Pit daddy s high school math problem, high school math problem, see if this question is wrong

Is it c, because the center of the circle is fixed at a distance of 1 from the straight line y=3 on y=2So when the shortest radius is the shortest chord length, the shortest radius is 2 (obtained by the formula on both sides of the equation), so the chord length is 2 times the square of 2 under the root number minus 1 to get c

Circle: x 2+y 2-2mx-4y+4m-4=0, which is reduced to the standard equation of a circle, yields: (x-m) 2+(y-2) 2=(m-2) 2+4.

The center of the circle is: (m,2), on the straight line: y=2;

The radius of the circle r,r= [m-2) 2+4].

The straight line: y=2, the distance from the straight line y=3 is 1, the chord length of the straight line y=3 is rounded: 2 (r 2-1), the shortest, then: r 2=(m-2) 2+4, the smallest;

And [(m-2) 2+4>=4, so the minimum value of r 2 is: 4.

The shortest chord length is: 2 3.

Therefore, C.

g(x)=-x²-3

f(x) quadratic function, f(x) + g(x) odd function.

Because if there are quadratic terms and constant terms, it cannot be an odd function.

So f(x)=x+ax+3

In this case, f(x) + g(x) = ax odd function.

min=1 when x [-1,2].

The axis of symmetry x=-a 2 is discussed on the left and right sides of the interval and in the interval to obtain a=2 or a=-2 2

f(x)=x +2x+3 or f(x)=x -2 2x+3

1.The problem of finding inequalities.

That is, find the range of x values when -x is 2-5 > 1-5x.

2.Draw your own axes.

Then b calculates the range of values with .

It can be seen from the coordinate axes.

3.It is divided into several areas.

When 2>x>-3.

When x>2.

When x<-3 are discussed separately.

lzI'm also a freshman in high school.,It's better to do it yourself、、、 otherwise you'll suffer.。。

Step 1: Solve the equation of the parabola y=x5 and the straight line y=1 5x to find x1=2, x2=3

Step 2: Draw an image of the parabola y=x5 straight line, y=1 5x, and the two images intersect in the x1=2 and x2=3 directions. The parabolic opening is downward.

It was learned in 2The point on the parabola y=x5 in the range of

2) Solution: Obtained from the known.

Because aub=, a=

So x in b is greater than or equal to -1

Because a=, anb=

So b in x 3

So x=3 and x=-1 are solutions to x2+ax+b=0] obtained by Vinda's theorem.

a=2,b=-3

Question 1 2 Question 2 A = -1

b=-6 I don't know the third question.

1.According to the title, -x 2-5>1-5x, find 22According to the problem, -2 and 3 are the roots of the equation x 2+ax+b=0, and we find a=1 and b=-6

1 is equivalent to -x 2-5> 1-5x constant.

x^2-5x+6>0 x<2 or x>32.From the drawing of the inscription, we can know that b=

so b=-3, a=-2

3.Due to |x+3|-|x-2|>=-|（x+3)-(x-2）|=-5 Deformation of trigonometric inequalities.

So a<-5

1. 1-5x+x^2+5<0

then x 2-5x+6<0

x-2)(x-3)<0

2-2}, then the other root intersects a at the same time between the interval (-2, -1) b={x|1a

Discussion, x<=-3, primitive=-5

32, original = 5

For the equation to hold, a<-5

1.What is the value of x, where the point on the parabola y=-x 2-5 is above the line y=1-5x?

Analysis: 1-5x=-x 2-5==> x 2-5x+6=0==>x1=2,x2=3

At x (2,3), the point on the parabola y=-x 2-5 is above the straight line y=1-5x (on the right).

2.Let a=, b=, aub=, a intersect b=, aub=, a intersect b={x|1a-b=1 (1)

a+√(a^2-4b)]/2=3==>3a+b=-9 (2)

1) (2) The simultaneous solution yields a=-2 and b=-3

3.If for any real number x, |x+3|-|x-2|> a has a solution, and the range of values of the real number a is obtained.

Let the function f(x) = |x+3|-|x-2|

When x (-3], f(x) = -x-3+x-2=-5

When x (-3,2), f(x) = x+3+x-2=2x+1

When x[2,+,f(x)=x+3-x+2=5

If for any real number x, |x+3|-|x-2|> A has a solution, then A <-5

The topic is not clear, the root number ends at the end of the **? And how can there be a decimal under the root number?

a(-1，-1)，b(-4，2)

Vector ab=(-3,3).

Vector ap|=(1/3)|Vector ab|

Vector ap=(1 3) vector ab=(-1,1)a(-1,-1).

p(-2，0)

c(3,0), point q is the midpoint of the line segment pc.

q(1/2，0)

Calculate the coordinates of p with the formula of the fixed score point, and then divide p+c by 2 to complete. There are two solutions. The formula is a compulsory four-vector for turning the book yourself.

x=(x1+λx2)/(1+λ)y=(y1+λy2)/(1+λ)

f(x) is an even function defined on r.

f(-x)=f(x)

x[f(x)+f(-x)]＜0

2xf(x)＜0

x＜0,f(x)＞0

then x -3x 0, f(x) 0

then 0 x 3

So the solution set is.

Hello, the ranges are (-infinity, -3) and (0,3).