It is proved that the 5th power of equation x minus 3x and then minus 1 is equal to 0, and there is at least one real root in the interval (1,2) Urgent solution
Let f (x) = x ^ 5-3x-1, it is obvious that f (x) is a continuous function on R
∵f(1)=-3,f(2)=25
And f (1) * f (2)
If the X-1 / 3rd power of equation (X-2) about x = 1 (x is not equal to 1)
(x-2)^(x-1/3)=1
X-1 / 3 = 0, x = 1 / 3 or X-2 = 1, x = 3
Prove that x to the fifth power - 3x = 1 has at least one root between 1 and 2
This equation is written as f (x) = x ^ 5-3x-1 = 0. F (x = 1) 0, so there is at least one root between 1 and 2
The root of this equation is x, y, Z. find x + y + Z = 0
The three roots are set to x1, X2, X3
Then x ^ 3 + PX + q = (x-x1) (x-x2) (x-x3) = x ^ 3 - (x1 + x2 + x3) x ^ 2 + (x1x2 + x2x3x3 + x3x1) x-x1x2x3
Comparing the coefficients of x ^ 2, X1 + x2 + X3 = 0
Given that the square of x minus 5x plus 1 equals 0, find the fourth power of x plus 1 divided by the square of X
From x ^ 2-5x + 1 = 0, both sides are divided by X to get x + 1 / x = 5
(x^4+1)/x^2=x^2+1/x^2=(x+1/x)^2-2=5^2-2=23
Given that the square of x minus 5x plus 1 equals 0, find the fourth power of x plus 1
From x-5x + 1 = 0, x + 1 / x = 5 (x ^ 4 + 1) / x = x + 1 / x = (x + 1 / x) - 2 = 25-2 = 23