Bài 1: Thực hiện các phép tính sau:
3x2(5x2 – 2x – 4)



(6x2+5y2)(2x2– y2)



(2x - y)(4x2 - 2xy + y2)
(6x5y2 - 9x4y3 + 15x3y4): 3x3y2
(2x3 - 21x2 + 67x - 60): (x - 5)

(x + y)2 - (x - y)2
(a + b)3 + (a - b)3 - 2a3
98.28 - (184 - 1)(184 + 1)

(x + 3)2 – (2x – 5)(x+ 3).
(3x + 2)2 +(3x-2)2 – 2 (9x2 – 4)
(3x + 4x2 ( 2)( (x2 +1 + 2x)



(x +3)(x2 –3x + 9) –(54 + x3)




(2x+y)(4x2–2xy+y2) –(2x –y)(4x2+2xy + y2)

(18x4y3 – 24x3y4 +6x2y5) : 6x2y3
(15x3y2 – 5x2y3 + 10xy4) : 5xy2

[3(x – y)5 – 2(x – y)4 + 3(y – x)2] : 5(x – y)2
(x4 + 2x2y2 + y4) : (x2 + y2)

(4x2 – 9y2) : (2x – 3y)
 (27x3 – 1) : (3x – 1)
(8x3 + 1) : (4x2 – 2x + 1)

(x4 + 2x3 +x - 25):(x2 +5)
(27x3 - 8): (6x + 9x2 + 4)
(x2 – 3x + xy – 3y) : (x + y)

(15+5x2–3x3 – 9x) : (5 – 3x)





(x4 (2x3 +4x2 (8x) : (x2 + 4)




Bài 3: Chứng minh biểu thức sau không phụ thuộc vào biến x,y
A= (3x - 5)(2x + 11) - (2x + 3)(3x + 7)
B = (2x + 3)(4x2 - 6x + 9) - 2(4x3 - 1)

C = (x - 1)3 - (x + 1)3 + 6(x + 1)(x - 1)


Bài 4: Phân tích các đa thức sau thành nhân tử:
A. x2 - y2 - 2x + 2y
2x + 2y - x2 - xy
3a2 - 6ab + 3b2 - 12c2

x2 - 25 + y2 + 2xy
a2 + 2ab + b2 - ac - bc
x2 - 2x - 4y2 - 4y

x2y - x3 - 9y + 9x
x2(x-1) + 16(1- x)
81x2 - 6yz - 9y2 - z2

xz-yz-x2+2xy-y2
x2 + 8x + 15
x2 - x - 12

81x4 + 4
x3 – 4x2 + 4x
x3 – 4x

2x2 + 3x- 2xy – 3y
x2 - 8x +7.
16 – 24y + 9y2

15x2y + 20xy2 ( 25xy
(x + y)2 ( 25
1 ( 2y + y2

4x2 + 8xy ( 3x ( 6y
27 + 27x + 9x2 + x3
1 ( 4x2

8 ( 27x3
16x3 + 54y3
64+x4

x2 - 8x +7
2x4+8y4



2x2 + 2y2 ( x2z + z ( y2z ( 2
3x2 ( 6xy + 3y2
 x2 ( 2xy + y2 ( 16

x6 ( x4 + 2x3 + 2x
x2 + 2x + 1 – y2
x2 + 2xy + y2 – 9z2

x3 - 10x2 + 25x - 16xy2
 3xy2– 2xy +12x



x2 + 2xy + y2 – xz – yz



x2 + 2x + 1
9x2 + y2 + 6xy
25a2 + 4b2 – 20ab

– x3 + 3x2 – 3x + 1
8 – 12x + 6x2 – x3
4 + 4xy + x2y2

125x3 – 75x2 + 15x - 1
8x3 – y3
x3 + 27

 x2 – x +

25 – 30x + 9x2
25 – x4

Bài 5: Tìm x biết:
2x(x-5)-x(3+2x)=26
5x(x-1) = x-1
2(x+5) - x2-5x = 0

(2x-3)2-(x+5)2=0
3x3 - 48x = 0
x3 + x2 - 4x = 4

36x2- 49 =0
x3-16x =0
(2x – 3)2 – (x + 5)2 = 0

(