30% with (x + 3) = 7x-6 solve equations

30% with (x + 3) = 7x-6 solve equations

3(x+3)=7x—6
3x+9=7x-6
9+6=7x-3x
15=4x
x=15/4

If the inequality MX & # 178; + MX + 4 ＞ 0 holds, find the value range of function M

In order to make the inequality MX & # 178; + MX + 4 > 0 hold, the following two conditions hold simultaneously
(1) M > 0 (ensure that the opening of the image is upward, and the image will not extend downward, so it will not always be less than zero)
（2）△

The reciprocal of 1 and 1 / 4 is 1 and 1 / 4 times of a certain number. How to find this number? It's a process! Thank you

One and a quarter is 5 / 4, and the reciprocal is 4 / 5
Let a number be x, so 5x / 4 = 4 / 5
x = 16/25
To sum up, this number is 16 / 25

The trajectory equation of curve 3x-4y + 3 = 0 with distance of 2 points is obtained

Let (x, y) be any point on the plane coordinate axis,
We obtain that | 3x-4y + 3 | / 5 = 2
Two straight lines, 3x-4y-7 = 0 and 3x-4y + 13 = 0, are obtained

Let f (x) defined on R satisfy f (x) · f (x + 2) = 13, if f (1) = 2, then f (99) = ()
A. 13B. 2C. 132D. 213

∵ f (x) · f (x + 2) = 13 and f (1) = 2 ∵ f (3) = 13F (1) = 132, f (5) = 13F (3) = 2, f (7) = 13F (5) = 132, f (9) = 13F (7) = 2, ∵ f (2n − 1) = 2 & nbsp; n is odd 132 & nbsp; n is even, f (99) = f (2 × 100 − 1) = 132, so C is selected

1) In the triangle ABC, the three sides are a, B, C. The equation 3x ^ 2 + 2 (a + B + C) x + (AB + BC + Ca) = 0
1) In the triangle ABC, the three sides are a, B and C respectively. The equation 3x ^ 2 + 2 (a + B + C) x + (AB + BC + Ca) = 0 has two equal real roots to determine the shape of the triangle ABC
2) If t is a non negative number and the quadratic equation (1 + T ^ 2) x ^ 2 + 2 (1-T) X-1 = 0 has two real roots, find the value of T and the roots of the corresponding equation

(1) Because the equation 3x ^ 2 + 2 (a + B + C) x + (AB + BC + Ca) = 0 has two equal real roots, so the discriminant
4 (a + B + C) ^ 2-12 (AB + BC + Ca) = 0; i.e
(a + B + C) ^ 2-3 (AB + BC + Ca) = 0
a^2+b^2+c^2-ab-bc-ca=0；
[0.5a^2-ab+0.5b^2]+[0.5a^2-ca+0.5c^2]+[0.5b^2-bc+0.5c^2]=0;
0.5(a-b)^2+0.5(c-a)^2+0.5(b-c)^2=0;
So a = b = C, ABC is an equilateral triangle
(2) Discriminant
4 (1-T) ^ 2 + 4 (1 + T ^ 2) > = 0
8-8t+8t^2>=0;
t^2-t+1>=0;
(t-0.5)^2+0.75>=0
The above inequality is always true

Quadratic function, various analytical symmetry axis
What are the three kinds of analytical formulas and their symmetry axis formulas
For example, the two formulas are: 2 / X1 + x2
What are the other two formulas for the axis of symmetry

General form
y=ax²+bx+c
The axis of symmetry is a straight line x = - B / 2A
Vertex type
y=a(x-h)²+k
The axis of symmetry is a straight line x = H
Intersection type
y=a(x-x1)(x-x2)
The axis of symmetry is a straight line x = (x1 + x2) / 2

Factorization in real number: a (a + 1) (a + 2) (a + 3) - 3
Urgent!

=a(a+3)*(a+1)(a+2)-3
=(a^2+3a)*(a^2+3a+2)-3
=(a^2+3a)^2 + 2*(a^2+3a) -3
=(a^2+3a+3)(a^2+3a-1)

For any real number a, B with incomplete zero, the equation 3ax2 + 2bx - (a + b) = 0 is in the interval (0,1) ()
A. No real root B. exactly one real root C. at least one real root D. at most one real root

(1) When a = 0, B ≠ 0, the equation is (2) when a ≠ 0, if a (a + b) ＜ 0, ∵ f (0) f (12) = - (a + b) · (- A4) = a (a + b) 4 ＜ 0, ∵ f (12) f (1) = - A4 · (2A + b) = - a24-a (a + b) 4 ＜ 0, the equation has at least one real root in the interval (0,1). If a (a + b) ≥ 0, ∵ f (12) f (1) = - A4 · (2a + b) = - a24-a (a + b) 4 ＜ 0, & nbsp; The equation has at least one real root in the interval (0,1)

Simplification: x2 − 16x2 + 8x + 16 + XX − 4

The original formula = (x + 4) (x − 4) (x + 4) 2 + XX + 4 = x − 4x + 4 + XX + 4 = 2x − 4x + 4