Three high school math problems, can you help me?

1(1) makes x=0, y=-1, f(-1+0)=1=f(0)+f(-1)-3=f(0)-2

Therefore f(0)=3

Let x=1, y=-1

f(1-1)=f(0)=f(1)+f(-1)-3

Therefore f(1) = 5

f(n)=f(n-1+1)=f(n-1)+f(1)-3=f(n-1)+2

So the sequence f(n) is a series of equal differences.

f(n)=2n+3

2)bn+1=bn+f(n-1)=bn+2n+1

bn=bn-1`+2n-1

bn-1=bn-2+2n-3

b2=b1+3

Accumulation, get. bn=(3)+(5)+-2n-3)+(2n-1)+b1=n(2n-1+1)/2=n^2

3) 1 b1+1 b2+--1 bn<( 2) 6<7 4 (infinite series).

2 (1) from the meaning of the title a(x 2-2x) <1

x^2-2x<0

Therefore, a>1 (x 2-2x) is always established.

a>-1

3(1)a1=1-2/3a1

a1=3/5

sn=1-2/3an

sn-1=1-(2/3)an-1

an=(2/3)an-1-(2/3)an

an=(2/5)an-1

Therefore, it is a proportional series with 2 5 as the common ratio and 3 5 as the first term.

2）an=(3/5)(2/5)^(n-1)

sn=(3/5)(1-(2/5)^n)/(1-2/5)=1-(2/5)^n

1: f(x+y)=f(x)+f(y)-3 can be seen to be a primary abstract function set f(x)=kx+b

Let x=-1 y=-1 be substituted.

f(x+y)=f(x)+f(y)-3 f(-2)=-1 to find f(n)=2n+3

bn+1=bn=f(n-1)？If you have a problem, please correct it.

2: What needs to be described.

f(1)≤0 ==>a≥-1

f(2)<0 ==> Heng is established.

3:sn=1-2/3an

sn-1=1-2/3an-1

Subtract to give an=-2 3an+2 3(an-1)=>an (an-1)=5 2

Therefore, an is a proportional series.

Another n=1 finds a1=3 5

sn=3/5*(5/2)^n-1

Let the linear equation y=kx+b, and bring in the point (2,0) to get y=kx+2, let (0,1) be brightened to the photographic point (x,y) of l

The formula for the distance between two points gets: x 2 + (y-1) 2 = 4 points (0, 1) and its projection points (x, y), (2, 0) form a right triangle, the length of each side can be found by the title, the auspicious number is 2, 1, the root number is 5, you can calculate it yourself, then the distance between (x, y) and (2, 0) = 1 can list a formula. Then you can get 3 equations, find the system of equations, and solve k.

I hereby judge.

(1) f'(x) = (e x)-x-a because a = 0, f'(x)=e^x)-x,f'(0)=1f(0)=0, let the tangent equation be y=f'(0) x+b, i.e. 0=0+b, b=0

Tangent equation: y=x

2) f'(x)=(e^x)-x-a

f''(x)=(e x)-1 because x>=1, so f''(x)>0 is constant, that is, the function f(x) is a concave function and is a constant increasing function, so the minimum value of the function is f(1)=e-1 2-a-1>=0, then a<=3 2-e

(1) f'(x)=e^x-x

2)f'(x)=e x-x-a and then find the second derivative f''(x)=e x-1 Because x>=1, the second derivative is everover to zero, that is, the original function is an increasing function, so the minimum function value of the original function is obtained at x=1 f(1)=e-1 2-a-1>=0, so a<=3 2-e

(1) a=-2 substitution:

f(x)=|2x-1|+|2x-2|, g(x)=x+3, the roots in the two absolute value signs, x=1 2, x=1, respectively, divide r into 3 intervals:

1>x 1: f(x)=2x-1+2x-2=4x-31 2 x 1,f(x)=2x-1+2-2x=1-2, to obtain the solution set [1 2,1).

3>x<1 2, f(x)=-2x+1-2x+2=-4x+30, and the solution set (0,1 2) is obtained

Merge: Solution set (0,2).

Knowledge. The cosine theorem translates into a and c relations. The cosine theorem proves that the square relationship between the four sides of the parallelogram and the diagonal is transformed into the square relationship between the three sides of the triangle and the upper midline of one side, and the relationship between a and c is obtained by using it.

The two equations are combined to obtain a and c. Then we found that ABC is a right triangle, and the answer was obtained.

Look at the process experience.

Satisfied, please promptly. Thank you!

Four-thirds times the root number three.

OX perpendicular oy, OX perpendicular to the plane oyz (a straight line perpendicular to the plane of two intersecting lines in the plane, then it is perpendicular to this plane), and because OX is in the plane such as no oxy, so the plane oxy is perpendicular to the plane oyz (one plane through the perpendicular line of the other plane, then the two planes are perpendicular to each other), the same reason can prove that the oxz is perpendicular to oyz, that is, the three planes are perpendicular to each other.

Let oa=x, of=y

x y=2, xy 2=4

Xie Zhaomo repents x=4, y=2

So a 4 = 2

So a=8, so y 2=8x

2) Straight line m: y=mx-3m, and y 2=8x upright.

Let b(x1,y1), c(x2,y2).

You just need to prove that the vector ob point multiplied by the vector oc is less than 0

This shows that BOC is obtuse.

Let oa=x, of=y

x y=2 and xy 2=4

The solution yields x=4 and y=2

So a transport beam 4 = 2

So khuohefhugw

2) Straight line m: y=mx-3m, and y 2=8x are combined.

Let b(x1,y1), c(x2,y2).

You just need to prove that the vector ob point multiplication vector oc is less than 0

Therefore, the manuscript can be quietly respected, and the BOC is obtuse.