X + y + 1 = 0, x + 4Y = 8 2. (x-3) quadratic + y = 9, x + 2, y = 0, 3.5 parts x + 4 parts y, 1, y = x minus 3, 4.9 parts (x + 1) quadratic minus 4 parts (Y-1) square equals 1, x minus y = 1 5, x plus y = 6, xy = 7.6, x plus y = 5 / 3, xy = minus 4, 7.7, x plus y plus 5 = 0, xy = minus 14, 8, y = x + 5, y = x + 5, xy = x + y = x + 5, X square + y square = 625

X + y + 1 = 0, x + 4Y = 8 2. (x-3) quadratic + y = 9, x + 2, y = 0, 3.5 parts x + 4 parts y, 1, y = x minus 3, 4.9 parts (x + 1) quadratic minus 4 parts (Y-1) square equals 1, x minus y = 1 5, x plus y = 6, xy = 7.6, x plus y = 5 / 3, xy = minus 4, 7.7, x plus y plus 5 = 0, xy = minus 14, 8, y = x + 5, y = x + 5, xy = x + y = x + 5, X square + y square = 625

I got it```
But the solution is not necessarily correct
I'm only in grade one``
The question should be:
1.X+Y+1=0 X^2+4*(Y^2)=8
2.(X-3)^2+Y^2=9 X+2Y=0
3.(X^2)/5+(Y^2)/4=1 Y=X-3
4.[(X+1)^2]/9-[(Y-1)^2]/4=1 X-Y=1
5.X+Y=6 XY=7
6.X+Y=5/3 XY=-4
7.X+Y+5=0 XY=-14
8.Y=X+5 X^2+Y^2=625
Questions 1, 2, 3 and 4 will not be solved
Question 5 x = ± (√ 2) - 3, y = 9 ± √ 2 (opposite sign)
Question 6 x = 3, y = - (3 / 4)
Question 7 x = 2, y = - 7
Question 8 x = 15, y = 20
I'm a junior high school student

Binary quadratic equations
1. Judge the root of the equation y ^ 2 + 3 (m-1) y + 2m ^ 2-4m + 4 / 7 = 0
2. When the value of the algebraic formula x ^ 2 + 4x-2 is 3, find the value of the algebraic formula 2x ^ 2 + 8x-5

（1）
Y & sup2; + 3 (m-1) y + (2m & sup2; - 4m + 4 / 7) = 0,4 / 7 means 4 / 7
Discriminant △
=b²-4ac
=9(m-1)²-4(2m²-4m+4/7)
=9m²-18m+9-8m²+16m-16/7
=m²-2m+47/7
=(m-1)²+40/7
>0
The discriminant is greater than 0, so the original equation has two unequal real roots
（2）
X & sup2; + 4x-2 = 3, so x & sup2; + 4x = 5
2x²+8x-5=2(x²+4x)-5=2×5-5=5

On the stability of binary quadratic equations
Is {x ^ 2-5x + 4 = 0; y ^ 2-3y + 2 = 0} a binary quadratic system?

Yes, there are two unknowns, the highest is twice
So it's a system of quadratic equations of two variables

The surface area of a cube is 36cm. How many cm is the area of each side and how many cm is its edge length

The area of each face is 36 △ 6 = 6 square centimeters
Its edge length is the arithmetic square root of 6 = root 6 cm

It is known that LXL = ax + 1 has only negative roots and no positive roots. Find the value range of A
Let's be more specific

We get - x = ax + 1
-1/（a+1）=x-1
If you have a positive heel
x=ax+1
-1/(a-1)>0
A-1 and a greater than 1
So a is greater than 1
There are two heels between - 1 and 1

What is the number of - 7 greater than - 4? Given two numbers 12 and - 18, what is the absolute value of the sum of these two numbers? What is the absolute value of the sum of these two numbers?
Urgent,
Given | x | = 3, | y | = 5, find the value of X + y
The sum of all integers whose absolute value is less than 2012 is

-4-7=-13;
6;
12+18=30
|X | = 3, x = 3 or x = - 3;
|Y | = 5, y = 5 or y = - 5
When x = - 3 and y = - 5, x + y = - 8;
When x = - 3 and y = 5, x + y = 2;
When x = 3 and y = - 5, x + y = - 2;
When x = 3 and y = 5, x + y = 8;

The function f (x) = x ^ 3 + SiNx + 1, X ∈ R, if f (a) = 2, find the value of F (- a)

f（x）=x^3+sinx+1,x∈R
=>f（a）=a^3+sina+1=2
=>a^3+sina=1
And f (- a) = (- a) ^ 3 + sin (- a) + 1
f（-a）=-a^3-sina+1
=-（a^3+sina）+1
=-1+1
=0

Given a, B ∈ R, the linear L1: x + A ^ 2Y + 1 = 0 and the linear L2: (a ^ 2 + 1) x-by + 3 = 0, if L1 / / L2, find the value range of B.)
The correct answer is (negative infinity, - 6) intersection (- 6, 0]

l1//l2
1/(a^2+1)=a^2/(-b)
-b=a^4+a^2
b=-a^4-a^2
=-(a^4+a^2+1/4)+1/4
=-(a^2+1/2)^2+1/4
a^2+1/2>=1/2 -(a^2+1/2)^2

For real numbers a and B, we can use min {a, B} to represent the smaller of a and B, such as min {6, - 2} = - 2
If the image of the function y = min {3x ^ 2, a (X-H) ^ 2} about X is symmetric about the line x = 3, then the values of a and h may be_____ .

y=a(x-h)^2
Should meet
3×3^2=a(3-h)^2
0=a(6-h)^2
So h = 6
a=3

Let's use Taylor's formula for the higher derivative of 1 / ax + B
Scores enough? Convenient, please pass the photos. OK, plus 20. Thank you

This problem can be solved by Taylor's formula at 0, but it can't be solved by Taylor's formula at other points. At x = 0, y = 1 / (AX + b): 1 / ax + B = (1 / b) - (A / b ^ 2) x + (a ^ 2 / b ^ 3) x ^ 2 - (a ^ 3 / b ^ 4) x ^ 3 + +(- 1) ^ (n) * [a ^ n / b ^ (n + 1)] x ^ n + O (x ^ n)