If the square of X + X-1 = 0, then the square of the algebraic expression 2x + 2x-6=

If the square of X + X-1 = 0, then the square of the algebraic expression 2x + 2x-6=

x^2+x-1=0,
——》2x^2+2x-6=2(x^2+x-1)-4=-4.

The square-x-6 = - 1 collocation method of 2x

2x²-x-6=-1
2x²-x-5=0
2（x²-1/2x-5/2+41/16-41/16）=0
2（x-1/4）²-41/8=0
（x-1/4）²=41/16
X-1 / 4 = ± (radical 41) / 4
X = (radical 41 + 1) / 4 or (- radical 41 + 1) / 4

If there are only two terms of the fourth term in the square expansion of [x-2x-1 / 2] n
If only the binomial coefficient of the fourth term is the largest in the square expansion of [x squared - 2x 1 / 2] n, how to find the sum of the coefficients in the expansion?

(x^2-1/2x)^n
The fourth term = t (3 + 1) = C (n, 3) * (x ^ 2) ^ (n-3) * (- 1 / 2x) ^ 3
The binomial coefficient of the fourth term is C (n, 3) the largest, and the fourth term is the middle term, n = 7
The sum of the coefficients in the expansion,
(x ^ 2-1 / 2x) ^ 7 let x = 1
=(1-1/2)^7=1/128

What is the method of transforming the square of X + 5x-14 = 0 into (X-2) (x + 7) = 0
Is rt a formula or something? I didn't learn it

The square of X + 5x-14 = 0 to (X-2) (x + 7) = 0 is a cross multiplication method
RT is the symbol of right triangle

The monotone increasing interval of function f (x) = 5x2-2x is______ ．

∵ y ′ = 10x-2, let y ′ > 0, the solution is: x > 15, so the answer is: (15, + ∞)

Monotone interval of F (x) = LG (2x ^ 2-5x-3)
RT

2x^2-5x-3>0
(x-3)(2x+1)>0
x> 3 or X3 or X5 / 4, monotonically increasing
X3} monotone decreasing interval {X / X}

Find monotone interval f (x) = 1 / 2x ^ 2-5x + 6 (Note: x ^ 2-5x + 6 is superscript)
For monotone interval f (x) = (1 / 2) x ^ 2-5x + 6 (Note: x ^ 2-5x + 6 is superscript), please write the detailed explanation

Is it: F (x) = (1 / 2) ^ (x ^ 2-5x + 6)
Because g (y) = (1 / 2) ^ y is a monotone decreasing function, so
The monotone interval of y = x ^ 2-5x + 6 is obtained as decreasing: X ∈ (- ∞, 5 / 2]
Increasing (x / 5 ∞)
That is, the decreasing interval of F (x) is: X ∈ (5 / 2, ∞), and the increasing interval is: X ∈ (- ∞, 5 / 2]
I'm wrong about the question upstairs

F (x) = logx (- 2x & sup2; + 5x-2), find the monotone interval of F (x)

f(x)=logx(g(x))
If the simple increasing interval is (1,4 / 5) and the simple decreasing interval is (4 / 5,2)
in summary
The simple increasing interval is (1,4 / 5)
The simple subtraction interval is (1 / 2,1), (4 / 5,2)

If f (2x-1) = the square of X, then f (5)=

Solution
Let 2x-1 = t
Then x = (T + 1) / 2
∴f(t)=(t+1)²/4
∴f(5)=(5+1)²/4=36/4=9

∮ (x-1) = the square of X - 2x + 7 then ∮ (x)=

Let a = X-1
Then x = a + 1
Then ∮ (a) = (a + 1) & sup2; - 2 (a + 1) + 7 = A & sup2; + 6
So ∮ (x) = x & sup2; + 6