Given X4 + X3 + x2 + X + 1 = 0, find the value of X100 + x99 + x98 + x97 + x96

Given X4 + X3 + x2 + X + 1 = 0, find the value of X100 + x99 + x98 + x97 + x96

∵ X4 + X3 + x2 + X + 1 = 0, ∵ X100 + x99 + x98 + x97 + x96 = x96 (x4 + X3 + x2 + X + 1) = 0; so the answer is: 0

Given x ^ 4 + x ^ 3 + x ^ 2 + x ^ 2 + 1 = 0, find the value of x ^ 100 + x ^ 99 + x ^ 98 + x ^ 97 + x ^ 96
See above

Because x ^ 4 + x ^ 3 + x ^ 2 + x ^ 2 + 1 = 0
therefore
x^100+x^99+x^98+x^97+x^96
=x^96(x^4+x^3+x^2+x^2+1)
=x^96*0
=0

Given x ^ 4 + x ^ 3 + x ^ 3 + x ^ 2 + x ^ 1 + 1 = 0, find the value of x ^ 100 + x ^ 99 + x ^ 98 + x ^ 97 + x ^ 96

x^4+x^3+x^2+x+1=0,
x^100+x^99+x^98+x^97+x^96
=x^96(x^4+x^3+x^2+x+1)=x^96*0=0

X ^ 4 + x ^ 3 + x ^ 2 + x ^ 1 + 1 = 0 find the value of x ^ 100 + x ^ 99 + x ^ 98 + x ^ 97 + x ^ 96

x^100+X^99+X^98+X^97+X^96
=x^96(x^4+X^3+X^2+X^1+1)
=k^96*0
=0

Let x1x2x3 Xn,Y1,Y2,Y3…… Yn (n > = 2) are all real numbers and satisfy X1 ^ 2 + x2 ^ 2 + X3 ^ 2 + Xn^2= (X1^2+X2^2+…… Xn^2-1)*(Y1^2+Y2^2+…… Yn^2-1)

For the convenience of narration, let X1 ^ 2 + x2 ^ 2 + +Xn^2=A ,Y1^2+Y2^2+…… +Yn^2=B ,X1Y1+X2Y2+…… Xnyn = s, a + B ≥ 2S, if a ≤ 1, it is proved that (s-1) ^ 2 ≥ (A-1) (B-1)
① If B ≥ 1, B-1 ≥ 0, because A-1 ≤ 0, then (A-1) (B-1) ≤ 0 and (s-1) ^ 2 ≥ 0 inequality holds
② If B ≤ 1, then 2 ≥ a + B ≥ 2S, s ≤ 1, then 1-s ≥ 1-1 / 2 (a + b) ≥ 0
So (1-s) ^ 2 ≥ (1-1 / 2 (a + b)) ^ 2 = 1-a-b + 1 / 4 (a + b) ^ 2 ≥ 1-a-b + AB = (1-A) (1-B) inequality holds
To sum up, the inequality holds

It is known that the perimeter of the rectangle is 20cm and the area is 10cm, then the diagonal length of the rectangle is 20cm

Let the rectangle be a in length and B in width
According to the meaning of the title
（a+b）×2=20
a+b=10
ab=10
Let the diagonal form be c
c²=a²+b²=（a+b）²-2ab=100-20=80
c=4√5

Find the maximum value of y = 3-x & # 178; - 9 / X & # 178; (x ≠ 0) and the corresponding value of X (using the mean value theorem)

Ask for a problem of partial integration, ∫ arcsinx / x ^ 2 DX
I'm sorry, thank you very much, good man

The original formula = - sarcsinxd (1 / x) = - 1 / X * arcsinx + S1 / xdarcsinx = - 1 / X * arcsinx + S1 / X * 1 / radical (1-x ^ 2) DXX = Sint, t = arcsinx, DX = costdt, S1 / X * 1 / radical (1-x ^ 2) DX = S1 / Sint * 1 / cost * costdt = scscsctdt = ln | CSCT Cott | + CT = arcsinx, and then

A square is divided into six rectangles of exactly the same size and shape, each with a circumference of 14 cm, which is the original square

Let the side length of the original square be 6x
Then the side length of each rectangle is:
（6x+x)×2 =14cm
.=.
x=1cm
So the side length of the original square is 6cm

Mathematics Evaluation Handbook page 90 question 5
As shown in the figure, translate △ ABO up three unit lengths to get △ A1 B1 O1; fold △ ABC along the Y axis to get △ A1 B1 O1
Rotate △ ABO clockwise 90 ° around point O to get △ A3 B3 O3
1) If points a (- 4,3), B (- 6, - 1), O (0,0), write the coordinates of points A1, B1, O1, A2, B2, O2, A3, B3, O3 respectively
2) If P (x0, Y0) is any point in △ ABO, P1, P2 and P3 are obtained respectively after the above translation, folding and rotation, and the coordinates of P1, P2 and P3 are written out

A1（-4,6） A2(4,3) A3(3,4)
B1（-6,2） B2(6,-1) B3(-1,6)
O1（0,3） O2(0,0) O3(0,0)
P1(x0,y0+3)
P2(-xo,y0)
P3(y0,-x0)