10－【－（－4）】＋（－3）－5 6－【－6－（－3）＋4】 8－【－11－（－4）＋8】 【－8＋（－3）】² ＋【－14＋（－6）】 【－15＋（－4）】² ＋【－19＋（－3）】 A more detailed formula

10－【－（－4）】＋（－3）－5 6－【－6－（－3）＋4】 8－【－11－（－4）＋8】 【－8＋（－3）】² ＋【－14＋（－6）】 【－15＋（－4）】² ＋【－19＋（－3）】 A more detailed formula

10－【－（－4）】＋（－3）－5=10-4-3-5=-26－【－6－（－3）＋4】=6-【-6+3+4】=6-1=58－【－11－（－4）＋8】=8-【－11+4+8】=8-1=7【－8＋（－3）】² ＋【－14＋（－6）】=-11²--20=-121-20=-141【－1...
1. =10-4-3-5
=-2
2. =6-【-6+3-4】
=13
3。 =8-【-11+4+8】
=7
4. =（-11）² -20
=101
5. =（-19）² -22
=339
Four mathematical problems
(1) Radical 13 - | 2 + radical 13|
(2) 2013 power of (- 1) + radical 12 + | radical 3-2|
(3) | - 2 | - (1 + radical 2) + radical 4
(4) The cube root of 1 / 2 x root sign 5 △ 125
Urgent need process. Thank you·
(1) Radical 13 - | 2 + radical 13|
=Radical 13-2-radical 13
=-2
(2) 2013 power of (- 1) + radical 12 + | radical 3-2|
=(- 1) + 2 radical 3 + 2-radical 3
=1-radical 3
(3) | - 2 | - (1 + radical 2) + radical 4
=2-1-radical 2 + 2
=3-radical 2
(4) The cube root of 1 / 2 x root sign 5 △ 125
=1 / 2 x root 5 ÷ 5
=Root of 10 5
1、-2
2. Root 3 + 1
3. 3-radical 2
4. Root 5 / 10
1,＝√13-2-√13
＝-2
2,＝-1+2√3+2-√3
＝1+√3
3,2-1-√2+2
＝3-√2
4,＝√5/2÷5
＝√5/10
Given the function FX = LG (a + 1) x + 1, find the domain of definition
Zero and negative have no logarithm
(a+1)x＞0
There is no solution when a = - 1;
When a ＜ - 1, the domain x ＜ 0;
When a ＞ - 1, the domain x ＞ 0
F (x) = log2 (3-2x-x2)?
The last x2 in brackets refers to the square of X, and the 2 after log is the small corner. This log is not a common logarithm. I hope you can teach me how to solve this kind of problem of calculating the range, not just the answer! I just can't do this kind of problem of calculating the range of logarithm
(1) First, find the range of 3-2x-x2 in brackets. This method can be changed into - (x-1) ^ 2 + 4, that is, the range in brackets is 0, so we know that the range in brackets is 0
If the definition field of function y = LG (KX ^ 2 + X + k) is (1 / 2,2), what is the value set of real number k
The left end of the range is LG (5K + 2) - LG4, the right end is LG (5K + 2), K ∈ (- 2 / 5, + ∞)
The range of y = log2 (x-x2)
What is the range
Domain: - x ^ 2 + x > 0,0
If the definition field of function f (x) = LG (x ^ 2 + KX + 2) is r, then the range of real number k (process,
It means that for the real number Y1 = x ^ 2 + KX + 2, when x takes any real number, the real number Y1 is always greater than 0, so there is no intersection between the image and the X axis, that is, the discriminant is less than 0, then:
k^2-8
The definition field of F (x) = LG (x ^ 2 + KX + 2) is r, which means that x ^ 2 + KX + 2 is always greater than 0, so K ^ 2-4 * 1 * 2
If the function y = x62-2ax has an inverse function when the domain of definition is [1,2], find the value range of a and the inverse function
The title is y = x ^ 2-2ax, the formula is y = (x-a) ^ 2-A ^ 2
So the axis of symmetry is x = a, the vertex coordinates are (a, a ^ 2), and the image passes through the origin
When the domain is [1,2], there is an inverse function, that is, when x = 1, Y1 = 1-2a, when x = 2, y2 = 4-4a,
When AY1, i.e. 4-4a-1 + 2A > 0, 3-2a > 0, A0, there are two cases, one is on the left side of the axis of symmetry, the other is on the right side of the axis of symmetry, so that it can be monotonic
① When [1,2] is on the left side of the symmetry axis, that is, a ≥ 2, then the function decreases monotonically on [1,2], that is, Y1 > Y2, a > 3 / 2,
Sum up a ≥ 2, inverse function is x = a - √ (a ^ 2 + y), domain of definition is [4-4a, 1-2a]
② [1,2] is on the right side of the axis of symmetry, i.e. a ≤ 1, then the function increases monotonically on [1,2], Y2 > Y1, a
The domain of the function y = LG (KX ^ 2 + 4x + K + 3) is r, then the range of the real number k?
A.(-∞,-4)∪(1,+∞),B.(-4,1),C.(-∞,-4),D.(1,+∞)
Study the function y = KX ^ 2 + 4x + K + 3
If k = 0, then y = 4x + 3, the domain of lgY is not R
So K ≠ 0, y = KX ^ 2 + 4x + K + 3 is a quadratic function
If y = KX ^ 2 + 4x + K + 3 is always greater than zero, there must be:
(1) K > 0. (parabolic opening upward)
(2)△=16-4k(k+3)=-4k^-12k+160
K1
Combined with (1), k > 1
The answer is: D
If the function y = x2-2ax + A has an inverse function in the domain [1,3], the range of a is obtained
Additional conditions: | A-3 | + | A-1 | less than or equal to 4
It has inverse function and is monotone function in interval
So the axis of symmetry is either less than 1 or greater than 3
That is: a = 3
Also: | A-3 | + | A-1|
|a-3|+|a-1|≤4
When a ≤ 1, the inequality is changed to - (A-3) - (A-1) ≤ 4, thus 0 ≤ a ≤ 1
When 1