High School Mathematics Ask for detailed solutions or methods

Let pc=b, the triangle abc becomes a, then ap= (a2-b 2), and rotate the bpc 60° counterclockwise around the point b'a, apparently p'bp=60,△bpc≌△bp'a, so bp'=bp

So bpp'is an equilateral triangle.

So bp'p=60, again p'=150°, so ap'p = 150° - 60° = 90°, using the Pythagorean theorem: Pa2 = AP'^2+pp'2, i.e. (A2-B2) = B2+PP'^2

pp'^2=a^2-2b^2=bp

4(a2-2b 2)=a2-b2, substitute a=1, solve b2=3 7, and then··· tan = two-thirds of the root number three.

In RT APC, if the midpoint of AC is D and PD is connected, PD=DC=PC=COS

In BPC, the sine theorem is applied.

sinbpc/bc=sin(π-bpc-bcp)/pctanθ=2√3/3

c•c-c•b-c•a-a•b=0

Because. a•b=0

c•c-c•b-c•a-=0

c•(c-a-b)=0

ac=1pc=cos angle pcb is 90-, then the sine theorem is expressed pb, and the cosine theorem is fine.

Which question? I'll see how to solve it.

It is known that the function f(x) = (1 3)x + (1 2)ax +2bx + c,(a,b,c r), and f(x) obtains a maximum value in the interval (0, 1), in the interval.

1, 2), then the range of Z=(A+3) +b is as follows

a.(√2/2, 2); b.(1/2, 4); c.(1, 2); d(1, 4)

Solution: Let f (x)=x +ax+2b=0; x, x is its two roots, x +x =-a; x₁x₂=2b.

And 0 is added to get 1 multiplied by two formulas, so 0<(a+3) +b <5 (1, 4) (0,5), so d. should be chosen

The derivative of f(x) = x +ax+2b It is easy to draw an image to obtain f"(2) 0 2a+2b+4 0

f”（1）＜0 1+a+2b＜0

f”（0）＞0 2b＞0

Linear programming makes the region z the square of the distance from point to point (-3,0) in the region and choose b

Point (-3,0) The distance from the point to 2a+2b+4=0 = 2 2 squared gives 1 2

Point (-3,0): The distance from point to point (-1,0) = 2 squared, resulting in 4

The answer should be C

The two upstairs have virtually expanded the range!

Dealing with the first formula is like simplifying it on the first floor, which is to find a derivative (if you don't know how to do this, it will become difficult!). Basically, you come across this kind of problem after you have learned to find derivations), refer to the inequalities on the first floor, and then build the coordinate system on the manuscript paper and draw the diagram (linear programming problems are often done!). This problem is also linear programming) to divide the area (shading or something, for this problem you should draw a triangle).

What makes me different is the handling of the second formula!

You can think of it as the expression of the circle whose center is (-3,0) radius squared and you want to find it, and then you can see it on the graph (you'd better find the intersection point or something like that so you can see the graph).

A(n+2)=4a(n+1)-3an gives a(n+2)-a(n+1)=3[a(n+1)-3an], which shows that (a(n+1)-an) is a proportional series proportional to 3, where the first term a2-a1=3

So a(n+1)-an=3 n

an=[an-a(n-1)]+a(n-1)-a(n-2)]+a2-a1)+a1

3^(n-1)+3^(n-2)+.3+1=(3 n-1) (3-1) (sum of proportional sequences) = (3 n-1) 2

(1) (an+1-an) is a proportional series with a ratio of 3

2）an=（1-3^n）/(-2)

Can you be more detailed, it's a little vague, cc

f(x)+g(x)=5x²-3x+1 (1)

f(-x)+g(-x)=5(-x)²-3(-x)+1＝5x²+3x+1

Then the left side changes according to parity, ie.

f(x)+g(x)=5x²+3x+1 (2)

1)+(2), then divide the two sides by 2 to get g(x)(1)-(2), and then divide the two sides by 2 to get f(x).

The focal point of the parabola x 2=4y is f(0,1), and the derivative of x 2=4y yields 2x=4y',y=x/2.

Let a(2a,a 2),b(2b,b 2),a≠b, the product of the slopes of the two tangents is known from the curve pa pb = ab=-1, and the equation for the straight line ab is y-a 2=(a+b)(x-2a) 2, i.e., y=(a+b)x 2+1, which passes the fixed point f(0,1).

2) By the symmetry of the parabola, when ab is perpendicular to the y-axis, ar*ab takes the minimum value, at this time, a(2,1),b(-2,1), then p(0,-1),r(0,0), ab*ar=(-4,0)*(2,-1)=8, is the request.