Several eighth-grade math problems (need to be solved)

Let the quadratic term of x+the quadratic term of y = t.

Then (quadratic of x+quadratic of y) (quadratic of x-quadratic of 65438+quadratic of 0 +y)

=t(t- 1)= 12

t^2-t- 12=0

(t-4)(t+3)=0

T=4, t=-3 (truncation)

So the square of x+the square of y =4.

When X>0, the equation becomes: x 2-5x+6 = 0.

(X-2)(X-3)=0

X=2，3

When x

X^2-5X-6=0

(X-6)(X+ 1)=0

X=6 (truncation). X=- 1

So the maximum value is 3, the minimum value is-1, and the product is -3.

X 1+X2=4，X 1X2=K

(x 1-x2)^2=(x 1+x2)^2-4x 1x2= 16-4k

x 1-X2 & lt； three

That is 16-4k < 3*3=9.

4K & gt； seven

K & gt7/4

The output of X-file products is: 76-(X- 1)*4, and the profit per piece is: 10+(X- 1)*2.

So y = (76-4x+4) (10+2x-2) = (80-4x) (8+2x) = 640+160x-32x-8x2.

y=-8x^2+ 128x+640( 1 = & lt； X = & lt 10)

When Y= 1080,

1080=-8x^2+ 128x+640

8X^2- 128X+440=0

X^2- 16X+55=0

(X- 1 1)(X-5)=0

X= 1 1 (exclusive), X=5.

So in order to produce five-grade products,