√(y+1)≤4÷(y-1)

√(y+1)≤4÷(y-1)

Both sides at the same time square (y + 1) ≤ 16 △ (Y-1) ^ 2 (y + 1) (Y-1) ^ 2 ≤ 16, because 0 ≤ √ (y + 1) ≤ 4 △ (Y-1), so Y > 1 for f (y) = (y + 1) (Y-1) ^ 2, it is easy to prove that when y > 1, f (y) monotonically increases. For equation (y + 1) (Y-1) ^ 2 = 16, the solution is y = 1 or y = 3 because y > 1, so y = 3, that is, when 1 < y ≤ 3

Solution set of inequality (3 - π) x > π - 3

(3-π)x>π-3
3 - π < 0, both sides divide by 3 - π:
x＜-1

Solve two mathematical problems about inequality (please write the detailed process,
(1) Given that m and N are real numbers, if the inequality (2m-n) x + 3m-4n4 / 9, find the solution set of the inequality (m-4n) x + 2m-3n > 0
(2) The equation about X is known: 4 / 3x-m = 8 / 7x-1. When m is some negative integers, the solution of the equation is negative integers. Try to find the maximum value of negative integer M

1. If the solution set is x > 4 / 9, then 2m-n ＜ 0. And 4 / 9 is the root of the equation (2m-n) x + 3m-4n = 0, substituting into M = 8N / 7, and (16N / 7) - n ＜ 0, n ＜ 0. Substituting into (m-4n) x + 2m-3n > 0, then 2m-n ＜ 0. Substituting into (m-4n) x + 2m-3n > 0, then 2m-n ＜ 0. Substituting into (m-4n) x + 2m-3n > 0, then 2m-n ＜ 0. Substituting into (m-4n) x + 2m-3n > 0, then 2m-n ＜ 0, then 2m-n ＜ 0. Substituting into ( ...