\(\frac{{9{x^2} + 5x + x}}{{3x}} = \frac{{9{x^2} + 6x}}{{3x}} = \frac{{9{x^2}}}{{3x}} + \frac{{6x}}{{3x}} = 3x + 2\)

\(\frac{{2{x^2} - 3x - 2}}{{2 - x}} = \frac{{2{x^2} - 3x - 2}}{{ - x + 2}} =  - 2x - 1\)

\(\left[\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2\right]:\left(x^2-6x+9\right)=\left[\left(3-x\right)^5-7\left(3-x\right)^4-4\left(3-x\right)^2\right]:\left(3-x\right)^2=\left(3-x\right)^2\left[\left(3-x\right)^3-7\left(3-x\right)^2-4\right]:\left(3-x\right)^2=\left(3-x\right)^3-7\left(3-x\right)^2-4=27-27x+9x^2-x^3-63+42x-7x^2-4=-x^3+2x^2+15x-40\)

\(\dfrac{\left(3-x\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{x^2-6x+9}\)

\(=\dfrac{-\left(x-3\right)^5-7\left(x-3\right)^4-4\left(x-3\right)^2}{\left(x-3\right)^2}\)

\(=-\left(x-3\right)^3-7\left(x-3\right)^2-4\)

\(=\left(3x^4-3x^3+x^3-x^2+8x^2-8x+9x-9\right):\left(x-1\right)\\ =\left(x-1\right)\left(3x^3+x^2+8x+9\right):\left(x-1\right)\\ =3x^3+x^2+8x+9\)

Tham khảo:

a) 4x²(x² - 5x + 2)

= 4x².x² - 4x².5x + 4x².2

= 4x⁴ - 20x³ + 8x²

b) (2x²  - 5x + 3) : (2x - 3)

= (2x² - 3x - 2x + 3) : (2x - 3)

= [(2x² - 3x) - (2x - 3)] : (2x - 3)

= [x(2x - 3) - (2x - 3)] : (2x - 3)

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

= x - 1

a, \(4x^2\left(x^2-5x+2\right)\\ =4x^4-20x^3+8x^2\)

b, \(\left(2x^2-5x+3\right):\left(2x-3\right)\\ =x-1\)

Vậy (2x2 – 9x2 + 10x – 3) : (x – 3) = (2x2 – 3x + 1)

11C 12D

Câu 11. Tìm các chữ số x, y biết 413x2y chia hết cho 5 và 9 mà không chia hết cho 2:

A. x = 9, y = 5          B.x = 0, y = 5                    C.x = 3, y = 5        D.x = 5, y = 3

Câu 12. Thực hiện phép tính 35. 68 + 68. 65. 4, kết quả là:

A.27 200                               B.6800                   C.6804                   D.20 060

1c

2d