Two high school math problems, everyone helps to do the next process

1.Decomposed, y=-(x-a) 2+a 2-a+1, when a>1, x=1, the maximum value of 2, and a=2

When a<0 and x=0, the maximum value is 2, and a=-1 is obtained

2.(1) x=1,y>0, the function has an image above the y-axis, (3a+2b+c)-(a+b+c)=2a+b>0,a+b<0

a>0, c>0 ,b<0

Can't ask for it. 2 Ibid.

3 seems to contradict the conclusion of 2).

Question 1: From f(x) we know that the derivative of f(x) is -a b*x-8 b, and the straight line is separated from the circle, so d > 8, i.e. (8 divided by square a plus square b under the root sign) >Question 2:

By sinacosc=8cosasinc, replace sina with the sine formula (i.e., sina = a 8r, sinc=c 8r), and replace cosc,c with the cosine formula....

y=-(x-a) 2+a 2-a+1

When a>1, x=1 obtains the maximum value 2, substituting a=2 when a<0, x=0 obtains the maximum value 2, substituting a=-1 can obtain the second question: 1) substituting x=1, you can get y>0, which means that the function has an image above the y-axis, (3a+2b+c)-(a+b+c)=2a+b>0, a+b<0

a>0, c>0 ,b<0

The 0 function has two distinct intersections with the x-axis.

2) [2a+b>0,a+b<0,a>0] -2 b a -1

3) From the equation 3ax 2+2bx+c=0, we can obtain:

x1-x2|= root number δ 3a = [2 root number (b 2+3a 2+3ab)] 3a [then substitute -2a b -a in].

There is a thirds of the root number three |x1-x2|2 3 It's so hard... I can't call the root number either...

Derivative, when x=a takes the maximum value, when 0 a 1, there is no solution.

When a is 1, it is 2

When a 0, it is -1

One. axis of symmetry x=-a

1.If -a<=0, a>=0

Then x=1 when y=a+2=2

a=<-a<1

a<0y=1-a=2

a=-1 in summary, a=-1 or 0

I don't know if it's right or not. I didn't learn derivatives, sorry.

(1) Focus coordinates (1,0)a(1,1)b(1,-1)(x-1) 2+y 2=1

2) Let the distance be a, b a>b

a-b=4a^2+b^2=20

a-b)^2+2ab=16

ab=2s=1

1 Solution: It is known by the definition of trigonometric functions.

n/m=tan600°=tan（360°+240°）=tan240°=tan60°=√3，∴ m/n=1√3=√3/3．

So the answer is: 3 3

m/n=√3/3

2.(1-sin squared) = cos squared ) = cos |,1-cos squared )=sin squared )=sin |, when the terminal edge of the angle a falls on the line y=x (first quadrant), the result is 1+1=2

When the terminal edge of angle a falls on the straight line y=x (third quadrant), the result is -1-1=-2

The answer to the first question is CTG60°, 1 3 (60°, you typed it wrong).

The second question you wrote is not very clear, is it sina ((1-sin a) + 1-cos a)) (cosa) = 2 2, your question is indeed too vague. I just looked at the upstairs, I'll revise it.

sina ( 1-sin a) + 1-cos a) (cosa) = 2, y = x means sina = cosa

The terminal side is actually a circle and then turns 240°, that is, the line where 60° is located is m:n=1:root number 3 according to the Pythagorean theorem

2.When a = 4+2k (k is an integer), the above equation gives 2

When a = 4+k (k is a non-zero integer), the above equation gives 0