If x ^ 2 + y ^ 2-6x + 5 = 0, calculate the maximum and minimum values of the following formula respectively 2) x ^ 2 + y ^ 2) y / X

If x ^ 2 + y ^ 2-6x + 5 = 0, calculate the maximum and minimum values of the following formula respectively 2) x ^ 2 + y ^ 2) y / X

X^2+Y^2-6X+5=0
( x-3)^2+y^2=4
X ^ 2 + y ^ 2 represents the square of the distance from the point to the far point. Draw a general diagram and you can see that the maximum value is 25 and the minimum value is 1
Y / x = k, y = KX. It can be seen from the graph that when y = KX is tangent to the circle, K gets the maximum and minimum
The distance from the center of the circle to the straight line = radius | - 3K | / √ (K & sup2; + 1) = 2
The solution is k = ± 2 √ 5 / 5

X-3 = 0 2x-5y = 9 x, what is y equal to?

X=3 Y=-3/5

If 2x + 5y-4 = 0, then the x power of 4 times the Y power of 32 is equal to?

2x+5y=4
4^x*32^y=2^(2*x)*2^(5*y)=2^(2x+5y)=2^4=16

What is the power series expansion of LN √ 1 + x ^ 2? How to find it?

=1/2·ln(1+x^2)
=1/2·Σ(-1)^(n-1)·x^2n/n
(n from 1 to + ∞)

The power series expansion of the following formula
sin(ax-a^3)

The first step,
sin(ax-a^3) =sinaxcosa^3-cosaxsina^3
Step two,
Using the power series expansion formula of SiNx and cosx
Sinax and cosax expansion
Step three,
The results are arranged in the power form of X

Expansion of arctanx function into power series of X

The expansion of F (x) = 1 / 1 + X into a power series of X

If the function is expanded into a power series, does it have to be f (x) = ∑ an (x ^ n)=
Is the form of (1 + x) / 2 ∑ an (x ^ n) ok?

It must be f (x) = ∑ an (x ^ n)
The form of G (x) = (1 + x) / 2 ∑ an (x ^ n) is not allowed, because it is not a power series
Of course, it can be easily transformed into f (x) = ∑ an (x ^ n)

Expand function into power series
The function f (x) = 1 / (X & # 178; + X-2) is expanded into a power series of X

f(x)=1/(x+2)(x-1)=1/3[1/(x-1)-1/(x+2)]=-1/3[1/(1-x)+0.5/(1+0.5x)]
=-1/3[1+x+x^2+.+0.5(1-0.5x+0.5^2x^2-...]

The expansion of function into power series
In this paper, we expand f (x) = ln [x / (x + 1)] into a power series of (x-1) - LN2 + (n = 1) ∑ (- 1) ^ (n + 1) / N multiplied by (1 - 1 / 2 ^ n) (x-1) ^ n, and the convergence interval is (0

When x = 2, we only need to see what follows ∑ becomes ∑ (- 1) ^ (n + 1) / N times (1 - 1 / 2 ^ n). This is a sign changing series. By using Leibniz's criterion, the general term (excluding the part of ∑ (- 1) ^ (n + 1)) is greater than or equal to 0, and it is monotonically decreasing towards 0, so it converges