Given that the m-th power of 3 is equal to 5, and the n-th power of 3 is equal to 2, find the value of 2m-3n + 1 power of 3

Given that the m-th power of 3 is equal to 5, and the n-th power of 3 is equal to 2, find the value of 2m-3n + 1 power of 3

2m-3n + 1 power of 3
=2m power of 3 △ 3N power of 3 × 1 power of 3
=(m power of 3) 2 × (n power of 3) 3 × 3
=5²÷2³×3
=75/8

Given that the m-th power of x = 5 and the nth power of x = 3, find the 2m-3n power of X

x^(2m-3n)
=x^2m÷x^3n
=(x^m)^2÷(x^n)^3
=5^2÷3^3
=25/27

On the third power of polynomial-5x of X - (2m-1) X-2 + (2-3n) X-1 does not contain quadratic term and first-order term. Find the value of M and n

-5x^3-(2m-1)x^2+(2-3n)x-1
=-5x^3-2mx^2+x^2+2x-3nx-1
Because the result of the original formula does not contain quadratic term and primary term
therefore
-2mx^2+x^2=0
-2m+1=0
2m=1
m=1/2
2x-3nx=0
2-3n=0
3n=2
n=2/3
X ^ 3 is the third power of X
X ^ 2 is the second power of X

(x-1) (xsquare + X + 1) = (x-1) (x-cubic + X + X + 1) = (x-1) (n-th power of X +. + x-squared + X + 1) = (x-1) (x-1) =

(x-1) (xsquare + X + 1) = x 3 - 1
(x-1) (X Cubic + X + X + 1) = x quartic-1
(x-1) (the nth power of X +. + the square of X + X + 1) = the (n + 1) - power-1 of X

The square of 1 + X + X (x + 1) + X (x + 1) + X (x + 1) +... + n-th power factorization factor of (x + 1)

1 + X + X (x + 1) + X (x + 1) 2 +... + (x + 1) ^ n = (1 + x) [1 + X + X (x + 1) +... + (x + 1) ^ (n-1)] = (1 + x) 2 [1 + X + X (x + 1) + X (x + 1) 2 +... +... + (x + 1) ^ (n-2)] = --- = (1 + x) to the N + 1 power of (1 + x)

If the n-th power of 2x + (m-1) x + 1 is a cubic binomial, find the square of M-N to the power of 2

Up to three times
therefore
N=3
Binomial
Then only (m-1) x = 0
M=1
m²-n²=1-9=-8

(1-x) (1 + X + X squared + + X to the nth power)

(1-x) (1 + X + X squared + + X to the nth power)
=1 + X + X squared + The nth power of + X - (x + X squared +...) + X to the nth power + X to the N + 1 power)
=The N + 1 power of 1-x

If the m power of 3 is 1 / 81 and the nth power of 1 / 2 = 4, find the value of (1 + X squared) to the 3N power

3^m=1/81 m=-4
(1/2)^n=4 n=-2
(1+m)^(3n)=(-3)^(-6)=3^(-6)=1/729

Given that 3 × 27 n + 1 × 81 n-1 = 3 to the 49th power, we find the solution of the equation system {x + y = 23x + NY = 10 Junior one math, thank you

Given that 3 × 27's N + 1 power × 81's n-1 power = 3's 49th power, then: 3 × 3's 3 (n + 1) power × 3's 4 (n-1) power = 3's (1 + 3N + 3 + 4n-4) power = 3's 7n power = 3's 49th power, so we can get: 7n = 49, get n = 7, then the equation system {x + y = 23x + NY = 10 can be written as: {x + y = 2 (1)

If 33x + 1 = 81, then X=______ If 642 × 83 × 2x = 42x, then x2=______ If x-3y = 4, then 2x △ 8y=______ .

∵81=34，
∴3x+1=4，
The solution is x = 1;
∵642×83×2x=（26）2×（23）3×2x=212×29×2x=221+x，
42x=[（22）2]x=24x，
∴21+x=4x，
The solution is x = 7,
∴x2=72=49；
2x÷8y=2x÷（23）y=2x÷23y=2x-3y，
∵x-3y=4，
∴2x÷8y=24=16．
So the answer is: 1; 49; 16