Given that the square of x plus x minus 1 equals 0, find the value of the third power of x plus the square of 2 x plus 3

Given that the square of x plus x minus 1 equals 0, find the value of the third power of x plus the square of 2 x plus 3

x³+2x²+3
=x³+x²-x+x²+x-1+4
=x(x²+x-1)+(x²+x-1)+4
∵x²+x-1=0
The original formula = 4

If the square of x minus x minus 1 equals zero, find the value of the fifth power of X and the fourth power of x plus 2x plus 1

X? - X - 1 = 0, i.e. x? = x + 1 denominator x ^ 5 = x? * x * x * x = (x + 1) * x * x * x = (x? + x) * x = (2x + 1) * x * x = (2x? + x) * x = (2 (x + 1) + x) * x = (3x + 2) * X. (1) = (3x? + 2x) = 3 (x + 1) + 2x = 5x + 3 numerator x ^ 4 + 2x + 1 =

Given that the a power of X is 32 and the B power of X is 4, find the power of a minus B of X

Given that the a power of X is 32 and the B power of X is 4, then:
The power of a minus B of x = a power of X △ (B power of x) = 32 ÷ 4 = 8

X of 3 plus the power of 1 times the power of 2 minus the power of X of 3 times the power of 2 plus the power of 1 equals the square of 2 times the square of 3, the value of ball X

X=2
3^(x+1)*2^x-3^x*2^(x+1)=2^2*3^2
2^x*3^x(3-2)=2^2*3^2
X=2

The quadratic equation x of one variable minus 7x minus 1 equals 0 to find X1 and x2

x²-7x-1=0
（x-7/2）²-1-49/4=0
（x-7/2）²=53/4
X-7 / 2 = ± 2 √ 53
X = 2 / 2 (7 ± √ 53)

Let X1 and X2 be the two real roots of the equation x squared minus x minus 5 equal to zero, then the third power of X1 plus 6 times x2 minus 5 equals to

The answer is 6
Weida theorem - A / b = X1 + x2 = 1 x2 = 1-x1
x1^3+6x2-5=x1^3-6x1+1=x1(x1^2-x1)+x1^2-6x1+1
Because we can get X1 ^ 2-x1 = 5
So the original formula is 5x1 + X1 ^ 2-6x1 + 1 = 5 + 1 = 6

It is known that the X of 2 plus the third power, and the X of 3 plus the third power is equal to 36 x minus the second power

2^(x+3)×3^(x+3)=36^(x-2)
(2×3)^(x+3)=(6^2)^(x-2)
6^(x+3)=6^^(2x-4)
x+3=2x-4
X=7

If the square of a plus 2A plus the square of B minus 8b plus 17 equals 0, then the value of a to the power of B is

a^2+2a+b^2-8b+1+16=0
(a^2+2a+1)+(b^2-8b+16)=0
(a+1)^2+(b-4)^2=0
If the square is greater than or equal to 0, the sum is equal to 0
If one is greater than 0, the other is less than 0
So both are equal to zero
a+1=0
a=-1
b-4=0
B=4
A^b
=(-1)^4
=1

The calculation of the second power of (1 / 3) - 10-2 | - 2014's 0-th power is an urgent process!

9+2-1-3+2=9

Nine eighths to the a power × 10 / 9 to the B power × 16 / 15 to the C power = 2000, calculate the values of a, B and C

(9/8)^a*(10/9)^b*(16/15)^c=2000
9^a*10^b*16^c=2000*8^a*9^b*15^c
3^(2a)*2^b*5^b*2^(4c)=2^4*5^3*2^(3a)*3^(2b)*3^c*5^c
2^(b+4c-4-3a)*3^(2a-2b-c)*5^(b-3-c)=1
b+4c-3a-4=0
2a-2b-c=0
b-3-c=0
a=33
b=23
c=20