\(a,\Leftrightarrow x\left(x-3\right)-2\left(x-3\right)=0\\ \Leftrightarrow\left(x-2\right)\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\\ b,\Leftrightarrow\left(2x-5\right)\left(2x+5\right)+\left(2x+5\right)^2=0\\ \Leftrightarrow\left(2x+5\right)\left(2x-5+2x+5\right)=0\\ \Leftrightarrow4x\left(2x+5\right)=0\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{5}{2}\end{matrix}\right.\)

2.

a. x³ - 4x² + 4x

= x(x² - 4x + 4)

= x(x - 2)²

c. 3x² + x - 4

= 3x² + 4x - 3x - 4

= x(3x + 4) - (3x + 4)

= (x - 1)(3x + 4)

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

b) \(x^2-y^2-6y-9=x^2-\left(y^2+6y+9\right)=x^2-\left(y+3\right)^2=\left(x-y-3\right)\left(x+y+3\right)\)

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

1F

2F

3F

4T

5T

6F

7F

8T

Câu 6: D

Cau 7: A

Câu 8: C

Câu 9: A

Câu 10: D

Câu 11: A

Câu 12: D

a: Xét tứ giác AKCI có

AK//CI

AI//KC

Do đó: AKCI là hình bình hành

\(a,=2x^3-2x-2x^3=-2x\\ b,=x^2-6x+9-x^2+4x+5=-2x+14\\ c,=8x^3+12x^2+6x+1-8x^3-12x^2+2x+3=8x+4\)