If the equation about x m 2 x squared 2 x 1 x m 0 has only one real root, 10

If the question is not copied incorrectly, then:

The equation about x (m-2) x squared -2 (x-1) x + m=0 has only one real root.

So. m-4) x 2 +2x +m = 0 has only one real root.

So. 4 - 4m(m-4) =0

m^2 +4m-1 = 0

m2 +4m = 1 m2 = 1-4m m = -2 + root number5 m = -2 - root number 5 equation mx squared - (m+2)x+(4-m)=0m+2) 2 - 4m(4-m).

m^2+4m+4 +4(m^2-4m)

1+4 + 4(1-4m -4m)

5+ 4(1-8m)

5+4 - 32 (-2 + root number 5).

9 + 64 -32 root number 5

So the equation has two unequal real roots.

Solution: There is only one real root from (m-2)x-2(x-1)x+m=0.

That is, (m-4)x+2x+m=0 has only one real root.

then =2 -4 (m-4) m=0

i.e. 16m-4m = -4

According to the equation mx -(m+2)x+(4-m)=0, we can see that:

-m+2）]²4×m×(4-m)

m+2）²-16m-4m²）

m+2）²-4）

m+2）²+4

Because (m+2) 0

Then =(m+2) +4 4

So the equation has two unequal real roots.

I think there is a problem with your first equation, otherwise, there are 3 m values to be found) because the equation (m-2) x -2 (x-1) x + m = 0 that is, (m-4) x +2x + m = 0 has only one real root, then the equation δ = 4-4m(m-4) = 0, that is, m -4m-1 = 0 can find m=2 5

Or m-2 = 0 i.e. m = 2

Then substitute m=2 5 into the original equation to see if there is only one real root.

The equation mx -(m+2)x+(4-m)=0, where δ=(m+2) -4m(4-m)=5m -12m+4 is brought into m.

δ≠ 0, but m=2 δ=0

So, find the value of x.

The first equation isn't right......

There are two cases if it is not (m-2) x 2-2 (m-1) x + m = 0.

1).m-2=0, i.e., m=2, the equation is -2x+2=0 and there is only one x=1, and the equation is 2x2-4x+2=0, i.e., x2-2x+1=0, and the root is x1=x2=1

2）.m-2≠0 is m≠2, then =[-2(m-1)] 2-4(m-2)m=0, no solution, no validity.

So there are two real roots, x1=x2=1

The first floor is embarrassing. . .

From the problem, if the equation has only one real root, then m=2, then the equation is: -2(x-1)+2=0 =>x=2 satisfies the requirements, then equation 2 is: 2x 2-4x+2=0 => x -2x+1=0 => x=1 , there are two equal real roots.

As an additional note, delta=0 does not have only one root, but two equal real roots.

2x is a flat imitation of pre-ignition + x-2m + 1 = 0 with a real number of root and void.

1^2-4×2×（1-2m)≥0

1-8+16m≥0

16m≥7m≥7/16

If b squared knows that -4ac>=0, there will be a real root, and the corresponding parameter in the equation will be substituted:

b square - 4ac> = (2m) square - 4 * 1 * 4 (m-1) 4m square - 16m + 16

2m-4) square >=0

Therefore, the equation must have a real root.

(m-2) x 2-2 (m-1) x + m + 1 = 0, note that the question does not say that the equation must be a quadratic equation, that is, m may be 2

1. The equation has only one real root, that is, when the equation is a one-dimensional equation, m=2, at this time, -2x=-3, and the solution is x=3 2

When m≠2 and the equation is quadratic, it should be satisfied.

4(m-1) 2-4(m-2)(m+1)=4(3-m)=0, i.e., m=3, the equation is x2-4x+4=0, and the solution is x1=x2=2

2. The equation has no real root, and it can be seen from the above question that if it is a one-dimensional one-dimensional equation, it must have a real root, so it can only be a quadratic equation, that is, m≠2, then.

4(m-1) 2-4(m-2)(m+1)=4(3-m)<0, and the solution is m>3

3. The equation has two unequal roots, and m≠2 at this time

4(m-1) 2-4(m-2)(m+1)=4(3-m)<0, and the solution is m>3

In summary: m<3 and m≠2

Untie; 1 has only one real root, then.

1) When m-2=0, that is, m=2, the original equation is the unary linear equation -2(2-1)x+2+1=0

The solution yields x=3 2, which satisfies the condition.

2) When m-2 is not 0, the equation has only one real root.

Then =4(m-1) 2-4(m-2)(m+1)=0 gives m=3

2.Equations do not have real roots.

Then =4(m-1) 2-4(m-2)(m+1)=0 gives m=3

3.The equation has two unequal real roots.

Then =4(m-1) 2-4(m-2)(m+1)=0 gives m=3

When an equation has only one real root, it is divided into 1 and is a unary equation, and m is equal to 2. Then there are the two cases of discriminant = 0.

There is no real root discriminant less than 0

Two unequal real root discriminants are greater than zero.

1) The coefficient of the quadratic term is zero or the discriminant formula is equal to 0

2) The coefficient of the quadratic term is not 0 and the discriminant formula is less than 0

x^2+mx+1=0……（1）

x^2-x-m=0……（2）

1) Vertical Clans - (2) De.

m+1)x+m+1=0

m+1)x=-(m+1)

Because there is only one real root with the same remainder, so m+1 is not equal to 0x=-1.

1)^2+1-m=0m=2

The original equation is. Equation about x, and.

There are two unequal real roots.

Voltain digging M-2

0, calendar =4-4(m-2) 0

m 3 and. m≠2

m is the largest integer in its range of values.

Nucleus-deficient m = 14 m -1 m = 4 - 1 = 3

[-m+(m+1)]x+2-m=0

x=m-2 for the equation to be solved as a common real number root.

x^2-mx+2=0

2-(m/2)^2=0

m=2 2 is x 2-mx+2=0 has m, and m is the same real root m value x 2-(m+1)x+m=0

xm-[(m+1)/2]^2=0

m^2+2m+1-4m=0

m-1)^2=0

m=1 is x 2-(m+1)x+m=0 has m and the same value of m as the root of the real number.

x-squared -mx+2=0 and x-squared -(m+1)x+m=0 have only one identical real root.

m²-8=0

m²=8m+1)²-4m=0

m²-2m+1=0

8-2m+1=0

m=9/2

b^2-4ac=(m-2)^2-4*(-2m)=(m+2)^2

After the root number, a hail Qing Dingyuan is a real number, so there must be a real number root.

I hope you are satisfied.

(1) The equation about x: mx squared + 2 (m + 1) x + m = 0 has two real number roots: δ = 4 (m + 1) -4m = 8m + 4 0 m -1 2

2) Let the two roots be a and b

Then according to the Vedic theorem.

a+b=-2(m+1)/m,ab=1

a²+b²=(a+b)²-2ab

4(m+1)²/m²-2

a²+b²=6

4(m+1)²/m²-2=6

m+1)²/m²=2

m²-2m-1=0

The solution of m=1+2 or m=1-2 is consistent with δ>0 m=1+ 2 or m=1-2