Find a high school math problem and a high school math problem

Let Z=4X-3Y make a set of straight lines L:4X-3Y=T parallel to 4X-3Y=0, then when L crosses the intersection of 4X+Y+10=0 and X+7Y-11=0, the T value is the smallest; When l crosses the intersection of 4x+y+10=0 and 7x-5y-23=0, the t value is maximum.

Therefore, the maximum value = 4 (-1) -3 (-6) = 14

If a maximum value of 4x-3y is required, a maximum value of x and a minimum value of y are required.

7x-5y-23<=0 a

x+7y-11<=0 b

4x+y+10>=0 c

From 7b-a=54y-54<=0=>y<=1=>x<=4

Substitute c ==>y>=-26

The maximum value of 4x-3y is 4*4+26=42

As a diagram, first draw the area enclosed by the three straight lines of 7x-5y-23=0, x+7y-11=0, 4x+y+10=0 and the focal point between them, and then draw a straight line with a slope of 4 3, translate the curve, and then the value of the two limit positions is the maximum and minimum value of 4x-3y.

Let z=4x-3y

Draw first. 5y-23<=0,x+7y-11<=0,4x+y+10>=0 to plot its feasible domain.

And then. Change z=4x-3y to.

y=4/3x+z/3

z 3 is the intercept of the line on the y-axis.

Then substitute for what can be sought.

First, the product number 1, 2, 3, 4, a, b (ab is defective), then there are , a, b

a、b、a、b

a、b、a、b

a、ba、b

a、b、a、b

a、b、a、b

a、b、a、b

a、b、a、b

a、b、a、b

a、b、a、b

a、b、a、b

a、b、a、b

A and B follow this law in turn, and the answer is 72

Analysis: Let the inclination angle of a known line be (90°) The equation for a straight line: xcos + 3

y+2=0 can be reduced to: y=-( 3 3)cos *x-(2 3) 3 then we can know the slope of the straight line k=tan =-( 3 3)cos Because -1 cos 1, so:- 3 3 -(3 3)cos 3 3 then there is - 3 3 tan 3 3 so:

When 0 tan 3 3, the solution is 0° 30°;

When -3 3 tan <0, 150° 180° is solved, and the range of inclination angles of the known straight line is [0°,30°] 150°,180°).

Passing the point M is ME AB, and crossing the point N is NE1 BC, respectively, passing BB1 to E, E1 because ME AB A1B1, so be BB1=BM BA1=ME A1B1=ME AB=PE PB (1).

And because ne1 bc b1c1, be1 bb1=bn bc1=ne1 b1c1=ne1 bc=pe1 pb (2).

The division of the two equations yields: be be1=pe pe1 (3) If be > be1, then there is e** on e1b1, then pe>pe1 is obtained from (3), that is to say, the point e is on the extension of pe1, the two statements here are contradictory, so be>be1 cannot be established, and similarly, be therefore the point e and the point e1 coincide.

So me ab, ne bc

So the plane that passes the point m, n, e plane abcd

Hence MN planar ABCD

1.From the product of vector points, f(x)=sqrt(3)sin(wx)cos(wx)-cos(wx) 2+1 2

sqrt(3)/2*sin(2wx)-cos(2wx)/2=sin(2wx-π/6).

The distance between the two adjacent axes of symmetry in the image is 4, which means that the minimum positive period is 4, i.e., 2 2w= 4==>w=4

2. f(x)=sin(8x-π/6).

If x belongs to (7 24, 5 12), then 4x belongs to (7 6, 5 3), in the third quadrant.

Let cos4x=t, there is sin4x=-sqrt(1-t 2).

by -sqrt(3)sqrt(1-t 2)t-t 2+1 2=-3 5

>t=sqrt(15) 5-sqrt(5) 10, or sqrt(15) 5+sqrt(5) 10.

3.If cosx 1 2, x belongs to (0, ) then x belongs to (0, 3), and 8x- 6 belongs to (- 6, 5 2).

f(x)=m has only one real root, which means that m is the maximum value of sin(8x- 6), m=1 or -1

and sin(8x- 6)=1==>8x- 6= 2 or 5 2==>x= 12 or 3, contradictory.

sin(8x-π/6)=-1==>8x-π/6=3π/2==>x=5π/24.

Therefore, m=-1 satisfies the requirement.

In the fraction after f(x), add 1 to the numerator and subtract 1, and the added 1 becomes sin(x) 2+cos(x) 2;Therefore, the molecule can be converted into (sin(x)+cos(x)) 2—1;Approximate the denominator by sin(x)+cos(x)+1;The remaining sin(x)+cos(x)-1;The fraction is preceded by a 1; So f(x) = sin(x) + cos(x).

The second fraction of g(x) is multiplied by cos(x) 2 at the same time, and the denominator becomes sin(x) 2-cos(x) 2;The molecule becomes cos(x) 2*(sin(x)+cos(x)); The numerator denominator is reduced to sin(x)+cos(x) at the same time, and the second fraction becomes cos(x) 2 (sin(x)-cos(x)); The first fraction is sin(x) 2 (sin(x)—cos(x)); Subtract the two, and then subtract a sin(x)—cos(x); Eventually, g(x) = sin(x) + cos(x), which is equal to f(x).

Parse: y=sinxcosx=1 2*sin2x when 2x=2k - 2, i.e. x=k - 4, k z,.

ymin=-1/2

When 2x = 2k + 2, i.e. x = k + 4, k z,.

ymax=1/2

y=sinxcosx=1/2(sin2x)

And you didn't say that the definition field of sin2x is generally infinite by default, so the minimum value of sin2x is [-1,1], so the minimum value is -1 2

Because y=sinxcosx=1 2sin2x

So the minimum value of y is -1 2

Solution: Let pf1 =m, pf2 =n, and the distance from point p to the left alignment of the hyperbola is d according to the second definition of the hyperbola: m d=e

According to the parabolic definition: n=d

These two formulas introduce m n=e

The equation is represented by letters: m n-2c m

e-2c/m

It is also required that m is defined according to the first hyperbola: m-n=2a

and m n = e = c a and subtract n

m=2c (e-1).

So the equation = e-(e-1) = 1

Because x+4/x = -a 1 x 4

Therefore, to get x2+ax+4=0, the meaning of the equation is greater than or equal to 0 for a2-16

The solution is greater than 4 or a less than -4

Solution: constructor f(x)=x+(4 x),(1 x 4)It is easy to know that the function f(x) is a "checkmark function", decreasing on [1,2] and increasing on [2,4].

f(1)=5,f(2)=4,f(4)=5.∴4≤-a≤5.===>-5≤a≤-4.

x+x4/x when 1 x 2 is a subtraction function, when 2 x 4 is an increasing function, 4 x+x/4 5, so -5 a -1

Use the algebraic expression of a to represent x, bringing in the arithmetic!!

Method 1: Transforming the original equation into a quadratic equation about x, which has a real root at 1 x 4, is transformed into a distribution problem of the root of the quadratic function.

Method 2: Derive x+4 x and evaluate the domain using the monotonicity of the function.

1.This is a power function, and the derivative gets:

f'(x)=(2n-n²)(2n²+3n-4)·x^(2n²+3n-5)

When x > 0, x (2n +3n-5) > 0

Therefore, (2n-n) (2n +3n-4) >0

i.e. n(n-2)(2n +3n-4)<0

It can be solved by threading: (-3-root number 41) 40 and then let the derivative = 0, i.e. g'(x)=0, x=-m, or m(rounded) omit the list.

then g on (-infinity, -m).'(x) >0, incrementing g over (-m,0).'(x) <0, decreasing.

There is a maximum value, which is taken when x=-m, and the maximum value is -2m if m<0, the same is true

Let the derivative = 0, i.e., g'(x)=0, x=-m, or m(rounded) omit the list.

then g on (-infinity, -m).'(x) >0, incremental.

g on (m,0).'(x) <0, decreasing.

There is a maximum value, which is taken when x=m, and the maximum value is 2m

2n^2+3n-4>0

2n-n^2>0

n belongs to (0,2).

n=1y=x

g(x)=x+m^2/x

When x belongs to (-infinity, -|.)m|], monotonically increasing.

When x [-m|, 0), monotonically decreasing.

When x=-|m|，ymax=-2|m|

n 1 gives f(x) and then the maximum value is 2 times the root number under m