Is there such a nonnegative integer m that m ^ 2x ^ 2 - (2m-5) x + 1 = 0 has two real roots? If it exists, find the value of M. if it does not exist, explain the reason

Is there such a nonnegative integer m that m ^ 2x ^ 2 - (2m-5) x + 1 = 0 has two real roots? If it exists, find the value of M. if it does not exist, explain the reason

Delta = (2m-5) ^ 2-4m ^ 2 = - 20m + 25 > = 0, get: M
Delta = (2m-5) ^ 2-4m ^ 2 = - 20m + 25 > = 0, get: M
(1) X ^ 2 + 4x-6 = 0
(2) 3x(x-1)=2-2x
(1) X & # 178; + 4x-6 = 0x & # 178; + 4x = 6x & # 178; + 4x + 4 = 6 + 4 (x + 2) & # 178; = 10x + 2 = ± √ 10, so x = - 2 + √ 10, or x = - 2 - √ 10 (2) 3x (x-1) = 2-2x3x (x-1) = 2 (1-x) 3x (x-1) - 2 (1-x) = 03x (x-1) + 2 (x-1) = 0 (x-1) (3x + 2) = 0, so X-1 = 0, or 3x + 2 = 0, so x = 1, or x =
(1)x²+4x-6＝0
x²+4x+4＝10
(x+2)²＝10
∴x1＝-2+√10,x2＝-2-√10。
(2)3x(x-1)＝2-2x
3x(x-1)＝-2(x-1)
(x-1)(3x+2)＝0
∴x1＝1,x2＝-2/3。
If real numbers x and y satisfy 3x + 4Y = 15, what is the minimum value of the square of X + the square of Y
15=3x+4y=15^2/25=9
The minimum value is 9
Given 1 + X + x2 + X3 + X4 = 0, find 1 + X + x2 + X3 + +The value of X2009______ .
1+x+x2+x3+… +x2009=（1+x+x2+x3+x4）+（x5+x6+… +x9）+… +（x2005+x2006+… +x2009）=（1+x+x2+x3+x4）+x5（1+x+x2+x3+x4）+… +x2005（1+x+x2+x3+x4）=（1+x+x2+x3+x4）（1+x5+x10+… +Since 1 + X + x2 + X3 + X4 = 0, the original formula is 0
Is there such a nonnegative integer m that the quadratic equation m2x2 - (2m-7) x + 1 = 0 with respect to X has two real roots? If so, ask for the value of M and solve the equation. If not, explain the reason
We know that m ≠ 0, △ = b2-4ac = 49-28m ≥ 0, m ≤ 74 and m ≠ 0, so that the equation has two real roots, and the non negative integer of M exists. At this time, M = 1, the equation can be reduced to x2 + 5x + 1 = 0, and the root formula is used to solve it: x = − 5 ± 212. Therefore, there exists such a non negative integer M
Is x ^ 2 + 1 / 5x + 5 = 0 a quadratic equation with one variable
It's a quadratic equation of one variable
Given that the real number x.y satisfies 3x + 4y-15 = 0, then the minimum value of x square plus y square
Y = (15-3x) / 4 is obtained from 3x + 4y-15 = 0
x*x+y*y=x^2+(15-3x)^2/16=(25x^2-90x+225)/16=(x-9)^2*25/16+225/16-25*81/16
When x = 9, the minimum value is - 112.5
If x is a point on the number axis, find the minimum value of | x-2009 | + | x-2007 | + | x + 2008 |
X is not 2009, 2008, 2007, right?
X is not - 2009, 2008, 2007, right?
When x is less than or equal to - 2008,
Original formula = 2009-x + 2007-x-2008-x = 2008-3x,
When x is - 2008, the minimum is 8032
When x is greater than - 2008 and less than or equal to 2007
Original formula = 2009-x + 2007-x + X + 2008 = 6024-x
When x is 2007, the minimum is 4017
When x is greater than 2007 and less than or equal to 2009
Original formula = 2009-x + x-2007 + X + 2008 = 2010 + X
When x is 2007, the minimum is 4017
When x is greater than 2009
Original formula = x-2009 + x-2007 + X + 2008 = 3x-2008
When x is 2009, the minimum is 4019
So when x is 2007, the minimum of the original formula is 4017
Is there such a nonnegative integer m that the quadratic equation MX - (2m-1) x + 1 = 0 with respect to X has two real roots,
(2m-1) - 4m ≥ 0, 4m-4m + 1-4m ≥ 0 - 4m + 1 ≥ 0, m ≤ 1 / 4  M = 0, but when m = 0, the original formula = x + 1 = 0 has only one root, so it does not exist
What is the solution of the quadratic equation x [X-1] = x?
x〔x-1〕=x
x-1=1
X=2
x〔x-1〕=x
Equivalent to
x(x-1)-x=0
x*(x-1-1)=0
x*(x-2)=0
So x = 0 or x = 2
X = 0 or x = 2
0 and 2
X(X-1)=X
X2-X=X
X2-2X=0
X(X-2)=0
So x = 0 or x = 2