Help with some math problems, thank you

I did the following, I don't know if it's correct.

The minimum value for the first question is -4

In the second question, choose your options c and d are the same, and there is no match in the options, that is, the question is wrong, and each correct answer is given, but the final result is given below.

My idea of solving the problem:

In the first question, if f(x) is moved down by two units, that is, the original function minus 2, which is an odd function centered on the origin, and it is monotonically increasing in r, then the maximum value of f(x)-2 on (0, m-2) is 8-2=6

The odd function satisfies f(-x)=-f(x), f(x)-2=-(f(x)+2)=f(-x)+2, and the minimum value of -6+2=-4 can be obtained on (-m+2,0).

The second problem, which is not easy to type, you can use the commutation method to do the second equation, so that the x square of logx=t,t=2 will become the first equation, that is, there is such a relationship between and : 2 to the power = , then bring this conclusion back to the first equation, you can get it.

+2=0, i.e. +=-2

Let's take a look at the last function, f(x), and open it up.

f(x)=xsquared+( x+( 1

Substituting x=0, 1, and 2 into f(x) respectively

f(0)=αβ+1

f(1)=1+(α1=1+-2+αβ1=αβ

f(2)=4+2(α+1=4-4+αβ1=αβ+1

It's easy to get: f(1).

1. Find the integer solution of the indefinite equation 117x1+21x2=38, 21x2=38 117x1, x2=38 21 (117 21)x138 21 (117 21)x1 is an integer, then (79 21) x1 is an integer.

But 79 21 is not an integer, x1 is not an integer, and there is no integer solution to this problem.

4. Find all positive integer solutions of the indefinite equation x+2y+3z=7, solution: x=2, y=1, z=1, only this set of positive integer solutions.

5. Find all positive integer solutions of the indefinite equation 15x1+10 x2+6x3=61, when x1=1, 10 x2+6x3=61 15=465x2+3x3=23

x2=1，x3=6；x2=4，x3=1。

When x1=3, 10 x2+6x3=61 45=16x2=1, x3=1.

There are 3 groups of all positive integer solutions, which are: x1=1, x2=1, x3=6;x1=1，x2=4，x3=1；x1=3，x2=1，x3=1。

7. Solve the congruence group 3x+5y 1 (mod7) (1) 2x 3y 2 (mod7) (2).

3 (1) + 5 (2) to get 19x 13 (mod7) when x = 4, 19x 13 = 63 0 (mod7) 3x + 5y = 12 + 5y 1 (mod7).

11+5y 0(mod7),y=2,11+10=21 (mod7)substitute x=4,y=2 into (1), (2) test, 3x+5y=12+10=22 1(mod7);

2x 3y=8 6=2 2 (mod7), correct.

The special solution is x=4, y=2;The general solution is x=4+7n, y=2+7n, and n is an integer.

1 The error of the instrument and the observation is a systematic error, so the average value is taken, a=(a1+a2+a3) 3

2f(x)=x2+px+5 image opening upwards 0 x +px+5 1 happens to have a real value as the solution.

i.e. p=4 or -4 <> with only one point =p 2-16=0 in [0,1].

2: You multiply a and b to get: a*b=-1, so choose c

3: I can't see the title clearly, you can make it clearer! Sorry!

4: Choose C, A you bring the answer in it is not equal to 0, so A is wrong, x = 0 or x = 3 in B, X in D can also be equal to 2 + 10 under the following sign

Big brother, it's hazy and can't see clearly.

After the simplest quadratic radical is formed, the number of squares is the same. Such a quadratic radical is called a homogeneous quadratic radical.

That's the right answer.

From the question a(-3,3).

s△aob=1/2×3×3=9/2=

Answer: The area of AOB is.

Hope it helps. Satisfied.

In the planar Cartesian coordinate system, it is known that the parabola y=-x2+2x+c passes through the point a(-1,0); The straight line l:y=- 34x+3 intersects with the x-axis at point b, with the y-axis at point c, and with the symmetry axis of the parabola at point m; The vertex of the parabola is d

1) Find the analytic formula of the parabola and the coordinates of the vertex d

2) Crossing point A as ap l at point P, P is perpendicular foot, find the coordinates of point P (3) If n is a moving point on a straight line L, and the perpendicular line of the x-axis intersects with the parabola at point E Q: Is there such a point n that makes a quadrilateral with points D, M, N, E as vertices a parallelogram? If it exists, find the abscissa of the point n; If not, please explain why

Let the linear equation be y-3=k(x+3), which forms a system of equations with the parabola. Eliminate y. x 2 + (2-k) x - (3k + 3) = 0

Because there is only one intersection. Then =0 solution k=-4. So the linear equation is y=-4x-9Then the coordinates of point b are (-9 4,0).

Area of AOB = 1 2 * 9 4 * 3 = 27 8

Quadratic curve, beginning up, xx2

where x1 and x2 are the solutions of the equation x 2-(a+a 2)x+a 3=0: x1=a 2, x2=a

So, XA