De cuong on tap va bo de kiem tra chuong I - Dai 8 (09-10)

ĐỀ CƯƠNG ÔN TẬP CHƯƠNG I THEO CHỦ ĐỀ – ĐẠI 8

Thu gọn:
2x(3x2y + 4xy2 – 3)
x2 – 4 – (x + 2)2
(x + 2)(x – 2) – (x – 3)(x + 1)
( x – 2 ) ( x + 2 ) – ( x – 2 ) ( x + 5 )
(6x + 1)2 + (6x – 1)2 – 2(6x + 1)(6x – 1 )



(2x + 1)2 + 2(4x2 – 1) + (2x – 1)2

Chứng tỏ biểu thức k0 phụ thuộc biến:
(2x +5)
- 30x (2x+5) -8x

(3x+1)2 + 12x – (3x+5)2 + 2(6x+3)
(x-3)(x+3) - 2x(x-7) + x2 - 14x + 19
(3x– 2)(9x2 + 6x + 4) – 3(9x3 – 2)
(3x + 5)2 + (6x + 10)(2 – 3x) + (2 – 3x)2
(x + 4) (x – 4) – 2x(3 + x) + (x + 3)2
(x2 + 4) (x + 2) (x - 2) - ( x2 + 3) (x2 - 3)
(x – 1)3- (x + 1)3+ 6(x + 1) (x – 1)
(3x – 5)(2x + 11) – (2x +3)(3x + 7)
(2x + 3)(4x2 – 6x + 9) – 2(4x3 – 1)


Phân tích đa thức thành nhân tử:
5x – 15y
5x2y2 + 15x2y + 30xy2
x3 – 2x2y + xy2 – 9x
y – x2y – 2xy2 – y3
x(x2 – 1) + 3(x2 – 1)
x
- 10x + 25
x
- 64






2xy – x2 –y2 + 16.
(x - 2)(x – 3) + (x – 2) – 1.


12y ( 2x-5 ) + 6xy ( 5- 2x)
ax – 2x – a2 + 2a
x
- 4(x
+5)- 25
a2 – b2 – 2a + 1

x2 – 2xy + y2 – xy + yz
x2 + 4x - y2 + 4
x4 - 1

a2 + 2ab + b2 – ac - bc





Tìm x:
(x – 2)2 – (x + 1)(x + 3) = -7
(3x + 5)(4 – 3x) = 0
3x(x – 7) – 2(x – 7) = 0
8x2 – 8x + 2 = 0
7x2 – 28 = 0
3x3 – 27x = 0


9x3 + 9 = 0
x3 - 9x = 0
(2x + 1) + x(2x + 1) = 0
2x3 – 50x = 0


9(3x - 2) = x( 2 - 3x)

25x2 – 2 = 0
x2 – 25 = 6x - 9
(2x – 1)2 – (2x + 5)(2x – 5) = 18
5x (x – 3) – 2x + 6 = 0


(2x + 3)2 – (x – 1)2 = 0
x
- 8 = (x - 2)



x3 – 4x2 + 4x = 0



25x2 – 10x + 1 = x2 + 4x +

Phân tích bằng pp thêm bớt hạng tử:
x4 + 4
x4 + 64
64x4 + 1
81x4 + 4
4x4 + 81
64x4 + y4
x4 + 4y4
x4 + x2 + 1
a3 + b3 + c3 -3abc
x5 + x4 + 1
x5 + x + 1

Phân tích bằng pp tách hạng tử:
x2 - 5x + 6
x2 + 3x - 18
x2 - 5x - 24
x2 + 8x + 7
x2 - 13x + 36
x2 - 7x + 12
4x2 - 3x - 1
2x2 - 5x - 12
3x2 - 16x + 5
5x2 - 8x - 4
7x2 + 4x - 3


Tính nhẩm:
20042 - 16
8922 + 892. 216 + 1082
362 + 262 – 52 . 36
37. 43
1132 + 252 - 132 - 1252
20,03 . 45 + 20,03 . 47 + 20,03 . 8
10,2. 9,8 – 9,8. 0,2 + 10,22 –10,