Current location - Training Enrollment Network - Mathematics courses - Math problem in the second day of junior high school [quadratic formula]
Math problem in the second day of junior high school [quadratic formula]
1 solution:

According to the topic:

1.3x+5y-2-m is greater than or equal to 0.

2.2x+3y-m is greater than or equal to 0

3.x- 199+y is greater than or equal to 0.

4. 199-x-y is greater than or equal to 0 (all numbers in the root sign are equal to or greater than 0).

Equations 3 and 4 show that x+y= 199.

So the root number (3x+5y-2-m)+ root number (2x+3y-m)=0.

So five. 3x+5y-2-m = 0。

6.2x+3y-m=0

Simplification: Equation 5- Equation 6 equals x+2y=2.

And because x+y= 199.

So x=396 y=- 197 is substituted into 1 and 2.

1 Type equals to m less than 20 1.

Equation 2 is equal to m less than 20 1.

Therefore, all numbers in the radical formula can be equal to m=20 1, greater than or equal to 0.

M= root number 10 1- root number 100= (root number 10 1- root number 100)//kloc-0.

Similarly, N= root number 99- root number 98= 1/ (root number 99+ root number 98)

Because the root number 10 1+ root number100 >; Root number 99+ root number 98

So m < N

Note: (root number 10 1- root number 100)/ 1

= (number of roots 10 1- number of roots 100) (number of roots10/+number of roots 100)/ (number of roots10/kloc)

= (101-100)/(root number 10 1+ root number100)

= 1/ (root number 10 1+ root number 100)