Given log3 (x power of log2) = 0, then the second power of x equals 0 A、8 B、6 C、4 D、2

Given log3 (x power of log2) = 0, then the second power of x equals 0 A、8 B、6 C、4 D、2

Log3 (x power of log2) = 0
The x power of log2 = 1
x=2
x² =4
The second power of X is equal to = 4

The value of 2 plus the bottom 5th power of log2 minus the bottom 3rd power of log2 multiplied by the bottom 5th power of log3

(2 + log2 bottom 5th power) - (log2 bottom 3rd power) of 2
=The bottom 5th power of log3 of 4 × 5-3
=20-5
=15

① Absolute value of (- 4) × (- 8) - (- 5) × (- 7)
② (- 1 / 2) × (- 1 / 3)=
③ 3 and 2 / 3 × (- 1 and 3 / 5)=
④ 0 × (- 11 / 99)=
⑤（-1）×125=

① Absolute value of (- 4) × (- 8) - (- 5) × (- 7)
= 4 x 8 - (-5) x 7
= 32 + 35
= 67

One of the original functions of F (x) is SiNx / X
Excuse me: a primitive function of F (x) is SiNx / x, integral ∫ (π / 2 to π) x * f '(x) DX = 4 / π - 1
Please do it. Note: π is pie, 180 degree angle

∫x*f'(x)dx=cosx-2(sinx)/x

Find the length of the common chord of circle x ^ + y ^ - 4 = 0 and circle x ^ + y ^ - 4x + 4y-12 = 0

Comparison of circle x ^ + y ^ - 4 = 0 and circle x ^ + y ^ - 4x + 4y-12 = 0
The intersection coordinates are (0,2) and (- 2,0)
And then with the formula

Let f (x) have f (x + y) = f (x) + F (y) for any real number x, Y. if x is greater than 0, f (x) is less than 0, find the maximum value of F (x) in the interval [a, b]

Let x = y = 0, then f (0) = 0
Let x = M0, then f (0) = f (m) + F (- M)
f(m)=-f(-m)>0
If X1 > x2
f(x2)=f(x1)+f(x2-x1)>f(x1)
So f (x) decreases monotonically
So the maximum value is f (a)

By using Taylor's formula, the 99th derivative of F (x) = x ^ 2sinx at x = 0 is obtained

Taylor expansion of SiNx: SiNx = x - x ^ 3 / 3! + x ^ 5 / 5! - x ^ 7 / 7 +x^97/97!+O(x^97)f(x) = x^2*sinx =x^3 - x^5/3!+ x^7/5!- x^9/7!… +The 99th derivative of x ^ 99 / 97! + O (x ^ 99) f (x) at 0 is equal to 99! * 1 / 97! = 99 * 98 = 9702

Let a > 0, f (x) = e ^ X / A + A / e ^ X be even functions on R
1. Find the value of A
2. Prove that f (x) is an increasing function on (0, + infinity)
Do you like to answer

f(x)=f(-x)
We get (e ^ x) / A + A / (e ^ x) = e ^ (- x) / A + A / [(e ^ (- x))]
(e^x)/a+a/(e^x)=1/（ae^x)+ae^x
That is, (e ^ x) (1 / A-A) + (A-1 / a) / (e ^ x) = 0
(a-1/a)[1/(e^x)-e^x]=0
Because of the arbitrariness of X, only A-1 / a = 0
That is, a ^ 2-1 = 0
From a > 0, a = 1
Next, we prove that f (x) = e ^ x + 1 / (e ^ x) is an increasing function
Let x1, X2 ∈ (0, + ∞), X1 < x2
f(x1)-f(x2)=e^x1+1/e^x1-（e^x2+1/e^x2）=e^x1-e^x2＋1/e^x1-1/e^x2=（e^x1-e^x2）（e^x1e^x2-1）/e^x1e^x2
x1,x2∈（0,＋∞）,e^x1e^x2-1＞0,e^x1-e^x2＜0
（e^x1-e^x2）（e^x1e^x2-1）/e^x1e^x2＜0
f(x1)＜f(x2)
F (x) is an increasing function from 0 to positive infinity

It is proved that the eigenvectors of different eigenvalues of real symmetric matrix must be orthogonal

I just passed the matrix test yesterday, but I forgot all about it today

Have or has for singular nouns? Have or has for plural nouns? Have or has for countable nouns? Have or has for uncountable nouns?
What's the difference between have and has?

Has for uncountable nouns, has for the singular and have for the plural