1. - 1.5t-6 (the square of T) - 6 (the third power of T) 2.2 × (square of X + square of Y) - 8xy (square of X + square of Y) + 8 (square of x) (square of Y) 3. (1-1 / (the square of 2)) (1-1 / (the square of 3)) (1-1 / (the square of 9)) (1-1 / (the square of 10)) 4. Given (the square of a + the square of B) (the square of a + the square of B-8) + 16 = 0, find the square of a + the square of B

1. - 1.5t-6 (the square of T) - 6 (the third power of T) 2.2 × (square of X + square of Y) - 8xy (square of X + square of Y) + 8 (square of x) (square of Y) 3. (1-1 / (the square of 2)) (1-1 / (the square of 3)) (1-1 / (the square of 9)) (1-1 / (the square of 10)) 4. Given (the square of a + the square of B) (the square of a + the square of B-8) + 16 = 0, find the square of a + the square of B

1.-1.5t-6t²-6t³=-1.5t（1+4t+4t²）=-1.5t（1+2t）².2.2（x²+y²）²-8xy（x²+y²）+8x²y²=2[（x²+y²）²-4xy（x²+y²）+4x²y&su...
Factorization: - 49 / 25 + 0.64=
Factorization: the fourth power of 2x - 8 / 1=
Factorization: 3x & # 178; - 30x + 75=
Factorization: 10a-5a & # 178; - 5=
Factorization: 25A (n + 2) - 10A (n + 1) + a (n)=
-49 / 25 + 0.64 = 0.8 & # 178; - (7 / 5) &# 178; = (0.8 + 7 / 5) (0.8-7 / 5) = (0.8 + 1.4) (0.8-1.4) = - 1.322x quartic power - 8 / 1 = 2 (x quartic power - 1 / 16) = 2 (X & # 178; - 1 / 4) (X & # 178; + 1 / 4) = 2 (x + 1 / 2) (x-1 / 2) (X & # 178; + 1 / 4) 3x & # 178; - 30x + 75 = 3 (X & #
1、-49/25+64/100=-49/25+16/25=-33/25
1. (4 / 5-7 / 5) (4 / 5 + 7 / 5) 2.2 (x-1 / 2) (x + 1 / 2) (square of X + 1 / 4) 3. (3x-15) (X-5) 4. (5x-5) (- x + 1) 5. The nth power of a multiplied by the square of (5a + 1)
Let Max {a, B} denote the greater of real number a and B, then the minimum value of function f (x) = max {x + 1 |, | X-2 |} (x ∈ R) is___
The image method is the simplest. When we draw | x + 1 | and | X-2 | images in the same coordinate system, we can directly see that f (x) has a minimum value of 1.5 when x = 1 / 2
More specifically, X
When | x + 1 | > | X-2 |, the range of X is obtained, which has the minimum value of X; f (x) = | x + 1|
When f (x) = |x-2 |; |x + 1|
Given the function f (x) = log2 (x + 2), find the value of f ^ - 1 (- 3)
Obviously, f (x) is a monotone function, so there exists F-1 (x) let y = log2 (x + 2) get x = 2 ^ Y-2, that is, F-1 (x) = 2 ^ X-2, F-1 (- 3) = - 17 / 9
f^-1(x)=2^x-2 f^-1(-3)=2^(-3)-2=1/8-2=-15/8
If f (x) = min {f (x), G (x)} is defined, then the maximum value of F (x) & nbsp; is ()
A. 0B. 1C. 2D. 3
From 2-x2 ≥ x, the solution is - 2 ≤ x ≤ 1. The maximum value of Min {2-x2, X} = x, - 2 ≤ x ≤ 12-x2, X ＜ - 2, or X ＞ 1 is 1 when - 2 ≤ x ≤ 1; the maximum value of Min {2-x2, X} = 2-x2 when x ≤ - 2 or X ≥ 1 is 1. In conclusion, the maximum value of Min {2-x2, X} = 2-x2 is 1
Given f (x) = lg1-x / 1 + X, find f ^ - 1 (LG2)
o(>_
f(x)=lg[(1-x)/(1+x)]
F ^ - 1 (LG2) = x is equivalent to f (x) = LG2
lg[(1/x)/(1+x)]=lg2
(1-x)/(1+x)=2
2/(1+x)=3
x=-1/3
It should be like this
Given the function f (x) = 2-x ^ 2, G (x) = x, if we define the function f (x) = min {f (x), G (x)}, then the maximum value of F (x) is
F (x) = g (x) - f (x) = x2 + X-2 = (x + 1 / 2) 2-9 / 4, so the minimum value of F (x) is - 9 / 4
Given the function f (x) = x ^ 2 / (AX + b), where a and B are constants, and the equation f (x) - x + 12 = 0 has two real roots, which are X1 = 3 and X2 = 4
(1) Finding the analytic expression of function f (x)
(2) Let k > 1, solve the inequality f (x) about X
(1) Substitute X1 = 3 and X2 = 4 into the equation f (x) - x + 12 = 0 respectively
9/(3a+b)-3+12=0,3a+b=-1
16/(4a+b)-4+12=0,4a+b=-2
A = - 1, B = 2
f(x)=x^2/(ax+b)=x^2/(2-x)
(2).f(x)(k+1)x-k
x^2-(k+1)x+k>0
(x-k)(x-1)>0
When k = 1, the solution is that x is not equal to 1
K2
2> When k = k > 1, the solution is x > 2
k> 2, the solution is x > K
X
If min {g (x), H (x)} is defined as the smaller of the median values of G (x) and H (x), then the maximum value of F (x) = min {2-x ^ 2, X} is
When 2-x ^ 2 > x, that is - 2
It is known that the function f (x) = (1 / 2) ^ ax, a is a constant, and the function image passes through the point (- 1,2)
If G (x) = (4 ^ - x) - 2 and G (x) = f (x), find the value of X satisfying the condition
F (x) = (1 / 2) ^ ax, a is a constant, and the function image passes the point (- 1,2), so 2 = (1 / 2) ^ - A, a = 1, f (x) = (1 / 2) ^ X
Because g (x) = f (x), so 4 ^ - X-2 = 2 ^ - x, let t = 2 ^ - x, T ^ 2-2 = t, T ^ 2-t-2 = 0, t = 2 or T = - 1, that is, 2 ^ - x = 2 or 2 ^ - x = - 1 (rounding off 0)
So x = - 1,