If the linear y = (m-1) x-n is parallel to y = - 2x + 3 and passes through points (1,4), then the analytic expression of the linear is 2. If in the linear function y = (2-3K) x - (K + 1), y follows The range of K is 0 If the perimeter of isosceles triangle is 24, the length of its base is x, and the length of its waist is y, then the functional relation of Y with respect to X is the independent variable, and the value range of X is

If the linear y = (m-1) x-n is parallel to y = - 2x + 3 and passes through points (1,4), then the analytic expression of the linear is 2. If in the linear function y = (2-3K) x - (K + 1), y follows The range of K is 0 If the perimeter of isosceles triangle is 24, the length of its base is x, and the length of its waist is y, then the functional relation of Y with respect to X is the independent variable, and the value range of X is

(1) Because the line y = (m-1) x-n is parallel to y = - 2x + 3, the slopes of the two lines are the same, that is, (m-1) = - 2, that is, y = -- 2x-n
Then we can get n = - 6 from point (1,4), so the analytic formula is y = -- 2x + 6
(2) Because y = (2-3K) x - (K + 1), y increases with the increase of X
So 2-3K > 0 so K

On the equation 3x & # 178; - 2 (3K + 1) + (3K & # 178; - 1) = 0
When the value of K is zero, there are two real roots of opposite numbers, and two reciprocal numbers

∵ B & # 178; - 4ac = 4 (3K + 1) & # 178; - 4 × 3 (3K & # 178; - 1) ≥ 0 ∵ K ≥ - 2 / 3 (1) when 3K & # 178; - 1 = 0, k = ± √ 3 / 3 ∵ when k = ± √ 3 / 3, this equation has a root of 0; 2) according to Weida's theorem, X1 + x2 = - B / a = 2 (3K + 1) / 3 = 0k = - 1 / 3 ∵ when k = - 1 /

When k takes what value, the equation 3x & # 178; - 2 (3K + 1) x + 3K-1 = 0
The following conditions are met:
One root is greater than 1 and the other is less than 1
Use Veda's theorem

x1-10
So (x1-1) (x2-1)

Let the proposition p: (3K-2) x ^ 2 + 2kx + k-1 ＜ 0. The proposition q: (k ^ 2-I / 12) x ^ 2 + KX + 1 ＞ 0. If at least one of P and Q holds, the value range of K is obtained
19 pages

Let's look at P: 3K-2 < 0, and at the same time [4 (3K-2) (k-1) - 4K ^ 2] / 4 (3K-2) < 0
The range of K is obtained
Q:(k^2-1/12)＞0 [4(k^2-1/12)-k^2]/4(k^2-1/12)＞0
The range of K is obtained
Put the two areas together