We define a new operation: a * b = the second power of a - B + AB to find the value of (- 2) * [2 * (- 3)] I need a procedure

We define a new operation: a * b = the second power of a - B + AB to find the value of (- 2) * [2 * (- 3)] I need a procedure

Original formula = 4-2 * (- 3) - 2 times 2 * (- 3) = 4-3 times (4 + 3-6) = 1
Find the range of function y = Log1 / 2 (- 2x ^ 2 + 5x + 3)
-I will find the range of 2x ^ 2 + 5x + 3, which is the domain of the original function, but how to find the range after that?
Because the base number is less than 1, the logarithm function is a decreasing function. When the real number is infinite and approaches 0, the value of the function approaches positive infinity. When the value of the quadratic function is the maximum, y is the minimum, and I simplify it to 3 + 2log1 / 27
First, define the formula, then compare the endpoint
Let t = - 2x & # 178; + 5x + 3,
t> 0 is x ∈ (- ∞, - ½) ∨ (3, + ∞)
In this case, t ∈ (0, 49 / 8]
Using the combination of number and shape, the image of y = Log & # 189; t is drawn
Take the part of (0, 49 / 8]
Then y ∈ [3 + Log & # 189; 49, + ∞)
A → B = (a-b) AB Power, a ← B = (a + b) AB Power, find the value of (2 → 3) + (2 ← 3)!
2→3=(2-3)^6=(-1)6=1
2←3=(2+3)^6=15685
So (2 → 3) + (2 ← 3) = 1 + 15685 = 15686
The range of function y = 5x + 3 / 2x-3,
It is known that y (2x-3) = (5x + 3), that is, 2xy-3y-5x-3 = 0, that is, x = 3 (y + 1) / (2y-5), so y ≠ 5 / 2
Set 0
loga(a^2x-2a^x-2)0；
That is: (T-3) (T + 1) > 0, get: T3;
Because t = a ^ x > 0, T3 is omitted
That is: A ^ x > A ^ (loga3)
Because 0
loga(a^2x-2a^x-2)0
Y3
Because y = a ^ x > 0
So, Y3, that is, a ^ x > 3
X
What is the range of the function y = 2x-3 + √ 13-4x?
y=2x-3+√(13-4x)
Let t = √ (13-4x), then t ≥ 0, and the solution is x = - T & # / 4 + 13 / 4,
So, y = 2 (- t-178 / 4 + 13 / 4) - 3 + T = - 1 / 2 (t-1) & 178; + 4
Because t ≥ 0, the range is (- ∞, 4]
We know the system of inequalities x-m ≥ 2n, 2x + m about X
The solution is x ≥ 2n + m from x-m ≥ 2n
From 2x + M
That is, M + 2n ≤ M
Find the range of function (1) y = 2 - √ (- x ^ 2 + 4x) (2) y = 2x-3 + √ (13-4x)
(1) First find - x ^ 2 + 4x
(2) Let t = √ (13-4x) be transformed into quadratic function
How to solve this inequality?
The same base and 0.3 ＜ 1, so the index before ＜ after, and then factorization, that is, (x-1) (3x-1) ＜ 0, the solution is 1 / 3 ＜ x ＜ 1
Given the function f (x) = log2 (3 + 2x-x ^ 2), find the range of the function
f(x）=log2（3+2x-x^2),
g(x) = -x^2+2x+3 = - (x-1)^2 + 4
g(x)＞0
∴0 ＜ g(x) ≤ 4
-∞ ＜ log2（3+2x-x^2) ≤ 2
Range (- ∞, 2]