(1) known a = - A, simplified 1-A + √ (A-2) & #178; + 2A 2) If real numbers a and B satisfy (a + b-2) &# 178; + √ b-2a + 3 = 0, find the value of 2b-a + 1

(1) known a = - A, simplified 1-A + √ (A-2) & #178; + 2A 2) If real numbers a and B satisfy (a + b-2) &# 178; + √ b-2a + 3 = 0, find the value of 2b-a + 1

(1) 1-A + √ (A-2) & # + 2A
=1-a+2-a+2a
=3
2) A + B-2 = 0, b-2a + 3 = 0
Add the two formulas to get 2b-a + 1 = 0
First simplify and then evaluate; [x + 1-15 / X-1] / x-4 / X-1, where x = 5 and root 2-4
The original formula = {[(x + 1) (x-1) - 15] / (x-1)} × [(x-1) / (x-4)]
=[(x+1)(x-1)-15]/(x-4)
=(x²-16)/(x-4)
=(x+4)(x-4)/(x-4)
=x+4
=5√2-4+4
=5√2
Root 12 * root 6 / root 8
Root 12 * root 6 / root 8
=2√3×√6÷2√2
=3√2÷√2
=3；
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You can do it step by step. It's three
Find the monotone interval of the function y = 3x cube-9x + 5 in the interval [- 2,2]
By deriving the function y = 3x cube-9x + 5, we get the derivative of y = 9 (x ^ 2-1), let it = 0,
We can get x = 1 or x = - 1
Monotone increasing interval of function y = Log1 / 2 (- X & # 178; + 2x)
because
Y = Log1 / 2, X is a decreasing function
therefore
-The minus interval of X & # 178; + 2x is
Monotone increasing interval of function y = Log1 / 2 (- X & # 178; + 2x)
and
-The decreasing interval of X & # 178; + 2x is: (1, + ∞)
also
-x²+2x>0
x(x-2)
The function f (x) = BX ^ 2 + 3x defined in the interval [3-A, 5] is an odd function, and the values of a and B are obtained
The domain of definition of odd functions is symmetric with respect to the origin
So 3-A = - 5
A=8
f(-x)=-f(x)
So BX & # 178; - 3x = - BX & # 178; - 3x
2bx²=0
So a = 8, B = 0
The definition interval of odd function is symmetric about the origin. So 3-A = - 5, so a = 8
Because the odd function satisfies f (x) = - f (- x), the solution B = 0
Domain symmetry is 3-A = - 5, a = 8
F is an odd function
f(x)=bx^2+3x=f(-x)=bx^2-3x
B=0
8, 0
Defined in the interval [3-A, 5], the domain of definition should be symmetric, so 3-A = - 5, a = 8
F (x) = - f (- x) is easy to get b = o
If a = 8, B = 0 is not carefully considered, its derivative function is 2bx + 3, so B = 0 ensures that this function is odd. Next, f (0) = 0, the premise is that x = 0 is differentiable, or symmetrically, f (3-A) = f (5)
So: not too hard
Find the maximum energy of the function f (x) = - (Log1 / 2x) ^ 2 - (Log1 / 4x) + 5 in the range of 2 ≤ x ≤ 4
First, change the formula, f (x) = - (log2x) ^ 2 + 1 / 2 (log2x) + 5, when 2 ≤ x ≤ 4, the range of log2x is 1 ≤ log2x ≤ 2, then log2x can be treated as t, the original formula becomes f (x) = - t ^ 2 + 1 / 2T + 5, t ∈ [1,2], and then so easy
Next, do it yourself
Judge the monotonicity of the following functions, and find the monotone interval, f (x) = sinx-x, X belongs to (0, π)
f(x)=sinx-x
If f '(x) = cosx-1, X belongs to (0, PAI), then there is - 14
(x+1/2)^2>17/4
x> - 1 / 2 + radical 17 / 2, or X
If the function y = f (x) has an inverse function, then the equation f (x) = m (M is a constant) ()
A. There is and only one real root B. there is at least one real root C. There is at most one real root d. There is no real root
Let B be the range of function y = f (x). When m ∈ B, the equation f (x) = m has a real root; when m ∉ B, the equation f (x) = m has no real root; so the equation f (x) = m has at most one real root, so C
Function f (x) = x ^ 3-ax ^ 2 + 3x + 6 if the tangent of function f (x) at x = 1 is parallel to the X axis, any x belongs to [- 1,4], and f (x) > F '(x) is in the range of F (0)
f(x)=x^3-ax^2+3x+b
f'(x)=3x²-2ax+3
The slope of the x-axis is 0
So f '(1) = 6-2a = 0
A=3
Let g (x) = f (x) - f '(x) = x & sup3; - 6x & sup2; + 9x + B-3
-1
f'(x)=3x²-2ax+3
The slope of the x-axis is 0
So f '(1) = 6-2a = 0
A=3
Let g (x) = f (x) - f '(x) = x & sup3; - 6x & sup2; + 9x + B-3
-1