The maximum value of the function y = sin ^ 2x + acosx + 5 / 8a-3 / 2, (x ∈ R) is 1

The maximum value of the function y = sin ^ 2x + acosx + 5 / 8a-3 / 2, (x ∈ R) is 1

To solve this problem with collocation method embodies the thought of reduction in mathematics
y = 1- cos^2x +acosx+5/8a-3/2
Simplification: y = - cos ^ 2x + a cos x + 5A / 8 - 3 / 2
(key steps below)
y = - ( cosx - a/2)^2 +a×a/4 +5a/8 - 1/2
Since the function y = sin ^ 2x + acosx + 5 / 8a-3 / 2, the maximum value of (x ∈ R) is 1
So: a × A / 4 + 5A / 8 - 1 / 2 = 1, the solution is: A1 = 3 / 2, A2 = - 4 (rounding off)
So when a = 3 / 2, that is, when cosx = 3 / 4, the maximum value of the function is 1
So a = 3 / 2

Find the minimum value of function f (x) = - SiNx square - acosx + 1 is - 6, and find the value of real number a

F (x) = - SiNx square - acosx + 1
=cos^2x-1-acosx+1
=cos^2x-acosx
=（cosx-a/2）^2-a^2/4
-a^2/4=-6
a^2=24
A = 2 radical 6 or a = - 2 radical 6

Mathematical equation, 3 yuan equation solution. 10 minutes reward 10 points!
x y z z y x
－+ － + － =51 ① － +－ +－ =53.4 ② x+y+z=3.3 ③
50 66 2 83 1 50 66 2 83 1
－ － － －
3 3 3 3

Item by item elimination

(x-1)^2=4(x-3)^2
^2: The meaning of the second power

(x-1)²=4(x-3)²
（x-1）²=[2(x-3)]²
Square on both sides
x-1=2x-6
x-2x=-6+1
-x=-5
x=5

8 (X-2) = 2 (x + 7)

8x-16=2x+14
6x=30
x=5

(x-1) + (x + 1) + (x multiplied by 1) + (x divided by 1) = 60. How can I solve this equation? I will hand it in tomorrow,

(x-1) + (x + 1) + (x multiplied by 1) + (x divided by 1) = 60
x-1+x+1+x+x=60
4x=60
x=15

1 / 3 [X-2 / 1 (x-1)] = 2 / 3 (X-2 / 3)
We need to write the process out, and I don't know how to calculate it

1 / 3 [X-2 / 1 (x-1)] = 2 / 3 (X-2 / 3) both sides at the same time × 31 × [X-2 / 1 (x-1)] = 2 × (X-2 / 3) X-2 / 1 (x-1) = 2x-3x-2 / 1 + 2 / 1 = 2x-32 / 1 + 2 / 1 = 2x-32x-2 / 1 = 2 / 1 + 32 / 3x = 2 / 7X = 2 / 7

The total workload of group A's four workers in March is 20 pieces more than that of the per capita quota of this month. The total workload of group A's five workers in March is 20 pieces less than that of the per capita quota of this month
1. If the actual per capita workload of the two groups of workers in this month is equal, what is the per capita quota of this month
2. If the actual per capita workload of group a workers in this month is 2 pieces more than that of group a workers, what is the per capita quota in this month?
3. If the actual per capita workload of group a workers in this month is 3 pieces less than that of group a workers, what is the per capita quota in this month?

Suppose that the per capita quota of this month is x pieces, it can be seen from the meaning of the question: the total workload of 4 workers in group A in March is 4x + 20, the per capita workload of this month is (4x + 20) / 4 = x + 5, the total workload of 5 workers in group A in March is 6x-20, the per capita workload of this month is (6x-20) / 5 = 1.2x-4 (1)

Solve the following equation
A.5（X-5)+2=-4 B.3（X-2)=(8-8X) C.3X-12+4(X-3)=2(X+1)-5 D.3X-2(X+2)=4(X-1)+7 E.3X-2[X-5(X+1)-4]=4 F.4/3[3/4(1/5X-2)-6]=1 G.3(X+5)+5[(X+5）-1]=7(X+5)-1 H.5(2-3X)=4(X+1) I.3-2（X-1)=2(X-3) J.1/3(6X-3)-1/2(8X-4)=5 k.(3X+2)+2[(X-1)-(2X+1)]=6 L.3X-2/5=X+1/2 M.4(X+2)/3-X-7/5=12 N.t-t-1/2=2-t+2/3 O.3x-2(x+3)/3=6-x-2/6 P.3/2[2/3(X/4-1)-2]-8=X Q.2X+1/3-5X-1/6=1 R.3(2X+1)/4-2(2X-1)/3=1 S.y-y-1/2=2-y+2/5

a:
5(x-5)+2=-4
5x-25+2=-4
5x=19
x=19/5
b:
3（X-2)=(8-8X)
3x-6=8-8x
11x=14
x=14/11
c:
3X-12+4(X-3)=2(X+1)-5
3x-12+4x-12=2x+2-5
7x-24=2x-3
5x=21
x=21/5
d:
3X-2(X+2)=4(X-1)+7
3x-2x-4=4x-4+7
x-4=4x+3
-3x=7
x=-7/3
e:
3X-2[X-5(X+1)-4]=4
3x-2[x-5x-5-4]=4
3x-2[-4x-9]=4
3x+8x+18=4
11x=-14
x=-14/11
f:
4/3[3/4(1/5X-2)-6]=1
(1/5x-2)-8=1
1/5x=11
x=55
g:
3(X+5)+5[(X+5）-1]=7(X+5)-1
3(x+5)+5(x+5)-5=7(x-5)-1
8(x-5)-7(x-5)=4
x-5=4
x=9
h:
5(2-3X)=4(X+1)
10-15x=4x+4
-19x=-6
x=6/19
i:
3-2（X-1)=2(X-3)
3-2x+2=2x-6
-4x=-11
x=11/4
j:
1/3(6X-3)-1/2(8X-4)=5
2x-1-4x+2=5
-2x=4
x=-2
k:
(3X+2)+2[(X-1)-(2X+1)]=6
3x+2+2(x-1)-2(2x+1)=6
3x+2+2x-2-4x-2=6
x=8
l:
3X-2/5=X+1/2
3x-x=1/2+2/5
2x=5/10+4/10
2x=9/10
x=9/20
m:
4(X+2)/3-X-7/5=12
(4x+8)/3-x-7/5=12
4x/3+8/3-x-7/5=12
x/3=12+7/5-8/3
x=36+21/5-8
x=28+21/5=161/5
n:
t-t-1/2=2-t+2/3
-1/2=2-t+2/3
t=8/3+1/2
t=16/6+3/6
t=19/6
o:
3x-2(x+3)/3=6-x-2/6
3x-2x/3-2=6-x-1/3
7x/3-2=-x+6-1/3
10x/3=8-1/3
10x=25
x=5/2
p:
3/2[2/3(X/4-1)-2]-8=X
2/3[x/6-2/3-2]-8=x
x/9-4/9-4/3-8=x
x-4-12-72=9x
-8x=88
x=-11
q:
2X+1/3-5X-1/6=1
12x+2-30x-1=6
-18x=5
x=-5/18
r:
3(2X+1)/4-2(2X-1)/3=1
9(2x+1)-8(2x-1)=12
18x+9-16x+8=12
2x=-5
x=-5/2
s:
y-y-1/2=2-y+2/5
-1/2=2-y+2/5
y=2+2/5+1/2
y=20/10+4/10+5/10
y=29/10

A store sets the price of a commodity at the purchase price plus 20%. Because the sales are not good, it decides to reduce the price by 20%. This is the sales price of 96 yuan / piece. If it is sold out soon, the profit and loss of this business is ()
A. Earn 6 yuan B. don't lose C. lose 4 yuan D. lose 24 yuan
(I don't write ideas,

Suppose the purchase price is x, the price is x (1 + 20%), and the sales price is x (1 + 20%) (1-20%), which should be equal to 96. The solution x is 100, that is, the purchase price is 100 yuan, and the loss is 4 yuan?