Find the real number k so that the roots of the equation KX ^ 2 + (K + 1) x + k-1 = 0 about X are integers

Find the real number k so that the roots of the equation KX ^ 2 + (K + 1) x + k-1 = 0 about X are integers

kx^2+(k+1)x+(k-1)=0
When k = 0, x = 1
When k is not equal to 0
The equation is reduced to
x^2+(1+1/k)x+(1-1/k)=0
Discriminant = (1 + 1 / k) ^ 2-4 (1-1 / k) = 1 + 1 / K ^ 2 + 2 / K-4 + K / 4
=1/k^2+6/k-3=(1/k+3)^2-12
Let 1 / K + 3 = a,
a^2-12=b^2
(a+b)(a-b)=12
Because a and B are all integers, there are:
A + B = 2, A-B = 6, a = 4, B = - 2, k = 1
A + B = 6, A-B = 2, the solution is a = 4, B = 2
A + B = - 2, A-B = - 6, a = - 4, B = 2, k = - 1 / 7
A + B = - 6, A-B = - 2, a = - 4, B = - 2
When k = 1,
The equation is
X ^ 2 + 2x = 0, two are respectively 0, - 2, which are integers
When k = - 1 / 7
The equation is reduced to
X ^ 2-6x + 8 = 0, two are 2 and 4, both are integers
therefore
K can be taken
-1/7,0,1
I wish you a happy study

On the two real roots of the equation x2-kx + 6x = 0 of X are greater than 1, the value range of real number k is obtained

If the two real roots of the equation x2-kx + 6x = 0 are greater than 1, then it is necessary to satisfy the following conditions: △≥ 0 (x1-1) (x2-1) > 0 (x1-1) + (x2-1) > 0 △ = B ^ 2-4ac = k ^ 2-4 * 1 * 6 ≥ 0 (1) (x1-1) (x2-1) = x1x2 - (x1 + x2) + 1 = 6-k + 1 > 0 (2) (x1-1) + (x2-1) = X1 + x2-2 = K-2 > 0 (3) k ≥ 2 √ 6 or K from (1)

1 of 8 + 1 of 24 + 1 of 48 +. + 1 of 248 + 1 of 360

1 of 8 + 1 of 24 + 1 of 48 +. + 1 of 248 + 1 of 360
=1/2×（1/2-1/4+1/4-1/6+…… +1/18-1/20）
=1/2×（1/2-1/20）
=1/2×9/20
=9/40

Change the following determinant into triangle determinant, and find its value 2-53113-13011-5-1-42-3

2 -5 3 1 1 3 -1 3 0 1 1 -5-1 -4 2 -3r1 + 2r4,r2 + r40 -13 7 -5 0 -1 1 0 0 1 1 -5-1 -4 2 -3r1 + 13r3,r2 + r30 0 20 -70 0 0 2 -5 0 1 1 -5-1 -4 2 -3r1 - 10r20 0 0 -20 0 0 2 -5 0 1 1 -5-1 -4 2 -3r1r4,r2r3...

Three eighths divided by 3 means to divide three eighths into () parts equally, and find out the number of () parts, that is, find out the number of () parts of three eighths, so three eighths divided by 3 = three eighths multiplied by ()

Given proposition p: for any x ∈ [1,2], x ^ 2-A ≥ 0, and proposition q: there exists x ∈ R, x0 ^ 2 + 2ax0 + 2 = 0, if proposition "P and Q" is true,
Find the value range of real number a

The proposition "P and Q" is true, that is, P is true and Q is true
x²-a≥0
x²≥a
X ∈ [1,2], then x & # 178; ∈ [1,4], if the inequality holds, a ≤ 1
X0 & # 178; + 2ax0 + 2 = 0, the equation has real roots, and the discriminant △≥ 0
(2a)²-4×1×2≥0
a²≥2
A ≥ √ 2 or a ≤ - √ 2
In conclusion, a ≤ - √ 2

Simple calculation of 2.5 × 7.8 + 7.4 × 7.8 + 0.78

2.5＊7.8+7.4＊7.8+0.78=7.8（2.5+7.4+0.1）=7.8*10=78

1/3x-(1/2x-4y-6z)+3(-2z+2y)
Simplify the whole process
Evaluate x = 6, y = 1, z = - 2

1/3x-(1/2x-4y-6z)+3(-2z+2y)
=1/3x-1/2x+4y+6z-6z+6y
=-1/6x+10y
=-1/6×6+10×1
=9

How to simplify the calculation: 840 divided by 35

=840÷7÷5
=120÷5
=24

Given that the equation of line L passes through point a (3,4), its inclination angle is twice of that of line 2x-y + 1 = 0, the equation of line L is solved

Let the inclination angle of the line 2x-y + 1 = 0 be a
Then TGA = 2
The inclination angle of line L is 2A
tg(2a)=2tga/(1-(tga)^2)=4/(1-2^2)=-4/3
Let the equation of line l be y = - 4 / 3x + B
Point a (3,4) is substituted into 4 = - 4 + B = 8
So the equation of line L is y = - 4 / 3x + 8 4x + 3y-24 = 0