Two High School Math Problems!! Urgent! A high school math problem, in a hurry!!

, so f(x-1) -f(3-2x)=f(2x-3) because the function is decreasing on (-2,2), so.

22.Because -3 log(1 2) x -1 2, 1 2 log(2) x 3

f(x)=(log(2)x-1)(log(2)x-2) uses the commutation method to make log(2)x=t obtainable.

f（x）=（t-1）（t-2）=t^2-3t+2=(t-3/2)^2-1/4

where t=log(2)x belongs to [1, 2, 3].

So when the minimum value of the function is t=3 2, the minimum value is -1 4, and when t=3, the maximum value of the function is 2

Then -2 x-1 2 and -2 3-2x 2 give 1 2 x 5 2 (1)g(x)=f(x-1)+f(3-2x) 0f(x-1) -f(3-2x).

f(x) is an odd function.

f(x-1)≤f(2x-3)

f(x) decreases monotonically over the defined domain.

x-1≥2x-3

Solution x 2 (2).

Take the intersection of (1) and (2).

1/2＜x≤2

f（x)=[log（2）x/2]·[log(2)x/4][log（2）x-log（2）2]*[log（2）x-log（2）4]

log(2)x-1]*[log(2)x-2] set log(2)x=t

then f(x)=[log(2)x-1]*[log(2)x-2]=(t-1)(t-2).

t²-3t+2

t-3/2)²-1/4

log（1/2）x=-log（2）x=-t-3≤-t≤-1/2

Then 1 2 t 3, -1 t-3 2 3 2t = 3 2 f(x) takes the minimum value -1 4

When t=3, f(x) takes the maximum value of 2

Swap with triangles.

x 2+(y-1) 2=1, let x=sin, y=1+cos x+y+c 0

c≥-x-y=-sinα-(1+cosα)=-sinα-cosα-1

The maximum value of 2sin(+4)-1,- 2sin(+4)-1 is 2-1 c 2-1.

The conditions satisfied by x and y can be drawn as a circle, and its x+y+c>=0, and c is unknown, and the range can be found as long as the range of x+y is known, and the value of x and y changes in a circle, let x+y=d deform to y=-x+d, and the limit value of d is required, as long as y=-x+d equation and the circle are tangent, the two ends of the tangent circle can obtain the d value, so the x+y limit value can be obtained. You should know what to do, right?

b+c=-7,bc=11

b 2 + c 2 = (b + c) 2-2bc = 27 according to the cosine theorem cosa = (b 2 + c 2-a 2) 2bc substitute the known condition: 1 2 = (27-a 2) 2*11 to obtain a = 4

a^2=b^2+c^2-2bccosa

b+c)^2-2bc-2bccosa

According to the relationship between the root and the coefficient.

b+c=7，bc=11

A 2 = 49-22-22 cos (60 degrees) = 49-33 = 16a = 4

Hello 1 a, b is the square of equation 4x - 4mx+m+2=0 of the two real roots =16m -4*4(m+2) 0, m 2, m -1 by Veda's theorem: a+b=m, ab=(m+2) 4a +b =(a+b) -2ab=m -(m+2) 2=(m-1 4) -17 16

The axis of symmetry m=1 4, so when m=-1, a +b obtains a minimum value of 1 2

2 ) A 2+b 2=(a+b) 2-2ab by Vedic theorem The above equation =(-m) 2-(m+2) 2=m 2-m 2-1=(m-1 4) 2-(17 16) So when m=1 4, a 2+b 2 takes the minimum value, which is -17 16 hehe, this solution is wrong.

Since a b is a real number, m itself is bound, and the range m can be derived from being greater than or equal to 0.

2 We already know that the interval is [-1,1], so the axis of symmetry must be less than or equal to 0, otherwise the maximum price is not 1 (you have already drawn a diagram of the opening of the function 2 times upward), so we put y=(x-a) 2+1 to get y=x 2-2ax+a 2+1, and then according to the formula of the axis of symmetry x=-2a 2, it is said that it should be less than or equal to 0, so a is less than or equal to 0

And when x=a is the minimum value, a should also be in [-1,1], and finally the range of a value is a[-1,0].

(1) a 2 + b 2 = (a + b) 2-2ab and then substitute according to the sum of the two roots and the product of the two roots.

2) You can ignore the +1 in the back, just look at (x-a) 2, draw a parabolic diagram, about x=a symmetry, first of all, the value range of a should be in, (x [-1,1]), which can be seen through the graph: -1

Direct substitution. f(1 x) = (1 + (1 x) )1-(1 x) side grinding) = (x +1) (x -1) = -f(x).

Or just substitute it.

f(x1+x2 stupid Kaiku2).

a((x1+x2)/2)+b

ax1/2+b+ a·x2/2

and f(x1)+f(x2) 2 = (ax1+b+ax2+b) 2=ax1 2+b+ a·x2 2

The proposition has to be proven.

What's wrong with this question? Isn't it okay to directly substitute the operation?