AX3 + ax2-2x3 + 2x2 + X + 1 is a quadratic polynomial of X. it is also necessary to find the value 2 of A2 + (1 / a) 2 + A to be the square of 2

AX3 + ax2-2x3 + 2x2 + X + 1 is a quadratic polynomial of X. it is also necessary to find the value 2 of A2 + (1 / a) 2 + A to be the square of 2

AX3 + ax2-2x3 + 2x2 + X + 1 is a quadratic polynomial of X, then the sum of the coefficients of the cubic term of X (A-2) = 0, a = 2 A2 + (1 / a) 2 + a = 4 + 1 / 4 + 2 = 25 / 4

How to solve the quadratic polynomial of X

The formula for finding roots of cubic equation with one variable can not be made by ordinary deductive thinking. The matching method similar to the formula for finding roots of quadratic equation with one variable can only formalize the standard cubic equation with one variable of type ax ^ 3 + BX ^ 2 + CX + D + 0 into a special type of x ^ 3 + PX + q = 0
The solution of the formula for solving the cubic equation of one variable can only be obtained by inductive thinking, that is, the form of the formula for finding the root of cubic equation of one variable can be concluded according to the form of the formula for finding the root of linear equation of one variable, quadratic equation of one variable and special higher order equation. The form of the formula for finding the root of cubic equation of one variable, such as x ^ 3 + PX + q = 0, should be x = a ^ (1 / 3) + B ^ (1 / 3), It is the sum of two open cubes. The form of the formula for finding the root of cubic equation with one variable is summed up. The next step is to find out the content in the open cube, that is, to use P and Q to express a and B
(1) Let x = a ^ (1 / 3) + B ^ (1 / 3) be cubic at the same time
（2）x^3=(A+B)+3(AB)^(1/3)(A^(1/3)+B^(1/3))
(3) Because x = a ^ (1 / 3) + B ^ (1 / 3), so (2) can be changed into
X ^ 3 = (a + b) + 3 (AB) ^ (1 / 3) X
(4) Compared with the univariate cubic equation and the special type of x ^ 3 + PX + q = 0, we can see that x ^ 3-3 (AB) ^ (1 / 3) x - (a + b) = 0
(5) - 3 (AB) ^ (1 / 3) = P, - (a + b) = Q
（6）A+B＝－q,AB＝-（p/3）^3
(7) In this way, the root formula of cubic equation with one variable is transformed into the root formula of quadratic equation with one variable, because a and B can be regarded as the two roots of quadratic equation with one variable, and (6) is the Veda theorem about the two roots of quadratic equation with one variable in the form of ay ^ 2 + by + C = 0
（8）y1＋y2＝－（b/a）,y1*y2=c/a
(9) Comparing (6) and (8), we can make a = Y1, B = Y2, q = B / A, - (P / 3) ^ 3 = C / A
(10) Because the root formula of quadratic equation of type ay ^ 2 + by + C = 0 is
y1＝－（b＋（b^2－4ac）^(1/2)）/(2a)
y2＝－（b－（b^2－4ac）^(1/2)）/(2a)
Can be transformed into
(11)y1＝－(b/2a)-((b/2a)^2-(c/a))^(1/2)
y2＝－(b/2a)+((b/2a)^2-(c/a))^(1/2)
Substituting a = Y1, B = Y2, q = B / A, (P / 3) ^ 3 = C / A in (9) into (11), we can get
(12)A＝－(q/2)-((q/2)^2＋（p/3）^3）^(1/2)
B＝－(q/2)+((q/2)^2＋（p/3）^3）^(1/2)
(13) Substituting a and B into x = a ^ (1 / 3) + B ^ (1 / 3), we get
(14)x=（－(q/2)-((q/2)^2＋（p/3）^3）^(1/2)）^(1/3)+（－(q/2)+((q/2)^2＋（p/3）^3）^(1/2)）^(1/3)
Equation (14) is only a real root solution of one variable cubic equation. According to Weida's theorem, one variable cubic equation should have three roots, but according to Weida's theorem, as long as one of the roots is solved, the other two roots can be easily solved

On the polynomial (A-4) x3-xb + X-B of X is a quadratic trinomial, then a=______ ，b=______ ．

∵ the polynomial (A-4) x3-xb + X-B is a quadratic trinomial, ∵ (1) does not contain X3 term, i.e. A-4 = 0, a = 4; (2) the degree of the highest term is 2, i.e. B = 2

Calculation: 3 / 2-5 / 6 + 7 / 12-9 / 20 + -197 / 9702 + 199 / 9900 (hint: 3 / 2 = 1 + 2 / 1 * 2, 5 / 6 = 2 + 3 / 2 * 3)
It's a process!
All right, score

It's very simple
Original formula = 1 + 1 / 2 - (1 / 2 + 1 / 3) + (1 / 3 + 1 / 4) - (1 / 4 + 1 / 5) -（1/98+1/99）+（1/99+1/100）
=1+1/2-1/2-1/3+1/3+1/4-1/4-1/5…… -1/98-1/99+1/99+1/100
=1+1/100
=101/100

191 + 193 + 195 + 197 + 199 + 201 + 203 = () * () = () how to fill in the following brackets to make this formula equal

191+193+195+197+199+201+203=(197 )*(7 )=(1379 )
Because 197 is the average of these numbers

Calculation: 197 × 198-196 × 199=______ ．

197 × 198-196 × 199 = (196 + 1) × 198-196 × (198 + 1) = 196 × 198 + 198-196 × 198-196 = 198-196 = 2