1.｛2X-1＜7,2X+5≤3（X+2） 2 2-X 1 ＿＿=1+＿＿ 3+X X+3 The second question is 2-X 1 ＿＿=1+＿＿ 3+X X+3

1.｛2X-1＜7,2X+5≤3（X+2） 2 2-X 1 ＿＿=1+＿＿ 3+X X+3 The second question is 2-X 1 ＿＿=1+＿＿ 3+X X+3

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On your question, the first question means to find the range of X: X is greater than or equal to - 1 and less than 4 (1)-
Several mathematical problems
1/2（7-4x)=6+3/2(4x-7)
1-y/3 -y=3- y+2/4
1.5x-1/3-x/0.6=0.5
1/2（7-4x)=6+3/2(4x-7)
-1/2（4x-7)=6+3/2(4x-7)
3/2(4x-7)+1/2（4x-7)=-6
2（4x-7)=-6
4x-7=-3
4x=7-3
4x=4
X=1
(1-y)/3 -y=3- (y+2)/4
12*(1-y)/3 -12*y=12*3- 12*(y+2)/4
4(1-y) -12y=36-3(y+2)
4-4y-12y=36-3y-6
4-16y=30-3y
16y-3y=4-30
13y=-26
y=-2
1.5x-1/3-x/0.6=0.5
3x/2-1/3-x/(3/5)=1/2
3x/2-1/3-x*5/3=1/2
3x/2-1/3-5x/3=1/2
6*3x/2-6*1/3-6*5x/3=6*1/2
9x-2-10x=3
9x-10x=3+2
-x=5
x=-5
If the function f (x) = 4x2-kx-8 is monotone on [5,8], then the value range of K is______ .
According to the properties of quadratic function, we know that the axis of symmetry x = K8 is a monotone function in [5,8], then the axis of symmetry can not be in this interval 〈 K8 ≤ 5, or K8 ≥ 8, K ≤ 40, or K ≥ 64. So the answer is: (- ∞, 40] ∪ [64, + ∞)
The inverse function of y = LG (x-1)?
x-1 = 10^y,x = 1+10^y
The inverse function is expressed as y = 1 + 10 ^ x, X ∈ R
The known functions f (x) = asin (KX + π / 3) and φ (x) = btan (KX - π / 3), k > 0
If the sum of their minimum positive periods is 3 π / 2, and f (π / 2) = φ (π / 2), f (π / 4) = - √ 3 φ (π / 4) + 1, the analytic expressions of F (x) and φ (x) are obtained
Let f (π / 2) = φ (π / 2), asin (π + π / 3) = btan (π - π / 3), - √ 3A / 2 = - √ 3b, a / 2 = B. Let f (π / 4) = - √ 3 φ (π / 4) + 1, asin (π / 2 + π / 3) = - √ 3btan (π
The definition and value range of (2x-x & sup2;) power of y = 1 / 3
The definition field of y = (1 / 3) ^ (2x-x & # 178;) is r, y = (1 / 3) ^ (2x-x & # 178;) = (1 / 3) ^ [- (x-1) & # 178; + 1], ∫ (x-1) & # 178; + 1 ≤ 1, ■ (1 / 3) ^ [- (x-1) & # 178; + 1] ≥ (1 / 3) & # 185;, that is, y ≥ 1 / 3, so the value field of y = (1 / 3) ^ (2x-x & # 178;) is [1 / 3, + ∞)
Given the function f (x) = lgkx − 1x − 1. (K ∈ R and K ＞ 0). (1) find the domain of definition of function f (x); (2) if function f (x) increases monotonically on [10, + ∞), find the range of value of K
(1) It is concluded that KX − 1x − 1 ＞ 0, i.e. (x-1) (kx-1) ＞ 0, ∵ K ＞ 0, ∵ should be divided into three cases. When 0 ＜ K ＜ 1, the domain of definition is (− ∞, 1) ∪ (1K, + ∞), when k = 1, the domain of definition is (- ∞, 1) ∪ (1, + ∞), when k ＞ 1, the domain of definition is (− ∞, 1K) ∪ (1, + ∞); (2
The value domain definition domain of x2-2x power of y = a
The value domain definition domain of x2-2x power of y = a
The domain is r
x2-2x
= x2-2x+1 -1
= (x-1) ^2 -1 ≥ -1
When a > 1
The definition field of x2-2x power of a is y ≥ 1 / A
When 0
If the domain of function f (x) = LG (KX ^ 2 + X + 1) is r, then the value range of K is r
The definition field of function f (x) = LG (KX ^ 2 + X + 1) is: KX ^ 2 + X + 1 > 0
That is to say, the solution set of equation KX ^ 2 + X + 1 > 0 is r
KX ^ 2 + X + 1 > 0 is constant on R
i) When k = 0, x + 1 ＞ 0, X ＞ - 1, it is obviously not suitable
II) when k ≠ 0, then K ＞ 0, Δ = 1-4k ＜ 0, the solution is k ＞ 1 / 4
In conclusion, the range of K is (1 / 4, + ∞)
If the definition field of function f (x) = LG (KX ^ 2 + X + 1) is r
Then KX ^ 2 + X + 1 > 0
Then K ＞ 0, and (denotes that the function y = KX ^ 2 + X + 1 opens upward)
Δ = B & # 178; - 4ac = 1-4k ＞ 0 (means that the function y = KX ^ 2 + X + 1 has no intersection with the X axis, so they are all above the X axis, y ＞ 0)
K < 1 / 4
Then 0 ＜ K ＜ 1 / 4
The range of function y = log2 (1 / x2-2x + 5) ==
If x is on R, then x ^ 2-2x + 5 is on [4, + infinity]
Take the reciprocal as (0,1 / 4]
Taking logarithm is (- infinity, - 2]
The answer is (- infinity, - 2]