² +2x - 3x - 6
= x(x + 2) - 3(x + 2)
= (x + 2)(x - 3)

g,2x^2(3x - 5)
= 2x^2 x 3x - 2x^2 x 5
= 6x^3 - 10x^2

\(\text{1) }\)

\(\text{a) }5x^2-10xy+5y^2-20z^2\)

\(=5\left(x^2-2xy+y^2-4z^2\right)\)

\(=5\left[\left(x^2-2xy+y^2\right)-4z^2\right]\)

\(=5\left[\left(x-y\right)^2-\left(2z\right)^2\right]\)

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

\(\text{b) }5x^2+5xy-x-y\)

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

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

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

\(\text{c) }2\left(x+4\right)-x^2+16\)

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

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

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

\(=\left(x+4\right)\left(6-x\right)\)

\(\text{d) }x^2+4x+3\)

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

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

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

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

\(\text{e) }x^2+5x-6\)

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

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

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

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

a) 3x³ + 6x²y

= 3x².(x + 2y)

b) 2x³ - 6x²

= 2x².(x - 2)

c) 18x² - 20xy

= 2x.(9x - 10y)

d) xy + y² - x - y

= (xy + y²) - (x + y)

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

= (x + y)(y - 1)

e) (x²y² - 8)² - 1

= (x²y² - 8 - 1)(x²y² - 8 + 1)

= (x²y² - 9)(x²y² - 7)

= (xy - 3)(xy + 3)(x²y² - 7)

f) x² - 7x - 8

= x² - 8x + x - 8

= (x² - 8x) + (x - 8)

= x(x - 8) + (x - 8)

= (x - 8)(x + 1)

a: \(3x^3+6x^2y\)

\(=3x^2\cdot x+3x^2\cdot2y=3x^2\left(x+2y\right)\)

b: \(2x^3-6x^2=2x^2\cdot x-2x^2\cdot3=2x^2\left(x-3\right)\)

c: \(18x^2-20xy=2x\cdot9x-2x\cdot10y=2x\left(9x-10y\right)\)

d: \(xy+y^2-x-y\)

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

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

e: \(\left(x^2y^2-8\right)^2-1\)

\(=\left(x^2y^2-8-1\right)\left(x^2y^2-8+1\right)\)

\(=\left(x^2y^2-7\right)\left(x^2y^2-9\right)\)

\(=\left(x^2y^2-7\right)\left(xy-3\right)\left(xy+3\right)\)

f: \(x^2-7x-8\)

\(=x^2-8x+x-8\)

\(=x\left(x-8\right)+\left(x-8\right)=\left(x-8\right)\left(x+1\right)\)

g: \(10x^2\left(2x-y\right)+6xy\left(y-2x\right)\)

\(=2x\cdot\left(2x-y\right)\cdot5x-2x\cdot\left(2x-y\right)\cdot3y\)

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

h: \(x^2-2x+1-y^2\)

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

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

i: \(2x\left(x+2\right)+x^2\left(-x-2\right)\)

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

k: \(-x^2+6x-9=-\left(x^2-6x+9\right)\)

\(=-\left(x^2-2\cdot x\cdot3+3^2\right)=-\left(x-3\right)^2\)

l: \(-2x^2+8xy-8y^2\)

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

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

m: \(3x^2+5x-3y^2-5y\)

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

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

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

g) 10x²(2x - y) + 6xy(y - 2x)

= 10x²(2x - y) - 6xy(2x - y)

= 2x(2x - y)(5x - 3y)

h) x² - 2x + 1 - y²

= (x² - 2x + 1) - y²

= (x - 1)² - y²

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

i) 2x(x + 2) + x² (-x - 2)

= 2x(x + 2) - x²(x + 2)

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

k) -9 + 6x - x²

= -(x² - 6x + 9)

= -(x - 3)²

l) 8xy - 2x² - 8y²

= -2(x² - 4xy + 4y²)

= -2(x - 2y)²

m) 3x² + 5x - 3y² - 5y

= (3x² - 3y²) + (5x - 5y)

= 3(x² - y²) + 5(x - y)

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

= (x - y)[3(x + y) + 5]

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

a, x⁴ + 2x³ +x² = x⁴ +x³ +x³ +x²  =(x⁴+x³ )+(x³ +x² ) =x³(x +1 ) + x² (x+1 ) =(x+1)(x³+x²⁾

a) x⁴ + 2x³ + x²

= x²(x² + 2x + 1)

= x²(x + 1)²

= [x(x + 1)]²

= (x² + x)²

b) 5x³ - 10xy + 5y² - 20z²

= 5(x³ - 2xy + y² - 4z²)

c) x²y - xy² + x³ - y³

= xy(x - y) + (x - y)(x² + xy + y²)

= (x - y)(x² + 2xy + y²)

= (x - y)(x + y)²

d) x² - xy + 4x - 2y  + 4

= (x² + 4x + 4) - (xy + 2y)

= (x + 2)² - y(x + 2)

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

d) x² - x - 6

= x² - 3x + 2x - 6

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

= (x + 2)(x - 3)

f) 3x² - 5x - 8

= 3x² + 3x - 8x - 8

= 3x(x + 1) - 8(x + 1)

= (3x - 8)(x + 1)

g) x³ + 3x² + 6x + 4

= (x³ + 3x² + 3x + 1) + (3x + 3)

= (x + 1)³ + 3(x + 1)

= (x + 1)[(x + 1)² + 3]

h) 3x³ - 5x² - 6x + 8

= 3x³ - 3x² - 2x² - 6x + 8

= 3x³ - 3x² - 2x² + 2x - 8x + 8

= 3x²(x - 1) - 2x(x - 1) - 8(x - 1)

= (3x² - 2x - 8)(x - 1)