Given the equation x2-5x + 1 = 0, then the value of X4 + 1x4 is () A. 521B. 527C. -521D. -527

Given the equation x2-5x + 1 = 0, then the value of X4 + 1x4 is () A. 521B. 527C. -521D. -527

If both sides of the equation x2-5x + 1 = 0 are divided by X, the solution is: X-5 + 1x = 0, then x + 1x = 5, the square of both sides is: x2 + 2 + 1x2 = 25, then x2 + 1x2 = 23, and the square of both sides is: X4 + 1x4 + 2 = 529, then X4 + 1x4 = 527
X2-5x + 1 = 0, find X4 + 1 / x2
x^2-5x+1=0
x^2+1=5x
(x^4+1)/x^2=(x^2+1)^2-2x^2/x^2
=(5x)^2-2x^2/x^2
=23x^2/x^2
=23
X2-5x-1 = 0, find the value of X4 + 1 / X4
X2-5x-1 = 0 divided by X on both sides
x-5-1/x=0
X-1 / x = 5 square
x2-2+1/x2=25
X2 + 1 / x2 = 27 square on both sides
x4+2+1/x4=729
x4+1/x4=727
I've solved it myself! A friend who meets this question will give you a perfect answer
x2-5x-1=0
Obviously, X ≠ 0
x-5-1/x=0
x-1/x=5
square
x2+1/x2-2=25
x2+1/x2=27
square
x4+1/x4+2=729
x4+1/x4=727
..
What is the point on the parabola y equals 2x squared - 3x + 1?
I want to ask which two points are the points on the x-axis. Factorization (2x-1) (x-1), so the two points are (05,0), (1,0)
How to transform a quadratic equation of two variables into a quadratic equation of one variable
For example, 3x + 5Y = 8 can be converted to y = - 3 / 5x + 8 / 5
What is the transformation process of the above question?
The main reason is that there is a conversion of scores
3x + 5Y = 8 (both sides of the equation add, subtract, multiply or divide by the same number )
5y=8-3x
y=(8-3x)÷5
y=-3/5x+8/5
If the parabola y equals the square of a x and the parabola y equals the square of negative 3x is symmetric about the X axis, then the value of a is
a＝3
The difference and connection between one variable first order equation and two variable first order equation
Don't be too wordy, cut to the topic directly, the answer is short and to the point!
The difference between the two kinds of equations is that there is an unknown number in the first-order equation of one variable, but there are two unknowns in the second-order equation of two variables. Moreover, the solution of the first-order equation of two variables is not unique. It is generally required to solve the first-order equations of two variables instead of a single equation. The connection is that when solving the first-order equations of two variables, the substitution elimination method is generally used (that is, using one equation in the equations, the elimination method is used), To express an unknown with another unknown, and then substitute it into another equation) to transform an equation in the system of equations into a linear equation of one variable to solve. Is there any doubt?
The axis of symmetry of the parabola y = 1 / 2 x squared + 3x + 5 / 2 is ()
The symmetry axis of parabola y = 1 / 2 x square + 3x + 5 / 2 is (straight line x = - 3)
Y = 1 / 2x square + 3x + 5 / 2 = 1 / 2 (X & # 178; + 6x + 5) = 1 / 2 (x + 3) &# 178; - 2. So the axis of symmetry of the parabola is x = 3
The square + 3x + 5 / 2 of the parabola y = 1 / 2 x can be reduced to y = 1 / 2 (X & # 178; + 6x + 5) = 1 / 2 (x + 3) &# 178; - 2, so the symmetry axis of the square + 3x + 5 / 2 of y = 1 / 2 x is x = - 3
What is the relationship between bivariate linear equation and univariate linear equation
Urgent need
They're just different equations
It is necessary to say that the solutions of bivariate linear equation are expressed by univariate linear equation respectively; univariate linear equation is the final expression of the solutions of all equations. Bivariate linear equation is composed of two univariate linear equations
Binary has two unknowns; one has only one
No,
Binary has two unknowns; one has only one. There is only one equation for one variable and two for two variables
Element refers to the number of unknowns, and secondary refers to the idempotent of unknowns. For example, x + y + 5 = 0 is a quadratic equation of two variables, while 2x-5 = 0 is a quadratic equation of one variable
The number of intersections of parabola y = x & # 178; - 3x + m and X axis is
A. There is and only one intersection B. There are two intersections C. There is no intersection D. It is uncertain
x²-3x+m=0
△=（-3）²-4m
=9-4m
Because 9-4m can be greater than, equal to or less than zero
So the number of intersections can not be determined
x²-3x+m=0
Discriminant △ = 9-4m
∵ m is an unknown number
The answer is d
The number of intersection points of parabola y = x & # 178; - 3x + m and X axis is X & # 178; - 3x + M = 0. There are several real roots,,, Δ = 9-4m,,, which can not be determined