{ x-y/7-x+y/10=1/2 {2(x-y)+5(x+y)=3

{ x-y/7-x+y/10=1/2 {2(x-y)+5(x+y)=3

x-y/7-x+y/10=1/2
Take 70 on both sides
10x-10y-7x-7y=35
3x-17y=35 (1)
2(x-y)+5(x+y)=3
2x-2y+5x+5y=3
7x+3y=3 (2)
(2)*3-(1)*7
9y+119y=9-245
128y=-236
y=-59/32
x=（3-3y)/7=39/32

LIM (x → 1) {3 / (1-x ^ 3-I / (1-x)} thank you very much

We know that three non negative numbers a, B and C satisfy 3A + 2B + C = 5, 2A + b-2a + b-3c = 1. (2) we get a-7c = - 3. (3) B + 11C = 7. (4) from (1) - (4)

The process of finding limit LIM (x → 4) (√ (2x + 1) - 3) / (√ (X-2) - √ 2)

Find limit LIM (x → 4) [√ (2x + 1) - 3] / [√ (X-2) - √ 2]
x→4lim [√(2x+1)-3]/[√(x-2)-√2]=x→4lim [√(2x+1)-3][√(x-2)+√2]/(x-4)
=x→4lim (2x-8)[√(x-2)+√2]/(x-4)[√(2x+1)+3)]=x→4lim 2[√(x-2)+√2]/[√(2x+1)+3]
=2(2√2)/6=(2/3)√2
(the numerator and denominator are all rationalized. They can be carried out in two steps (I do it in two steps), or they can be carried out at the same time (i.e. in one step)
To the factor (x-4) that causes the denominator to be zero at the same time

How to find the asymptote of a function?

This is a mathematical problem. First, the image method. Second, the basic function method. We can see how the function is transformed by the basic function. We can get it by combining with the original function. In addition, the asymptote can be divided into three categories: vertical, horizontal and oblique. At first, I had a good grasp of it when I taught myself, but Years make people old

What aspects should be considered to find the asymptote of a function image?
For example: how many asymptotic lines is the number of y = 2x + LNX / (x-1) + 4?

Firstly, we consider whether there is an asymptote perpendicular to the x-axis at the discontinuity of the function,
Secondly, we consider two different cases when x tends to positive infinity and negative infinity
For this problem, there are two asymptotes, one is x = 1, the other is y = 2x + 4

F (x) = (- x ^ 2 + 2x + 4) / (x-3) for this function image, what are the key points and asymptotes?

f(x)=-x-1+1/(x-3)
Key points: (- 1, - 1 / 4)
The asymptote is x = 3

How to find the asymptote of function f (x) = lnlnx?

Because Lim [x → 1 +] lnlnx = - ∞, x = 1 is the vertical asymptote of the function
Because Lim [x → + ∞] (lnlnx) / X
=lim【x→+∞】1/(xlnx)
=0
So Lim [x → + ∞] lnlnx = ∞, so there is no oblique asymptote or horizontal asymptote

The Real Coefficient Quadratic Equation with - 1 + 2I as root is? Detail

According to the pairing of complex roots, if the other root is - 1-2i, it has (- 1-2i) + (- 1 + 2I) = - 2 (- 1-2i) x (- 1 + 2I) = 5
The Real Coefficient Quadratic Equation with root - 1 + 2I is x ^ 2 + 2x + 5 = 0

If 3 + 2I is a root of the equation 2x2 + PX + q = 0 (P, Q ∈ R) about X, then the value of Q is () (a) 3-2i (b) 13

The imaginary roots of the real coefficient equation are conjugate pairs, so the other root of the quadratic equation is 3-2i
According to Weida's theorem, two products = (3 + 2I) (3-2i) = 3 ^ 2 + 2 ^ 2 = 13 = q / 2
Q = 26

If the prime numbers P and Q are the roots of the equation x ^ 2-13x + M = 0, then the value of Q / P + P / Q is

Weida theorem:
p+q=13
And PQ are prime numbers
So one is 2 and the other is 11
2/11+11/2=125/22