The square + 4x + 8 of cube-6x of polynomial x is written as the power of (x-3)

The square + 4x + 8 of cube-6x of polynomial x is written as the power of (x-3)

Let f (x) = x ^ 3-6x ^ 2 + 4x + 8, then: F (3) = - 7. F '(x) = 3x ^ 2-12x + 4, so: F' (3) = - 5F '' (x) = 6x-12, so: F '' (3) = 6F '' '(x) = 6. So: F' '' (3) = 6. So: F (x) = f (3) + F '(3) (x-3) + F' '(3) / 2 (x-3) ^ 2 + F' '' (3) / 6 (x-3) ^ 3. = - 7-5 (x-3) + 3 (x-3) ^ 2 + (x-3) = - 6
N + 2 power of 4x-n power of 9x + n-1 power of 6x-n-2 power of X
The factorization of n is more than 2
First, we put forward the N-2 power of x ^ (n-2) * (4x ^ 4-9x ^ 2 + 6x-1) and observe 4, - 9, + 6, - 1. We find that the sum of four numbers is 0, so we can definitely decompose an X-1. This is a skill. Then we divide the polynomial division of X-1 by 4x ^ 4-9x ^ 2 + 6x-1, where x ^ (n-2) * (x-1) * (4x ^ 3 + 4x ^ 2-5x + 1) = x ^ (n-2) * (x-1) * (4x
If the line ax & sup2; + 2BY-2 = 0 (a > 0, b > 0) always bisects the circumference of the circle X & sup2; + Y & sup2; - 4x-2y-8 = O, then the minimum value of 1 / A + 1 / B is
If the circle is bisected, the straight line passes through the center of the circle (2,1),
So 2A + 2b-2 = 0, that is, a + B = 1
1/a+2/b=(a+b)(1/a+2/b)
=3 + 2A / B + B / a > = 3 + 2 radical 2 (if and only if B = (radical 2) a, a = radical 2-1 takes equal sign)
That is, the minimum value is 3 + 2 root sign 2
The solution set of inequality | X & # 178; - 2 | 2
From | X & # 178; - 2 | 2
Get - 2
-2 to 2, hope you adopt! Questioning: process
If the line ax + 2BY-2 = 0, (a > 0, b > 0) always bisects the circumference of the circle: x ^ 2 + y ^ 2-4x-2y-8 = 0, find the maximum of a and B
That is, the line passes through the center of the circle (2,1)
2a+2b-2=0;
a+b=1;
a>0,b>0
So 0
Find out the center point of the circle (2.1) and get 2A + 2b-2 = 0
The solution set of inequality 2 Λ (X & #178; + 1) ≤ (1 / 4) Λ (X-2) is?
The equation of a line is x-2y + 5 = 0, and the slope of the line is?
The slope of the straight line ax + by + C = 0 is - A / B, i.e. 1 / 2
4-x & # 178; ≤ 0 to solve quadratic inequality of one variable
And 3x & # 178; - 4 < 0
∵4-x²≤0
∴x²≥4
Ψ x ≥ 2 or X ≤ - 2
The solution set of the original inequality is {x │ x ≥ 2 or X ≤ - 2}
3x²-4＜0
x²＜4/3
∴-2√3/3＜x＜2√3/3
The solution set of the original inequality is {x │ - 2 √ 3 / 3 ＜ x ＜ 2 √ 3 / 3}
4-x & # 178; ≤ 0 X & # 178; - 4 ≥ 0 x ≤ - 2 or X ≥ 2
3x²-4＜0 - 2√3/3
Find the linear equation with known slope 2 and circle x2 + y2-2y-4 = 0 tangent
Let the linear equation be y = 2x + B and be changed to 2x-y + B = 0
Circle: x ^ 2 + (Y-1) ^ 2 = 5, Center (0,1), radius √ 5
The distance from the center of a circle to a straight line is equal to the radius
|0-1+b|/√(1^2+2^2)=√5
B = 6 or - 4
The linear equation is 2x-y + 6 = 0 or 2x-y-4 = 0
The solution set of inequality | x2-x | 2 is______ .
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