The value of proposition algebra X & # 178; - 4x + 7 must not be less than 3. Is this a true proposition or a false proposition? Please explain the reason

The value of proposition algebra X & # 178; - 4x + 7 must not be less than 3. Is this a true proposition or a false proposition? Please explain the reason

True proposition
x²-4x+7
=x²-4x+4+3
=（x-2）²+3
≥3
So it's a true proposition
This is a real tomorrow, because X & # 178; - 4x + 7 = (X-2) &# 178; + 3 ≥ 3
It is a true proposition
Hope to adopt
x^2-4x+7=(x-2)^2+3>=3
So this is a true proposition
True proposition x2-4x + 7 = (X-2) 2 + 3 > = 3
The equation of the line parallel to the line 3x + 4y-12 = 0 and the distance from it is 7 is______ .
Let the linear equation be 3x + 4Y + C = 0, ∵ the distance between two parallel lines be 7, | - 12 − C | 32 + 42 = 7, which is reduced to | - 12 + C | = 35, the solution is C = 23 or - 47, so the linear equation is 3x + 4Y + 23 = 0 or 3x + 4y-47 = 0. So the answer is 3x + 4Y + 23 = 0 or 3x + 4y-47 = 0
On the quadratic equation with known square root of X + M
It is known that the square + 7x + 11 ˉ M = 0 of the quadratic equation of one variable x has real roots
Find the value range of M
When m is a negative number, we can find two following equations
(1) If the equation x & # 178; + 7x + 11-M = 0 has real roots, then Δ = 49-4 (11-M) = 5 + 4m ≥ 0, then m ≥ - 5 / 4
(2) If M ＜ 0, then - 5 / 4 ≤ m ＜ 0
From the root formula, x = [- 7 ± √ (5 + 4m)] / 2
If x * x-4y * y = - 15, x + 2Y = 3, find the values of X and y
You don't seem to know math at all. X-2y = - 5, x + 2Y = 3,
Given the circular equation x ^ 2 + y ^ 2 + ax + 2Y + A ^ 2 = 0, a certain point a (1,2), if there are two tangents passing through the fixed point a (1,2) as elements, then the range of a
Given the circular equation x ^ 2 + y ^ 2 + ax + 2Y + A ^ 2 = 0
The standard equation is: (x + A / 2) ^ 2 + (y + 1) ^ 2 = 1-3a ^ 2 / 4 > 0
So - 2 √ 3 / 3 ＜ a ＜ 2 √ 3 / 3. (1)
If there are two tangents of a circle made through a fixed point a (1,2), then there are two tangents
It shows that point a (1,2) is outside the circle
So the distance between the fixed point and the center of the circle is greater than the radius
So (- A / 2-1) ^ 2 + (- 1-2) ^ 2 ＞ 1-3a ^ 2 / 4
So a ^ 2 + A + 9 > 0
So a ∈ R. (2)
From (1) and (2) - 2 √ 3 / 3 ＜ a ＜ 2 √ 3 / 3
It is proved that the value of the algebraic formula 2x + x-3 is not less than 25 / 8
It is proved that: ∵ 2x + x-3 = 2 (x + 1 / 2x) - 3 = 2 × [x + 2 × 1 / 4 × x + (1 / 4) - (1 / 4)] - 3 = 2 [(x + 1 / 4) - 1 / 16] - 3 = 2 (x + 1 / 4) - 1 / 8-3 = 2 (x + 1 / 4) - 25 / 8 ∵ no matter what real number x is, all (x + 1 / 4) ≥ 0 ∵ 2 (x + 1 / 4) - 25 / 8 ≥ - 25 / 8
Given that the line L: 3x + 4Y-2 = 0, the distance between line a and line L is 1, then the equation of line a is
Obviously parallel
So it's 3x + 4Y + a = 0
Take a point on 3x + 4Y-2 = 0, such as (2, - 1)
Then the distance to 3x + 4Y + a = 0 is 1
|6-4+a|/√(9+16)=|a+2|/5=1
a=-7,a=3
So 3x + 4y-7 = 0 and 3x + 4Y + 3 = 0
32.4×2÷4.5
=14.4
A: this side is 14.4 meters long
(1-5/8)x=15
3/8x=15
x=15÷3/8
x=40
Y=5/X=5X^(-1)
Y'=5*(-1)*X^(-1-1)
=-5X^(-2)
=-5/X^2
The largest integer C with real roots of quadratic equation x2 + 7x + C = 0 is ()
A. 8B. 10C. 12D. 13
According to the meaning of the question, △ = 49-4c ≥ 0, that is, C ≤ 1214, the largest integer C is 12
Given x2 + 4y2 = 4xy, then the value of X + 2yx − y is______ .
∵ x2 + 4y2 = 4xy, ∵ x2 + 4y2-4xy = 0, that is, (x-2y) 2 = 0, the solution is x = 2Y, then x + 2yx − y = 2Y + 2y2y − y = 4
It is known that the equation of a circle is x2 + Y2 + ax + 2Y + A2 = 0, a certain point is a (1,2), so that there are two tangent lines of a circle passing through the fixed point a, and the value range of a is obtained
x^2+y^2+ax+2y+a^2=0
(x+a/2)^2+(y+1)^2=1-3a^2/4
Then a ^ 2
A: just outside the circle
C;(x+a/2)2+(y+1)2=1-3/4a2
C(-a/2,-1) r2=1-3/4a2
[CA]2=(1+a/2)2+9>1-3/4a2
Can get a
α∈( -2√3/3，2√3/3)
Negative 2 / 3 root sign 3, 2 / 3 root sign 3