On X equation (a-6) * x ^ 2-8x + 6 = 0 with real roots, what is the maximum value of integer a?

On X equation (a-6) * x ^ 2-8x + 6 = 0 with real roots, what is the maximum value of integer a?

∵ the equation has real roots
That is (- 8) ^ 2-4 (a-6) × 6 ≥ 0
64-24a＋144≥0
-24a≥-208
a≤26/3
The maximum value of a is 26 / 3

It is known that the equation 2kx + 3x − 1-7x2 − x = 4kx about X has exactly a real solution. Find the value of K and the solution of the equation

If k = 0, 3x-7 = 0, x = 73, if K ≠ 0, we know that △ = (3-4k) 2-8k (4k-7) = 0, and the solution is K1 = 94, K2 = - 14, when K1 = 94, X1 = x2 = 23, when K2 = - 14, x1 = x2 = 4, if the equation has two unequal real roots, then one of them is an increasing root, when X1 = 1, k = 2, X2 = 14, when X1 = 0, k = 74, X2 = 87

Given that real numbers a and B satisfy a2-6a + 4 = 0 and b2-6b + 4 = 0 respectively, and a ≠ B, then the value of Ba + AB is ()
A. 7B. -7C. 11D. -11

According to the meaning of the question: A and B are two parts of the equation x2-6x + 4 = 0, a + B = 6, ab = 4, then the original formula = (a + b) 2 − 2abab = 36 − 84 = 7

If x = ay = B is a solution of the equation 2x + y = 1, then 6A + 3B + 2=______ ．

Substituting x = ay = B into the equation 2x + y = 1, we get: 2A + B = 1, then 6A + 3B + 2 = 3 (2a + b) + 2 = 3 × 1 + 2 = 5, so the answer is: 5

If the power of x minus log a X of inequality 2 is less than 0, it holds when x belongs to (0,1 / 2)

Because 2 ^ x is always positive, so logax is also positive at (0,1 / 2), so 0 ＜ a ＜ 1, so f (x) = 2 ^ 2 is an increasing function and G (x) = logax is a decreasing function. Let the two function images intersect P (m, n), and m ≥ 1 / 2 according to the meaning of the problem. Let x = 1 / 2, 2 ^ x = logax, then 2 ^ (1 / 2) = loga (1 / 2), then a = 2 ^ (- 2 / 2), so the

2 hectares = () square meters 50000 square meters = () hectares 2 square kilometers = () hectares

12000
two point five
3. Two hundred

2 times the square of x minus 8 divided by x minus 2 is factorized

(2x^2-8)/(x-2)
=2(x+2)(x-2)/(x-2)
=2(x+2)
=2x+4

How can the tangent equation of the curve Xe ^ y + XY + y = 1 at x = 0 be derived to the Y power of E

The implicit function is used to obtain the derivative of X on both sides
e^y+xe^y y'+y+xy'+y'=0
y'=-(y+e^y)/(xe^y+x+1)
When x = 0, y = 1 is obtained by substituting it into the original equation
Therefore, y '(0) = - (1 + e ^ 1) / (0 + 0 + 1) = - (1 + e)
The tangent is y = - (1 + e) x + 1

2.79 + 8 / 5 + 7.21 + 160% 6 / 5 x 13 / 1 + 2 / 1 x 3 / 5 + 13 / 1
It's easy to use,
X stands for multiply

2.79 + 8 / 5 + 7.21 + 160%
=(2.79+7.21)+(1.6+1.6)
=10+3.2
=13.2
5 / 6 x 13 / 1 + 1 / 2 x 13 / 5 + 1 / 13
=6 / 5x13 / 1 + 2 / 5x13 / 1 + 13 / 1
=1 x 13 (5 / 6 + 5 / 2 + 1)
=1 / 13 x 3 / 13
=1 / 3

What is the reciprocal of a number equal to 3 / 5 of it plus the sum of 20
What's the sum of the product of a number and its reciprocal plus five sixths?

This number is 20 (1-3 / 5) = 50
So the reciprocal is 1 / 50
The sum is 1 + 5 / 6 × 5 = 5 and 1 / 6