(254m-1011) divide by 375 to get the minimum M,

Let 254m-1011 = 375k = 3x125k
Divide the two sides by 3, and 85m-m / 3-337 = 125k
So m = 3p
254p-337 = 125k
Divide the two sides by 125 to get 2p-3 + 2 (2P + 19) / 125 = K
That is: 2p + 19 = 125 (2t + 1)
p=125t+53
So m = 3P = 375t + 159
The minimum value of M is 159

Given the equation x ＾ 2 + (2k-1) x + K ＾ 2 = 0, a necessary and sufficient condition for the equation to have two roots greater than 1 is obtained
Use the relation between root and coefficient to solve this problem! I can't understand the analysis!
There are two x1, x2
{△>=0
{(x1-1)+(x2-1)>0 =====>>>> k0
1. Why is x minus a 1 here?
2. Why doesn't the following one work?
{x1+x2>2
{x1×x2>1
The general question is: why should we subtract 1 from X1 and X2

A:
1. Make sure there is a root
2、
Two larger than 1 = > x1-1 > 0, x2-1 > 0,
So (x1-1) + (x2-1) > 0; and (x1-1) × (x2-1) > 0
If it is
{x1+x2>2
{x1×x2>1
If x 1 = 1 / 2, x 2 = 3, it is also satisfied, but x 1 = 1 / 2

Given the equation x ^ 2 + (2k + 1) x + K ^ 2 = 0, find the necessary and sufficient conditions for the equation to have two roots greater than 1

If △≥ 0: 4K + 1 ≥ 0, K ≥ - 1 / 4
The opening is upward, if both are greater than 1,
-(2k + 1) / 2 > 1, that is: K0, (K + 1) ^ 2 + 1 is always greater than 0, K takes any real number
To sum up: no solution, can not find such a condition, so that the equation has two roots greater than 1

A necessary and sufficient condition for the two real roots of the equation x ^ 2 + (2k-1) x + K ^ 2 = 0 to be greater than 1?

X & # 178; + (2k - 1) + K & # 178; = 0 △≥ 0 (2k - 1) &# 178; - 4K & # 178; ≥ 0 - 4K + 1 ≥ 0k ≤ 1 / 4 according to WIDA's theorem, X1 + x2 = 1 - 2K X1 * x2 = K & # 178; because two real roots are greater than 1, X1 - 1 > 0, X2 - 1 > 0, X1 - 1 + X2 -

(254m-1011) divide by 375 to get the minimum M,