Given the function f (x) = SiNx + 5x, X ∈ (- 1,1), if f (1-A) + F (1-a2) < 0, then the value range of a is______ .

Given the function f (x) = SiNx + 5x, X ∈ (- 1,1), if f (1-A) + F (1-a2) < 0, then the value range of a is______ .

Function f (x) = SiNx + 5x, X ∈ (- 1, 1), so the function is an increasing function, odd function, so f (1-A) + F (1-a2) < 0, we can get - 1 < 1-a2 < A-1 < 1, the solution is 1 < a < 2, so the answer is: 1 < a < 2
Given the vector a = (1,2), B = (x, - 1), if a ⊥ B, then the value of real number x is ()
A. 2B. -2C. 1D. -1
∵ a ⊥ B ≁ a · B = 0, that is, X-2 = 0, the solution is x = 2, so a
Given that 3-radical 2 is a root of the equation x + MX + 1 = 0, then M is equal to? And the other root is?
Let the other be x1,
∵X1(3-√2)=1,∴X1=1/(3-√2)=1/7(3+√2),
∴m=-[X1+(3--√2)]=-1/7(24-6√2).
If the equation AX & # 178; + BX + C = 0 has a real solution, then X1 + x2 = - B / A, x1 × x2 = C / A
Equation x & # 178; + MX + 1 = 0 x1 × x2 = 1 X1 + x2 = - M
∵ X1 = 3-following sign 2 ∵ x2 = 1 / X1 = (3 + radical sign 2) / 7
The number of M = - (x1 + x2) is calculated by yourself
Let the other root be X
From Veda's theorem: (3 - √ 2) * x = 1, x = (3 + √ 2) / 7
m=-[（3-√2）+(3+√2)/7]=-(24-6√2)/7
A mathematics problem in grade one of junior high school
There are 28 people working in department a and 18 people working in department B. now another 20 people are transferred to support. How many people should be transferred to department a and department B in order to make the number of people in department a double that in department B?
We should transfer x people to a
28+x=2【18+（20-x）】
28+x=2(18+20-x)
28+x=36+40-2x
x+2x=36+40-28
3x=48
x=16
20-16 = 4 (person)
A: 16 people should be transferred to department a and 4 people should be transferred to department B
Factorization of x2-8x + 7x & # 178; + 7x-18
x²-8x+7
=（x-1）（x-7）
x²+7x-18
=（x+9）（x-2）
x2-8x+7
=（x-1)(x-7)
x²+7x-18
=(x+9(x-2)
The function f (x) = - 5x + SiNx defined on (- 1,1), if f (1-A) > F (a ^ 2), then a range
f'(x)=-5+cosxf(a^2)
1-A
Let vector a = (x, - 3), B = (- 1,1), if | A-B | = 5, then the value of real number x is
a-b=(x-(-1),-3-1)=(x+1,-4)
|a-b|=√（x+1）^2+(-4)^2=√（x^2+2x+10）=5
x^2+2x+10=25
x^2+2x-15=0
(x-3)(x+5)=0
x=3 or x=-5
X = 8 or - 2
Root equation
√80+x² =2x+4√5
(80 + X & # 178; is in the root, 5 is in the root)
solution
The square of the two sides of the original equation can be obtained
x²+80=4x²+(16√5)x+80
You can get it
x[3x+16√5]=0
∴x1=0,x2=-(16√5)/3
When x = - (16 √ 5) / 3,
2X + 4 √ 5 = [- (32 √ 5) / 3] + [(12 √ 5) / 3] < 0
The root of the original equation is x = 0
solution
The square of the two sides of the original equation can be obtained
x²+80=4x²+(16√5)x+80
You can get it
x[3x+16√5]=0
∧ = (x 2 = 3) / x 15
The square of both sides of the original equation, we can get, X & # 178; + 80 = 4x & # 178; + (16 √ 5) x + 80
After finishing, we can get x [3x + 16 √ 5] = 0
∴x1=0, x2=-(16√5)/3
When x = - (16 √ 5) / 3, 2x + 4 √ 5 = [- (32 √ 5) / 3] + [(12 √ 5) / 3] < 0
The root of the original equation is x = 0
The equation has a root of 0, not no solution
The female students in the extracurricular group originally accounted for 13% of the total number of students in the group. After four female students were added, the female students accounted for 12% of the total number of students in the extracurricular group. What was the original number of students in the extracurricular group?
A: the original number of extracurricular group is 24
Make 2x & # 178; - 7x-4 = 0 with matching method
(2x + 1) (x-4) = 0 for x = - 1 / 2 or x = 4