On the equation of X (M & # 178; - 1) x & # 178; + 2 (M + 1) x + 1 = 0, there are two real roots, so we can find the value range of M

On the equation of X (M & # 178; - 1) x & # 178; + 2 (M + 1) x + 1 = 0, there are two real roots, so we can find the value range of M

4（m+1）^2-4(m^2-1)>=0
4m^2+8m+4-4m^2+4>=0
8m>=-8
m>=-1
Because m ^ 2-1 is not equal to 0
So m is not equal to + 1, - 1
So the value range of M is: M > - 1, and M is not equal to 1
There are two real roots △ 0 or more
⊿=4（m-1)²-4(m²-1)≥0
m²-2m+1-m²+1≥0
-2m≥-2
m≤1
∵m²-1≠0
∴m≠±1
The value range of M is m ＜ 1 and m ≠ - 1 ⊿ = 4 (m-1) &# 178; - 4 (M & # 178; - 1) ≥ 0 should be ⊿ = 4 (M + 1) &# 178; - 4 (M & # 178; - 1) ≥ 0. Finally, the answer is m ＞ - 1... Expansion
⊿=4（m-1)²-4(m²-1)≥0
m²-2m+1-m²+1≥0
-2m≥-2
m≤1
∵m²-1≠0
∴m≠±1
The value range of M is m ＜ 1 and m ≠ - 1. Question: ⊿ = 4 (m-1) &# 178; - 4 (M & # 178; - 1) ≥ 0 should be ⊿ = 4 (M + 1) &# 178; - 4 (M & # 178; - 1) ≥ 0. Finally, the answer is m ＞ - 1, right
If x is any real number, the value of quadratic trinomial x2-6x + C is not less than 0, then the condition of constant C is ()
A. c≥0B. c≥9C. c＞0D. c＞9
∵ x2-6x + C = (x-3) 2 + C-9 ≥ 0, and because (x-3) 2 ≥ 0, so C-9 ≥ 0, so C ≥ 9
Equation 3x + 18 = 5x
3x+18=5x
18=5x-3x
18=2x
X=9
3x-3x+18=5X-3X
18 = 2X
X= 9
3X+18=5X
18=2X
9=X
18=2x
X=9
X=9
3x+18=5x
3x+18=5x
18=5x-3x
18=2x
x=18/2
X=9
In the range of real number x & # 178; - 2 √ 3 x + 3
solution
x²-2√3x+3
=x²-2√3x+(√3)²
=(x-√3)²
If the equation (M & # 178; - 1) x & # 178; - 2 (M + x) x + 1 = 0 about X has real roots, the value range of M is obtained
（m²-1）x²-2（m+x）x+1=0 ∴(m²－3)x²－2mx＋1＝0
(1) If the equation has roots, then it can be transformed into a real equation
(2) When M & # 178; - 3 ≠ 0, i.e. m ≠ ± √ 3, the equation ⊿ = (- 2m) &# 178; - 4 × (M & # 178; - 3) × 1 = 12 ＞ 0 has real roots
M is all real numbers
A:
(m^2-1)x^2-2(m+x)x+1=0
(m^2-1)x^2-2mx-2x^2+1=0
(m^2-3)x^2-2mx+1=0
1) When m ^ 2-3 = 0, the equation has real roots, M = ± √ 3
2) When m ^ 2-3 ≠ 0, the discriminant = (- 2m) ^ 2-4 (m ^ 2-3) = 12 > = 0 holds
There is a real solution to the equation
To sum up, M is the real range R, and the equation has real roots.
If M is a real number, can the square of the algebraic formula - 2m + 6m-5.5 be a positive number? Can it be 0? Can it be a negative number?
-2m square + 6m-5.5
=-2（m-3/2)^2-1
Because (M-3 / 2) ^ 2 > = 0
Then - 2 (M-3 / 2) ^ 2-1
7x minus six is one minus four and two-thirds of the solution
-7 / 6x = 1-4 and 2 / 3
-7 / 6x = - 3 and 2 / 3
6 x 3 / 3
x=22/7
Ha ha, come on!
Decompose factor X & # 178; - 2 √ 3 x + 3 in the range of real number
x²-2√3 x+3
=x²-2√3 x+（√3）^2
=(x-√3)^2
If the equation A & # 178; - x-a (a > 0, a ≠ 0) about X has two roots, then the value range of the real number a?
If the equation a ^ x-x-a = 0 (a > 0, a ≠ 0) about X has two roots, then the value range of real number a?
You must have the wrong number. If the equation AX & # 178; - x-a (a > 0, a ≠ 0) about X has two roots, then the value range of real number a?
This is a quadratic equation about X, so delta > 0 is OK
Have you revised it
If the equation a ^ x-x-a (a > 0, a ≠ 0) about X has two roots, then the value range of real number a?
This is a zero point problem of mathematics compulsory 1 in senior high school
Let f (x) = a ^ x, G (x) = x + A, the equation a ^ x-x-a (a > 0, a ≠ 0) has several roots, that is to see how many intersections these two functions have
It can be seen from the drawing that a must be a > 1 to make f (x) and G (x) have two intersections
This question is wrong. There is only one evaluation of X. please confirm the original question: if the equation a ^ x-x-a (a > 0, a ≠ 0) about X has two roots, then what is the value range of real number a?
When the polynomial 8kx ^ 2 - (8K + 1) x + 2K (K ≠ 0) can not be factorized in the range of real number, the value range of real number k is ()
When (8K + 1) ^ 2-4 * 8K * 2K = 16K + 1 > 0 K > - 1 / 16, the real number can not be factorized