Xsquare - 2x + 1 greater than or less than 0

Xsquare - 2x + 1 greater than or less than 0

(x-1)^2≥0
See x ∈ R

The square of 2x-x is greater than or equal to - 1 Solving inequalities

2x-x^2≥-1
-x^2+2x+1≥0
x^2-2x-1≤0
(x-1)^2-2≤0
-√2≤x-1≤√2
-√2+1≤x≤√2+1
So x ∈ [1 - √ 2, √ 2 + 1]

X is a number greater than zero. When x, the square of X is less than 2x

If x ^ 2 = 2x, x = 2 (x > 0) is obtained

System of mathematical inequalities: X (square of X + 1) > equal to (x + 1) (square of X - x + 1), 1-2x is greater than the integer solution of (3x-9)

1. X * (x ^ 2 + 1) > = (x + 1) * (x ^ 2-x + 1) left formula = x * (x ^ 2 + 1) = x ^ 3 + X right formula = (x + 1) * (x ^ 2-x + 1) = x ^ 3 + 1 because of the left formula > = right formula, then x ^ 3 + x > = x ^ 3 + 1, then x > = 1, the integer solutions are 1,2,3 2. If 1-2x > 3x-9, then 1 + 9 > 3x-2x, then XX > = 1, then x is 1,2,3,4,5,6,7,8,9

If y + X is a real number, y + X is a real number

Square of (2x + 3) + Y - 6y + 9 = 0
Square of (2x + 3) + (Y-3) under root sign = 0
If one is greater than 0, the other is less than 0
So both are equal to zero
So 2x + 3 = 0, Y-3 = 0
x=-3/2,y=3

Let X and y be real numbers and (root X-2) + the square of Y + 6y + 9 = 0, find the value of 2x-y

√(x-2)+(y+3)²=0
If one is greater than 0, the other is less than 0
So both are equal to zero
So X-2 = 0, y + 3 = 0
x=2,y=-3
2x-y=4+3=7

Given that x, y are real numbers and (root 2x + 3) + the square of Y - 6y + 9 = 0, find the value of 2x-y

Because (root 2x + 3) + the square of Y - 6y + 9 = √ (2x + 3) + y ^ 2-6y + 9 = √ (2x + 3) + (Y-3) ^ 2 = 0 because X and y are real numbers, then √ (2x + 3) and (Y-3) ^ 2 are greater than or equal to 0, so x = - 3 / 2, y = 3, so 2x-y = - 3-3 = - 6

If the real number x satisfies x-radical (1 / 9-3 / 2x + x-squared) = 1 / 3, find the value range of X

x-√(x-1/3)²=1/3
x-|x-1/3|=1/3
∴|x-1/3|=(x-1/3)
∴x-1/3>=0
That is, x > = 1 / 3

Given that the value of the algebraic expression x2 + X + 3 is 8, then the value of the algebraic expression 9-2x2-2x is______ .

∵ the value of x2 + X + 3 is 8, that is, X2 + X + 3 = 8, X2 + x = 5,
∴9-2x2-2x，
=9-2（x2+x），
=9-2×5，
=-1．
So the answer is: - 1

Given that the value of the square of the algebraic formula x + X + 3 is 6, find the value of the square of the algebraic formula 2x + 2x-9 emergency

Because x 2 + X + 3 = 6
So x 2 + x = 3
So 2x? 2x + 9 = 2 (x? 2 + x) - 9
=-3