x2 - 6xy + 9y2 = 1 - ②

From ①: x2 = 4y2 = (2y)2.

∴ x = 2y or x =-2y

From ②: (x-3y) 2 = 1。

∴ x-3y = 1 or x-3y =- 1.

Then the following four equations can be obtained: x = 2 y x =-2.

X-3y = 1 solution: Y =- 1.

x = 2y x = 2

X-3y =- 1 solution: y = 1.

x = - 2y x = 2/5

X-3y = 1 solution: Y =- 1/5.

x = - 2y x = - 2/5

X-3y =- 1 solution: y = 1/5.

So the original equations have the above four groups of solutions.

Second, the solution: let the original profit of each block be X yuan.

Then the profit per unit after the price reduction is (x-0.5) yuan.

According to the meaning of the question, the original sales volume is (1000/x)

Then the sales volume after the price reduction is [( 1000/x)+200] pieces.

From the countable equation of "the total profit is more than the original 200 yuan":

[( 1000/x)+200]×(x-0.5)= 1000+200

It becomes an integral equation: 2x2-3x-5 = 0.

The solution is x = 2.5x =- 1 (truncation).

After testing, x = 2.5 is the root of the original fractional equation (don't forget to check the root of the fractional equation! )

So the original sales volume is:1000/x =1000/2.5 = 400 (pieces)

A: The original profit per piece was 2.5 yuan, and the original sales volume was 400 pieces.

Third, from the meaning of the question, the quadratic coefficient of the original equation is 1, which is a quadratic equation with one variable.

① When m = 0, the equation becomes x2+2x = 0, and there are two different real roots x 1 = 0x2 =-2, which satisfies the meaning of the question.

② When m = 2, the equation becomes X2- 1 = 0, and there are two different real roots X 1 = 1x2 =- 1, which satisfies the meaning of the question.

③ When m takes all other real numbers where m is not equal to 0 and m is not equal to 2, the discriminant of the root of the original equation.

△ = b2 - 4ac

= [ - ( m - 2 )]2 - 4 × 1 × ( - m2 / 4)

= ( m - 2 )2 + m2

∵ m is not equal to 0 ∴ m2 > 0

∵ m is not equal to 2 ∴ (m-2) 2 > 0.

The discriminant of the root of the original equation △ > 0

No matter what real value M takes in the original equation, there are always two different real roots. The square is too big, please forgive me. )