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A tolerance d 0 equal difference series, bn is the common ratio q is positive, and a reasonable proportional series is less than 1.

If a1 = d a1 2 = d 2 = 4d a2 2 a3 2 = 9d (the nature of the difference series).

If b2 b1 = d 2 = q d 2 b3 = q 2 d 2 (properties of the geometric sequence 2 a1 + a2 + a3 2 b1 + b2 + b3 = 14).

Q2 + Q +1 times.

The q numerator 14 with the title to 14Q2 + Q+1 is a positive integer value, and it is a positive integer denominator preferably 1,2,7,14

Q2 + Q + 1 time = 1

q = 0, or -1 (non-standard question, 0 2 2 + q + 1 time = 2

q =（-1 - 5）/ 2（1 +√5） / 2

Because (0 q 2 + q + 1 time = 7, q = 2 or 3 (non-standard questions, rounded).

q 2 + q +1 = 14, the same question gets the meaning of the nonconformity.

For the above reasons, q = (-1 + 5) 2

So, do the math answers, it's not clear at the moment and then ask me again.

Alpha. alpha

The five answers are, in order.

1/(x^2)-3/x+2

y-2=2(x-1)

Horizontal asymptote y=1

Solution: The polar coordinate system is established with the center of the earth o as the pole and the straight line where the elliptic Changshensen axis is located as the polar axis.

Let a be the perigee and b be the apogee, and let the polar equation of the elliptical orbit be =ep (1-ecos).

oa=6378+200=6578

ob=6378+350=6728, the coordinates of point A are (6578, ) and the coordinates of point B are (6728,0), and E is obtained by EP (1-E)=6728 and EP (1+E)=6578

Therefore, the polar coordinate equation for the elliptical orbit of the early filial cong is .

Extend Ca to H through A so that AH=AC.

Because acb= ecd. And ha=ac=ba, so hbc=90°=ced, so hbc is similar to dec.

So hc:cd=bc:ec. And because HC=2AC, 2AC:CD=BC:EC, so BC CD=2AC CE

As af bc is in d, afc is similar to dec so ac*ce=cd*cf and bc=2cf so it is proven.

Use the hypotenuse of the adjacent edge of the trigonometric function cosine:cos = to make a guide line aq perpendicular bc

Because ac=ac aq perpendicular bc so triangle abq congruent triangle acq so bc=2qc cd=ce cos dce qc=ac cos acq

bc×cd=2×ac×cos∠acq×ce/cos∠dce∠acq=∠dce

So bc cd=2ac ce

The answer to the first question is a=1, b=2The answer to the second question is 0

1. I don't know if your question is like this: (3x-4) = a (x-1)+b (x-2), if so, reduce the right side to the same as the left side, then compare the x coefficient and the constant term.

2、x2+1/x2= x2+2(x2×1/x2)+1/ x2-2=(x+1/x)2-2=2

2(x^2+2x+1)/x+x+1/x-3=02x+4+2/x+x+1/x-3=0

x+1/x+1/3=0

x^2+1/3x+1=0

b^2-4ac=1/9-4<0

So there is no real solution, there is a complex solution, and the solution is -1 6 i 35 6

2（x^2+2x+1)+x^2+1-3x=02x^2+4x+2+x^2+1-3x=0

3x^2+x+3=0

The general form of a quadratic equation is: ax 2 + bx + c = 0a = 3 b = 1 c = 3

First, find the discriminant formula: =b 2-4ac=1-4*3*3=-35<0 If 0, then the equation has a solution, <0, then it is misunderstood.