Ask a few math questions, senior year of high school, thank you for your help

Question 1: y=(x+1 2)+1 (x+1 2)+1 2>=2+1 2The condition for the equal sign to be valid is x+1 2=1 (x+1 2), combined with x>0, know, x=1 2So the maximum value of y is 5 2

Question 2: a+b=2 root number 3, ab=cosc=-1 2=(a 2+b 2-c 2) 2abThe solution is c=3That's ab=3

Question 3: First of all, the equation can be reduced to a 2sinc=(a 2+c 2-b 2)sinb, which is written as (a).sinc=c 2r,sinb=b 2r,r is the radius of the circumscribed circle of the triangle.

Substituting simplification yields a 2(c-b)=b(c-b)(c+b), and since b is not equal to c, b-c is not equal to 0So a 2 = b(b + c).b/(b+c)=b^2/a^2=(sinb/sina)^2.

In addition, by dividing 2ac on both sides of the equation at the same time, sina=sin2b. can be obtained according to the cosine theorem and the metaphysical theoremApplying this equation to the above equation gives b (c+b) = (1 4)*(secb) 2, and 01 4, it is obvious that b (b+c) < (b+c) (b+c) = 1

To sum up, 1 4 Question 4: According to the intuitive geometric definition of a parabola, the trajectory sought is a parabola. Can be set to:

x 2=-2p(y-h), in this problem p 2=1, and the parabola passes (0,2), and the equation for the parabola is obtained by combining these two conditions: x 2=-4(y-2)Beg.

1.(1) Subtract (-1, increase You can find the axis of symmetry, and draw a diagram

2.Take any 00

The third question is ao( oha! Done it yesterday).

The first question: 3a+2b+c=315,

a+2b+3c=285.②

4a+4b+4c=600

a+b+c=150

Problem 2 (Solve inequality x 6.)

The third question x=m-2 is a negative number, so if you imitate your hand, you should talk about m 2

<> "Question 1, Question 2, I copied it wrong, it's right, I can't count it."

Proof: (1) When the macro rolls when n=1.

ln(1+1)/1=ln2

Because ln2>0 the original formula holds when n=1.

2) When n>1.

In order to prove that the original formula holds, it is only necessary to prove that ln[(n+1) n]>1 n holds.

Let f(x)=ln[(n+1) n]-1 n, then f(x)=ln(n+1)-ln(n)-1 n, because the car is f'(x)=1 (n+1)-1 n+1 n 1 [n (n+1)]>0

So f(x) is an increasing function at n>1, and f(2) >0, so f(n) >0 when n>1

So ln[(n+1) n]>1 n holds.

To sum up, the evidence of the question stem is closed.

|a+c|.|a-c|.1=1

a^2-c^2=1

Because c=1a=root number 2

So the elliptic equation is.

x^2/2+y^2=1