1. Equation about x (x+m) 2 = m If there is a real number solution, the condition that should be met is ().

2. In the equation AX 2 (the square of X) +bx+c=0(a is not equal to 0), when B 2-4ac = 0, the solution of the equation is ().

3. If the equation kx 2 (the square of x) -6x+ 1=0 has two real roots, then the value range of k is ().

4. If the unary quadratic equation 2x (kx-4)-x 2+6 = 0 has no real root, then the minimum integer value of k is ().

5. If the unary quadratic equation (k- 1) x 2+2x+ 1 = 0 has a real root, what conditions should k satisfy?

Answer: 1. The equation about X (x+m) 2 = m has a real number solution, so the condition that should be satisfied is ().

Solution: Because (x+m) 2 ≥ 0,

Therefore, the equation (x+m) 2 = m about x has a real number solution, and the condition that should be satisfied is m≥0.

2. In the equation AX 2+BX+C = 0 (A is not equal to 0), when B 2-4ac = 0, the solution of the equation is ().

Solution: when b 2-4ac = 0

The solution of equation x 1=x2=-b/(2a)

3. If the equation KX 2-6x+ 1 = 0 has two real roots, then the value range of k is ().

Solution: if the equation KX 2-6x+ 1 = 0 has two real roots.

Then k should satisfy: k≠0 and △ = b 2-4ac = (-6)? -4k≥0

So k≤9, k≠0

4. If the unary quadratic equation 2x (kx-4)-x 2+6 = 0 has no real root, then the minimum integer value of k is ().

Solution: 2x (kx-4)-x 2+6 = 0.

(2k- 1)x？ -8x+6=0

Because the quadratic equation of one variable has no real root.

So b 2-4ac is less than 0,2k-1≠ 0.

That is (-8)? -4 (2k- 1) * 6 < 0 and k≠0.5.

So k > 1 1/6 and k≠0.5.

So k > 1 1/6.

So the smallest integer value of k is 2.

5. If the unary quadratic equation (k- 1) x 2+2x+ 1 = 0 has a real root, what conditions should k satisfy?

Solution: One-variable quadratic equation has real roots, and k should be satisfied.

2? -4(k- 1)≥0 and k- 1≠0.

The solution is k≤2, k≠ 1.

Give you five!