When a satisfies what conditions, the equation lx-1l-lx-3l about X has one solution? There are countless solutions? On the equation lx-1l-lx-3l = a of X

When a satisfies what conditions, the equation lx-1l-lx-3l about X has one solution? There are countless solutions? On the equation lx-1l-lx-3l = a of X

The equation with unknowns but no equal sign is not an equation at all

Find the value of X in the following expressions: LX - radical 2L = radical 5, l2x - 1L = radical 2

LX radical, 2L = radical 5, X - √ 2 = ± √ 5, x = √ 2 + √ 5, or x = √ 2 - √ 5;
L2x-1l = radical 2,2x-1 = ± √ 2,2x = 1 ± √ 2, x = (1 + √ 2) / 2 or x = (1 - √ 2) / 2

Mathematics problem of Junior Three: known a = √ X & # 178;, B = lx-1l (1) if x ≥ 0, simplify a + B (2) if x < 1, simplify a + B (detailed solution process)

1. From the definition of absolute value, step by step, one by one analysis can be done
First of all, the purpose of simplification is to remove the absolute value and make it become a conventional formula. How to remove the absolute value? We need to judge the sign of the number in the absolute value through the condition. If it is positive (or 0), we can directly remove the absolute value; if it is negative (or 0), we need to fill in a negative sign to remove the absolute value
① A - | a | = 0 to get a = | a |, what kind of absolute value is equal to itself? Only non negative numbers can be used, so a ≥ 0
② B + | B | = 0 to get | B | = - B, what kind of number absolute value is equal to its own opposite number? It is a non positive number, so B ≤ 0
③|a|

It is known that 3 (5x + 2) + 5 is less than 4x-6 (x-1)

3（5x+2)+5

Given 2x-4-1, simplify lx-2l + LX + 5L
Given 2x-4-1, simplify I X-2 I + I x + 5 I ("I" denotes absolute value)

2x-40
|x+5|=x+5
So | X-2 | + | x + 5 | = (2-x) + (x + 5) = 7

If the equation 9x + (4 + a) 3x + 4 = 0 about X has a solution, then the value range of real number a is
X is power

If there is a solution, there is a discriminant Δ≥ 0, that is, B ^ 2-4ac ≥ 0, which is simplified to a (a + 8) ≥ 0, so a ≤ - 8 or a ≥ 0
If they are all quadratic, the solution is a ≥ - 7

If the two real roots of the equation x2 + 3x-m = 0 are greater than - 2, then the value range of the real number m is______ ．

Let f (x) = x2 + 3x-m = 0, two real number roots of the equation x2 + 3x-m = 0 are greater than - 2, m satisfies △≥ 0 − 32 ＞ − 2F (− 2) > 0, the solution is − 94 ≤ m ＜ 2, and the value range of real number m is [− 94, − 2). So the answer is [− 94, − 2]

If the equation x3-12x + a = 0 has three different real roots, what is the value range of real number a

Y = x & # 179; - 12x + ay '= 3x & # 178; - 123x & # 178; - 12 = 03x & # 178; = 12x & # 178; = 4x = ± 2 when X1 = - 2, y = (- 2) & # 179; - 12 (- 2) + a = - 8 + 24 + a = 16 + a when x2 = 2, y = (2) & # 179; - 12 (2) + a = 8-24 + a = - 16 + a when x2 = 2, y = (2) & # 179; - 12 (2) + a = 8-24 + a = - 16 + a equation has three different real roots, two extremums, one positive and one negative

If the solution of the equation k-2x = 3x + 2 about X is a positive real number, then the value range of K is?

k-2x=3x+2
The result is: K-2 = 3x + 2x
K-2 = 5x
If both sides divide by 5: x = (K-2) / 5
The solution of ∵ equation is a positive real number
∴(k-2)/5＞0
Multiply both sides by 5: K-2 ＞ 0
Add 2: k > 2 on both sides at the same time
A: the range of K is (2, + ∞)
I don't know how to ask~

The length of one side of the triangle is 10, and the length of the other two sides is the two roots of the equation x2-14x + 48 = 0

10+6+8=24