Solution equation (6 / 5x + 0.5) ^ - (1.2x-1 / 2) ^ 2 = - 24

Solution equation (6 / 5x + 0.5) ^ - (1.2x-1 / 2) ^ 2 = - 24

(1.2x+0.5)^2-(1.2x-0.5)^2=-24
(1.2x+0.5+1.2x-0.5)(1.2x+0.5-1.2x+0.5)=-24
2.4x=-24
x=-10
f(x)=cos2x+1+√3sin2x
=2(sin2xcosπ/6+cos2xsinπ/6)+1
=2sin(2x+π/6)+1
When sin (2x + π / 6) increases, 2K π - π / 2
First, merge the similar terms, and find the value of the algebraic formula: X & # 178; - 2x & # 178; + 5 - X & # 178; - 3, in which x = 1 / 2
Write the merge and answer
x² -2x² +5 -x² -3
= 2 - 2x²
When x = 1 / 2
The original formula = 2-2 × (1 / 2) ² = 1.5
It is known that the domain of the function f (x) = ㏒ 2 (X & # 178; - 2mx + 2m & # 178; + 1 / (M & # 178; - 2)) is a set of real numbers
1. Find the set M composed of all allowed values of real number M;
2. Proof: for all m ∈ m, f (x) ≥ 2
(1) Logarithmic formula x ^ 2-2mx + 2m ^ 2 + 1 / m ^ 2-2 = (x-m) ^ 2 + (m-1 / M) ^ 2 > 0
So (m-1 / M) ^ 2 > 0
The solution is m ≠ 1, m ≠ - 1, m ≠ 0
To sum up, M = {m | m ≠ 1, m ≠ - 1, m ≠ 0, m ∈ r}
(2) Prove: logarithm formula x ^ 2-2mx + 2m ^ 2 + 1 / m ^ 2-2 = (x-m) ^ 2 + (M + 1 / M) ^ 2 + 4
So the logarithm is ≥ 4
So f (x) ≥ 2
Minimum value of polynomial x ^ 2 + 4Y ^ 2 + 2x-4y + 7
x^2+4y^2+2x-4y+7
=（x^2+2x+1)+(4y^2-4y+1)+5
=(x+1)^2+(2y-1)^2+5
≥5
The minimum value of polynomial x ^ 2 + 4Y ^ 2 + 2x-4y + 7 = 5
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x^2+4y^2+2x-4y+7
=(x^2+2x+1)+(4y^2-4y+1)+5
=(x+1)^2+(2y-1)^2+5
When x = - 1, y = 1 / 2
X ^ 2 + 4Y ^ 2 + 2x-4y + 7 min = 5
x^2+4y^2+2x-4y+7=(x+1)^2+(2y-1)^2+5
When the first two square terms are 0, the,
That is, when x = - 1, y = 1 / 2, there is a minimum value of 5
x^2+4y^2+2x-4y+7
=(x²+2x+1)+(4y²-4y+1)+5
=(x+1)²+(2y-1)²+5
Because (x + 1) & sup2 ≥ 0, (2y-1) & sup2 ≥ 0
So when x = - 1, y = 1 / 2, the minimum value of polynomial x ^ 2 + 4Y ^ 2 + 2x-4y + 7 is 5
x^2+4y^2+2x-4y+7
=(x^2+2x+1)+(4y^2-4y+1)+5
=(x+1)^2+(2y-1)^2+5
>=5
If x + 1 = 0,2y-1 = 0, take equality
x=-1,y=1/2
The minimum is 5
x^2+4y^2+2x-4y+7
=(x^2+2x+1)+(4y^2+4y+1)+5
=(x+1)^2+(2y+1)^2+5
The minimum value of two square terms is 0
So the minimum value of the polynomial is 5
Original formula = (x ^ 2 + 2x + 1) + (4Y ^ 2-4y + 1) + 5
-------------=（X+1)^2+(2Y-1）^2+5
So: the minimum value of the original formula is: 5 (at this time, x = - 1, y = 1 / 2)
x^2+4y^2+2x-4y+7
= x^2+2x+1+4y^2-4y+1+5
=(x+1)^2+(2y-1)^2+5
(x + 1) ^ 2 + (2y-1) ^ 2 is greater than or equal to 0
So the minimum is 5
0.5x-2 = 24
0.5x-2=24
0.5x=24+2
0.5x=26
x=26÷0.5
x=52
x=52
0.5x-2=24
0.5x=24+2
0.5x=26
x=26/0.5
x=52
0.5x-2=24
0.5x=24+2
0.5x=26
x=26÷0.5
x=52
0.5x=24+2
0.5x=26
x=52
0.5x-2=24
0.5x=26
x=26/0.5
x=52
First merge the similar terms, and then find the value of the algebraic formula: x ^ 2-2x ^ 2 + 5-x ^ 2-3
X ^ 2-2x ^ 2 + 5-x ^ 2-3, first merge the similar terms, and then find the value of the algebraic formula, where x = negative half
X^2-2X^2+5-X^2-3
=(1-2-1)x^2+2
=-2x^2+2
x=-1/2
Original formula = - 2 * / 2
=-2*1/4+2
=-1/2+2
=3/2
Given that the function f (x) is an increasing function in the domain of definition [- 3,3], the value range of the real number m satisfying f (1-m) > F (1 + 3M) is obtained
2. Let f (x) = (m Square-1) x + (m square-3m + 2) be a function of degree 1
Find: (1) if f (x) is a decreasing function and f (1) = 0, find the value of M
(2) If f (x) is an odd function, find the value of M
1. Firstly, 1-m1 + 3M must be in the domain of definition. Secondly, according to the properties of increasing function, we can get the
Solving the system of inequalities - 3 ≤ 1-m ≤ 3
-3≤1+3m≤3
1-m ＜ 1 + 3M, 0 ＜ m ≤ 2 / 3
2. (1) if y = ax + B is a decreasing function, then a < 0
There are m ^ 2-1 < 0 and - 1 < m < 1
F (1) = (m ^ 2-1) + m ^ 2-3m + 2 = 2m ^ 2-3m + 1 = 0 = M = 1 / 2
(2) The necessary condition for f (x) to be an odd function is that f (0) = 0
Then f (0) = m ^ 2-3m + 2 = 0, M = 2 or M = 1 (rounding off after test)
So m = 2
Two polynomials A.B, B is 4x square minus 5x-6, Xiaogang takes a + B as A-B in calculation, and the result is 10x-7x square
A-B=A-(4x^2-5x-6)=10x-7x^2
A=10x-7x^2+(4x^2-5x-6)=-3x^2+5x-6
A+B=-3x^2+5x-6+(4x^2-5x-6)=x^2-12
A + B = square X-12 a = 5x-3x Square-6
Use "(wrong result) 10x-7x square" plus "(b) 4x square minus 5x-6" to get a, and then subtract B from a to get the correct result
3.5-5x = 2 solution equation
3.5-5x=2
5x=3.5-2
5x=1.5
x=0.3
3.5-5X=2
5X=3.5-2
5X=1.5
X=0.3
5x=3.5-2=1.5
x=0.3
3.5-5x=2
1.5-5x=0
5x=1.5
x=0.3
3.5-5x=2
5x=3.5-2
5x=1.5
x=0.3
3.5-5x=2
Solution 5x = 3.5-2
5x=1.5
X = 1.5 divided by 5
x=0.3
3.5-5x=2
5x=3.5-2
x=0.3
The equation made by the authentic fifth grade students!!!
If the equation x & # 178; + MX + 2 = 0 and X & # 178; + 2x + M = 0 have the same root and m ≠ 2, then the value of M is
Let a be the same root of the equation x & # 178; + MX + 2 = 0 and X & # 178; + 2x + M = 0} {A & # 178; + Ma + 2 = 0A & # 178; + 2A + M = 0} of X. by subtracting the two equations, we can get ma-2a + 2-m = 0A (m-2) - (m-2) = 0 (m-2) (A-1) = 0 ∵ m ≠ 2} A-1 = 0A = 1 and substituting a = 1 into X & # 178; + MX + 2 = 0, we can get 1 + m + 2 = 0m = - 3
Let the common root of the two equations be x1, then X1 & # 178; + MX1 + 2 = 0, X1 & # 178; + 2x1 + M = 0, and the subtraction of the two equations to (m-2) * (x1-1) = 0
And m ≠ 2, so X1 = 1, substituting the former formula, we get m = - 3