10 câu Trắc nghiệm Toán 8 Bài 7: Phân tích đa thức thành nhân tử bằng phương pháp dùng hằng đẳng thức có đáp án (Nhận biết)

(x + 2y)³

(2x + y)³

(2x – y)³

(8x + y)³

(5x – 4)² – 49x² = -8(3x + 1)(x + 2)

(5x – 4)² – 49x² = (3x – 1)(x + 2)

(5x – 4)² – 49x² = -8(3x - 1)(x - 2)

(5x – 4)² – 49x² = -8(3x - 1)(x + 2)

(3x – 2y)² – (2x – 3y)² = 5(x – y)(x + y)

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

(3x – 2y)² – (2x – 3y)² = (x – y)(x + y)

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

4x² + 4x + 1 = (2x + 1)²

9x² – 24xy + 16y² = (3x – 4y)²

$\frac{x^2}{4}+2xy+4y^2=(\frac{x}{2}+2y{)}^2$

$\frac{x^2}{4}+2xy+4y^2=(\frac{x}{4}+2y{)}^2$

x² – 6x + 9 = (x – 3)²

4x² – 4xy + y² = (2x – y)²

$x^2+x+\frac{1}{4}=(x+\frac{1}{2}{)}^2$

-x² – 2xy – y² = -(x – y)²

(a – 3)²(a + 3)²

(a + 3)⁴

(a² + 36a + 9)(a² – 36a + 9)

(a² + 9)²

2x² + 8x + 8

2x² + 8x + 16

4x² – 8x+ 16

4x² + 8x + 16

9x² + 0,03x + 0,1

9x² + 0,6x + 0,01

9x² + 0,3x + 0,01

9x² – 0,3x + 0,01

$(\frac{x}{2}+2y)(\frac{x^2}{2}+xy+2y^2)$

$(\frac{x}{2}+2y)(\frac{x^2}{4}-xy+4y^2)$

$(\frac{x}{2}+2y)(\frac{x^2}{2}-xy+4y^2)$

$(\frac{x}{2}+2y)(\frac{x^2}{4}-2xy+4y^2)$