The number of real roots of equation 2 ^ x = x ^ 2 is discussed

The number of real roots of equation 2 ^ x = x ^ 2 is discussed

There are three real roots
X = 2, x = 4, and a real root X

(x ^ 2-1) ^ 2 - (x ^ 2-1) + k = 0 when k takes different values, the number of real roots of the equation is discussed

The original formula is as follows:
(x² - 1)² - (x² - 1) + k = 0
Let y = x & # 178; - 1, (X & # 178; = y + 1)
Where (1) y > - 1, X: 2
(2) Y = - 1, X: 1
(3) Y < - 1, X: 0
Substituting the above formula,
Then y & # 178; - y + k = 0
Discriminant: △ = (- 1) &# 178; - 4 * 1 * k = 1 - 4K
And the root of Y: Y1 = [1 + √ (1 - 4K)] / 2, y2 = [1 + √ (1 - 4K)] / 2
(1) When △ > 0, y has two roots
That is, 1 - 4K > 0, K < 1 / 4
Among them,
a) Y1 > 0 > - 1, if Y2 > - 1, (k > - 2)
When - 2 < K < 1 / 4, X: 2 + 2 = 4
b) Y1 > 0 > - 1, if y2 = - 1, (k = - 2)
When k = - 2, X: 2 + 1 = 3
c) Y1 > 0 > - 1, if Y2 < - 1, (k < - 2)
When k < - 2, X: 2 + 0 = 2
(2) When △ = 0, y has one root
That is 1 - 4K = 0, the solution is k = 1 / 4
Among them,
y1 = y2 = 1/2 > -1
That is, when k = 1 / 4, X: 2
(3) When △ < 0, y has 0 roots
That is 1 - 4K < 0, the solution is k > 1 / 4
When k > 1 / 4, X: 0
Synthesis (1), (2), (3),
When - 2 < K < 1 / 4, the number of real roots is 4
When k = - 2, the number of real roots of the equation is three
When k < - 2 or K = 1 / 4, the number of real roots of the equation is two
When k > 1 / 4, the number of real roots is 0

Find the number of roots of equation (x ^ 2-1) ^ 2 - | x ^ 2-1 | + k = 0 and the range of corresponding K
I only have 10 points, but if you answer this question correctly, if you are an 18-year-old boy, I will consider being your GH!
THESE ARE WRONG!

dalte=(|x^2-1|)^2-4*(x^2-1)^2*k
=(x^2-1)^2*(1-4k)
Because (x ^ 2-1) ^ 2 > = 0 holds
therefore
(1)dalte>0
1-4k>0
k

Is y = 1 / x ^ 3 derivative? What is it?

y=x^(-3)
So y '= (- 3) y ^ (- 2)

How to find the derivative of ^ (2) = (x / 1)?

LG2 = a find log2 ^ 25 equal to (please write the process)

25a
Idempotents can be transformed into log coefficients
lga^n=nlga

Let LG2 = a, then log2 ^ 25 is equal to (1-A) times of? A

Using the logarithm exchange formula:
log2^25
=(1g25)/(1g2)
=(1g25)/a
=[1g(100/4)]/a
=(1g100-lg4)/a
=(2-lg4)/a
=(2-lg2^2)/a
=(2-2lg2)/a
=(2-2a)/a
=2(1-a)/a

How is 9 1 equal to 100?

(9÷9+9÷1)²=100
99.9+.1=100

How much is 1 to 100

5050,(1+100)*100/2,

1 + 2.100 = how much?

It's 5050