1, (the simplest method) solution: x-3=x(x-3)

x(x-3)-(x-3)=0

(x-3)(x- 1)=0 Description: Extract the common factor (x-3).

Solution: x =3 or x= 1.

2. Solution: 2(x+3)? =x(x-3)

2(x？ +6x+9)=x？ -3 times

2x？ + 12x+ 18=x？ -3 times

x？ + 15x+ 18=0

Solution: solve by formula method.

X=(- 15-3√ 17)/2。 Or x=(- 15+3√ 17)/2.

3.( 1) The first scheme: It is meaningful to assume that the price of the product after the price increase is X yuan:

(x-8)×[200-(x- 10)/0.5× 10]= 700

(x-8)(400-20 times) =700

20x^2-560x+3900=0

Simplified as: x 2-28x+ 195 = 0.

(x- 13)(x- 15)=0

Solve; X = 13 or x= 15.

The second scheme: suppose the price of this product drops to Y yuan.

(y-8)×[200+( 10-y)/0.5× 10]= 700

Simplified as: (y-8)(400-20y)=700.

x^2-28x+ 195=0

(x- 13)(x- 15)=0

X= 13 or 15 is greater than 10, which contradicts my hypothesis, so it is not valid.

To sum up: when the price rises to 13 yuan or 15 yuan, the daily profit will reach 700 yuan.

Formula analysis (x-8) × [200-(x-10)/0.5 ×10] = 700: (x-8) represents the profit of each commodity; 200-(x-10)/0.5×10 represents the quantity of goods sold after the price increase; [profit per commodity] x[ quantity of commodities sold after price increase] = [daily profit] is the following formula: (x-8) × [200-(x-10)/0.5×10] = 700.

(2) If the selling price is set at X yuan per piece, and the profit gained on this day can be Y yuan, then the meaning of the question is:

y =(x-8)×[200-(x- 10)/0.5× 10]

=(x-8)(400-20 times)

= - 20(x^2-28x+ 160)

= -20[(x- 14)^2-36]

=720-20(x- 14)^2

When x= 14, ymax=720.

That is, when the selling price is set at 14 yuan, you can make the biggest profit on this day and 720 yuan.

The correct answer is as above, welcome to adopt! Thank you very much.