High school math problems, find the master speed

Left: The numerator is sin 2+cos 2+2sincos=(sin+cos) 2 The denominator is (sin+cos)(sin-cos) The numerator denominator is deleted from the common factor (sin+cos) to get: the numerator is (sin+cos), the denominator is (sin-cos) Right:

Numerator sin cos+1, then divided: (sin+cos) cos denominator sin cos-1, redivided: (sin-cos) cos numerator denominator deletes the common factor cos to obtain:

The numerator is (sin+cos) and the denominator is (sin-cos) Left = Right, so the equation holds.

Left =(1+2sinxcosx) (sinx 2-cosx 2) =(sinx 2+cosx 2+2sinxcosx) (sinx+cosx)(sinx-cosx) =(sinx+cosx) 2 (sinx+cosx)(sinx-cosx) =(sinx+cosx) (sinx-cosx)] The numerator and denominator are divided by cosx, resulting in =(tanx+1) (tanx-1). ) = right so the original tense holds.

(1) By the sine theorem, bsinc=csinb

So cos[(b+c) 2]=sina

cos(π/2-a/2)=sina

sin(a/2)=sina

a 2 = a or a 2 + a =

a=0 (rounded) or a=2 3

2) From the co-string theorem, a = b + c -2bccosa i.e. 16 = b + c + bc 2bc + bc

BC 16 3, take the equal sign if and only if B = C.

s=1/2*bcsina≤1/2*16/3*√3/2=4√3/3

(1) Solution: (1).

Simultaneous x+y=1

x^2/a^2+y^2/b^2=1

Get (a 2+b 2) x 2-2a 2x+a 2(1-b 2) = 0 Vedic theorem, get.

x1+x2=2a^2/(a^2+b^2)

X1X2=A2(1-B2) A2+B2Vector OP Vector OQ

x1x2+y1y2=0

2a^2b^2=a^2+b^2

1/a^2+1/b^2=a^2+b^2/a^2b^2=2 ..1)

2)√3/3≤e≤√2/2

1/3≤e^2=1-b^2/a^2≤1/22/3≥b^2/a^2≥1/2

1) b 2 a 2 = 1 2.

Substituting (1) yields a = 6 2

2) b 2 a 2=2 3.

Substituting (1) yields a = 5 2

The value range of the major axis of the ellipse is [ 5, 6].

2) sn n=(a1+an) 2=2a1+(n-1)*d because a1 d is a constant.

So it's a linear problem.

So pn is on the same straight line.

2.Let the total loss be w

w=50x+125n+250*4n

50x+1125000/（x-2）

50 (x-2) + 1125000 (x-2) + 100 is obtained from the mean inequality wmin=15100, x=152

1. Water seepage area = repair area.

200+4n=2nx

n=100 (x-2) where x is greater than 2

2. Let the total loss be y, and the total loss = the total seepage area of the embankment * 250 + the total labor cost + the total clothing cost y = (200 + 4n) * 250 + (75 + 50) nx + 50x and bring the result of 1 into the collation to find the maximum value.

(1) A total of x workers were dispatched, and the repair was completed in n days to get 2nx=200+4n, and the relationship was found. (2) Let the total loss be y,50( +x-2+1252).

50 (2 50 + 1252) = 67600 When =x-2 and x=52, it holds.

52 workers should be sent to rush to repair, and the total loss is the smallest.

g(x)=(2x 1) is about the parabola of the straight line x=1 2 symmetrical opening upward, let f(x)=ax then the problem is that the image of f(x) is higher than the image of g(x) There are only three integer values of x. Make an image of two functions... Discover:

It is only hoped when x starts from 1, so these three integers should be 1, 2, 3. Therefore, the f(3), g(3) and f(4) image solutions are the best solutions to this problem.

The inequality conversion becomes (1 x-2) 2-infinity(1 x-2) 2 tends to be close to 4, so x takes the negative number (1 x-2) and 2 is also less than 4

When x is a positive integer, (1 x-2) 2 must be less than 4, so x must be a positive integer and x is a positive integer(1 x-2) 2 increments, so the positive integer value of x should be 1,2,3

Therefore, argument a should be such that x can get 3 but not 4

i.e. (1 3-2) 2 to solve the answer (25 9, 49 16).

∵0≤(2x-1)²＜ax²，∴a＞0

Original inequality ---2x-1) -bx) = [(2+ a)x-1][(2- a)x-1] 0

The solution set of integers in the solution set is finite (3), (2+ a)(2- a) 0---0 a 2--- the solution set of inequalities is m=(1 (2+ a),1 (2- a)), where there are exactly 3 integers.

2 2 + a 4, 1 4 1 (2+ a) 1 2, i.e. the smaller root of the equation is between 1 4 and 1 2.

If there are exactly 3 integers in m, it must be

i.e.: 3 1 (2-a) 4

->1/4≤2-√a＜1/3

->5/3＜√a≤7/4

--25/9＜a≤49/16

Here's the answer:

0≤(2x-1)²＜ax²，∴a＞0

The original inequality can be converted to (2x-1) ax) so the inequality (2x-1) ax) can be reduced to 2x-1< ax and 2x-1>-ax or 2x-1> ax and 2x-1<-ax

So the former 1 (2 + a) 0 contradictory.

Therefore, it can be inferred that if there are exactly 3 integers in the solution set of the inequality (2x-1) ax about x, then the range of values of the real number a is .

25/9＜a≤49/16

From the original inequality we know a>0

The original inequality shifts are combined and decomposed to obtain: [x-(2+a) (4-a)]*x-(2-a) 4-a)]<0

So its solution is: ...Because the solution set has exactly three integers, a under the root number is also an integer, and the latter is greater than the former from the solution set, which is derived from this column.

Count B -4ac Town Cave 0 is to ensure that M ≠ Fiber Brigade is ruined for years n

It doesn't matter at all, it's not a quadratic function, ** come to b 2-4ac, you have a problem with your previous answer.