When k is an integer value, the square of the quadratic equation KX with respect to X is - 4x + 4 = 0 and the square of X is - 4kx + 4K is - 4k-5= fast

When k is an integer value, the square of the quadratic equation KX with respect to X is - 4x + 4 = 0 and the square of X is - 4kx + 4K is - 4k-5= fast

Is the equation like this?
kx^2-4x+4=0
x^2-4kx+4k^2-4k-5=0

When m is an integer, the solutions of the quadratic equations mx2-4x + 4 = 0 and x2-4mx + 4m2-4m-5 = 0 are integers?

If the quadratic equation mx2-4x + 4 = 0 and x2-4mx + 4m2-4m-5 = 0 of X have solutions, then m ≠ 0, then m ≠ 0, m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ m ∫ M

When k is an integer, the following equations of X: kx-4x + 4 = 0 and x-4kx + 4k-4k-5 = 0 are integers

From the question: x-4kx + 4k-4k-5 = 0 (1) kx-4x + 4 = 0 (2) 1: when k = 0, (1) equation is: X-5 = 0, root is not an integer, so K ≠ 0 2: when k ≠ 0, from (1) to (x-2k) = 4K + 5, we can get: K ≥ - 5 / 4 x = radical (4K + 5) + 2K or x = 2K radical (4K + 5) from (2) to △ = radical (16-16k) ≥ 0, so K ≤ 1, x = (4 + radical (16-16k)) / 4 = 1 + radical (1-k) or x = 1-radical (1-k) when k = 1 Satisfy the condition when k = - 1 does not meet the condition, so when k = 1, the roots of the two equations are integers!

On the univariate quadratic equation KX ^ 2 - (4K + 1) x + 2K ^ 2-1 = 0 of X, when k is a value, the equation has two unequal real roots
Is the discriminant of root: - (4K + 1) ^ 2-4k * (2k ^ 2-1) > 0

B ^ 2-4ac = (4K + 1) ^ 2-4k (2k ^ 2-1) = (4K + 1) ^ 2-8k ^ 3 + 4K + 1-1 = (4K + 1) (4K + 2) - (8K ^ 3 + 1) = 2 (2k + 1) (4K + 1) - (2k + 1) (4K ^ 2-2k + 1) = (2k + 1) (- 4K ^ 2 + 10K + 1) > 0 and - 4K ^ 2 + 10K + 1 > 0, the solution of (K-5 / 4) ^ 2 is obtained

Rational number: (13 / 3-7 / 2) * - 2 - (7 / 2-13 / 3) * 2=

Put the minus sign in (- 2) in the preceding bracket, that is:
(13/3-7/2)*(-2)-(7/2-13/3)*2=(7/2-13/3)*(2)-(7/2-13/3)*2=0

Rational number: (2 / 7-13 / 3) * - 2 - (7 / 2-13 / 3) * 2=

It's 205 / 21, which is 205 / 21

-How many rational numbers are there between 2.4 and + 3.5?

There are innumerable rational numbers, because rational numbers include integers and fractions, which are actually finite decimals and infinite circular decimals. Let alone - 2.4 to + 3.5, there are innumerable rational numbers from + 3.4 to + 3.5: 3.41, 3.4111111, and so on

Prove that 2 is a rational number

Any number that can be written as N / M (m, n is an integer) is rational, 2 = 2 / 1, so 2 is rational

How to prove that root 3 is not a rational number?

Certification:
Let root 3 = P / Q, that is, rational number, and (P, q) = 1
3=p^2/q^2
So p is a multiple of 3
Let P = 3K
3=9k^2/q^2
3q^2=9k^2
therefore
q^2=3k^2
So q is also a multiple of 3
And (Q, P) = 1, so root 3 is irrational
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But I want to know how to prove that root 4 is a rational number

a. B and C are rational numbers. The square root of a plus the square root of B equals C. how to prove that the square root of a and the square root of B are both rational numbers!

The Convention sqrt (x) is to calculate the square root of X
sqrt(a)+sqrt(b)=c
Obviously, if C = 0, otherwise a = b = 0, of course, sqrt (a) = sqrt (b) = 0 is a rational number
If C is not 0,
sqrt(a)=c-sqrt(b)
Square on both sides
a=c^2-2*c*sqrt(b)+b
So: sqrt (b) = (C ^ 2 + B-A) / (2C)
Obviously, the right is a rational number, so the left must also be a rational number··