Ask for help with 3 high school math problems, and 3 high school math problems Urgently ask for onli

1.Using the square of a square + b square" = 2ab (a + b) square < = 2 * (a square + b square) is too simple, I won't write the process.

2.It turns out that inequality can be equated with x*(x-a)+b<0....1) or x*(a-x)+b<0....2)

Because the title tells us that for any x it is between [0,1] and that it is constant, that is, we can think of it as the value of the original function in the interval [0,1] is always less than 0

When x-a>0, the original function is x*(x-a)+b=0(3) opening upward, the value is less than 0, that is, when x=0,1, (3)=0 brings x=0,1 into (3) to get 0*(0-a)+b=0 and 1*(1-a)+b=0 to get b=0, a=0Because x-a>0, x belongs to [0,1] so a<0 and because a=0 so a<=0

When x-a<0, the original function is x*(a-x)+b=0(4) The opening is downward, and the value is less than 0, but [0,1] is a closed interval, so the compromise is not valid.

The answer is a<=0

3.In fact, finding the minimum value of a fraction means that the denominator is the largest and the numerator is the smallest, that is, x y+z t is equivalent to 1 y+y 100 (because 1<=x<=y<=z<=t<=100), so the original formula = 1 y + y 100>=2*, under the root number = 2*, that is, when y=10, the minimum value is taken.

1. Let the coordinates of p point (x, y), then the vector ap=(x,y-1), the vector pb=(-x,-1-y), the vector pc=(1-x,-y), and according to the conditions, -x 2-y 2+1=kx 2-2kx+k+ky 2. (k+1)x 2+(k+1)y 2-2kx+k-1=0.

x 2 0, so -1 x Shishan 1, and x+k 0, so k -x. Hence k 1.

x 2+2kx+k 2=1-x 2 δ=4k 2+8 0, and the root number of Chaju -2 k annihilated the root number 2.

To sum up, 1 k root number 2.

Let the equations of the straight lines on the other two sides of the parallelogram be x+y+a=0 and 3x-y+b=0, and the distance from the intersection of the diagonal of the parallelogram to the parallelogram is equal.

3+0+1|/√1+1）=|3+0+a|/√1+1）4=|3+a|

a=7 or 1 (rounded) so a=-7

Similarly. 3*3+0+4|/√3^2+(-1)^2]= 3*3+0+b|/√3^2+(-1)^2]

13=|9+b|

b = 22 or 4 (rounded) so b = -22

The coordinates of the midpoint of ab are (2-2) 2=0, (-3-5) 2=-4, and the coordinates of the midpoint are (0, -4).

The slope of the line segment ab is [-3-(-5)] 2-(-2)]=1 2 Let the slope of the vertical square line of ab be k, then k*1 2=-1k= Let the equation of the vertical bisector of ab be y=

2x+b passes through point c(0,-4).

4=-2*0+bb=

y=2x-4

The center of the circle is also on x-2y-3=0.

Bring y=2x-4 in.

x-2(-2x-4)-3=0 is solved to x=

Bringing in the above equation gives y=

So the coordinates of the center of the circle are (-1, -2).

Let the circular equation be (x+1) 2+(y+2) 2=r 2(r>0), and bring the coordinates of point A (2,-3) into the solution to obtain r= 10

So the circular equation is (x+1) 2+(y+2) 2=10

Solution: 1. The other two are parallel to them.

are x+y+a=0 and 3x-y+b=0

The intersection of two straight lines (-3 4, 7 4) is known

The parallelograms are bisected diagonally.

So (-3 4, 7 4) and the midpoint of his diagonal vertex is m, so the diagonal vertex coordinates are x=3*2+3 4=27 4y=3*2-7 4=17 4

He is on x+y+a=0 and 3x-y+b=0.

So a=-11, b=-16

So if x+y-11=0 and 3x-y-16=02, over ab, the center of the circle is on the perpendicular bisector of ab.

ab slope = (-3+5) (2+2) = 1 2 so ab perpendicular bisector slope = -2

ab midpoint (0,-4).

So the ab bisector is 2x+y+4=0

and x-2y-3=0 is the center of the circle c(-1,-2)r =ac =3 +1 =10

So (x+1) +y+2) = 10

x²+y²=4

x+y=b is obtained from the above two equations.

2x²-2bx+b²-4=0

When a line and a circle are tangent, Eq. (1) has two equal real roots, so =0 i.e. 4b -4 2(b -4) = 0

The solution yields b = plus or minus 2 times the root number 2

That is, when b = 2 2, the line is tangent to the circle, and the coordinates of the intersection point are.

Solution: By. x+y+1=0

3x-y+4=0

The intersection of the two adjacent edges is (-5

4. The intersection point of the diagonal is m(3,0), which is obtained by the midpoint coordinate formula:

The intersection of the other two edges is (2 3-(-5.)

That is, (294, and the slope of the straight line where these two sides are located is equal to the slope of the straight line x+y+1=0 and 3x-y+4=0 respectively, which is -1 and 3, and the equation for the two straight lines is y-29

X+1 and Y-29

3(x+14, i.e., x+y-7=0 and 3x-y-22=0 Let the equation be: (x-x0) +y-y0) =r then: 2-x0) +3-y0) =r

2-x0)²+5-y0)²=r

2x0)4+(-8-2y0)2=0

Subtract the two equations).

2x0+y0+4=0

Again: x0, y0 at x-2y-3=0

x0-2y0-3=0

Simultaneous solution, obtain: x0=-1, y0=-2

r²=3²+(1)²=10

Equation: (x+1) +y+2) =10

Be what you want. Reduced to a general formula:

x²+y²+2x+4y-5=0

Question 1: First find the intersection point of two straight lines (-5 4, 1 4), find the intersection point of two straight lines (29 4, -1 4) according to the diagonal (3, 0), and let the branches of x+y+c=0 and 3x-y+d=0 respectively, put .

29 4, -1 4) to find c=-7, d=-22

That is, the two straight lines are x+y-7=0, and 3x-y-22=0 to be continued).

1. g(x)=-x 2-3, f(x) is a quadratic function, and f(x)+g(x) is an odd function.

Therefore, the sum of g(x) and f(x) quadratic terms is 0, (-3) + c=0 (monotonicity and over origin).

So f(x)=x2+bx+3

When x [-1,2], the minimum value of f(x) is 1b = (2 * root number 2) (vertex method) b = 2 * root number 2 rounded off (interval maximum).

So f(x)=x2-2(root number2)x+3

1: Let f(x)=ax 2+bx+c, then f(x)+g(x)=(a-1)x 2+bx+c-3, because f(x)+g(x) is an odd function, so a=1, c=3, f(x)=x 2+bx+3, because when x [-1,2], the minimum value of f(x) is 1, so f(-1)>=1, f(2)>=1, the solution is -3<=b<=3, so the axis is x=-b 2 in the range of (, closed interval, the sign is not easy to play), discuss when in (-1, (at this time the minimum value of 1 is its lowest point), find b; When (,-1), the minimum value is f(-1)=1, find b.

2: f(x)=ax 2+bx+c, if a>b>c, and f(1)=0, so a+b+c=0, so b 2-4ac=(a+c) 2-4ac=(a-c) 2>0 (because a>c), then two solutions, i.e., two zeros.

3: log2f(a)=a, and because a≠1, so a=2, because f(log2a)=-b, so b=-2, so f(x)=x 2-x+2, the minimum value is 7 4, at this time, log2x=1 2, x=root number

To be honest, I really want to help you, but can't do anything about it. Actually, I'm very interested in mathematics, but because I haven't touched mathematics for 2 years, and I don't need to study mathematics in college, I'm completely unfamiliar, and I don't know what some meanings are, such as odd functions, I don't know what they are, and I failed!

=-sin(∏+7)/8sin∏/7

f(x)=(1+cos2x+8sin^2x)/sin2x=[2(cosx)^2+8(sin)^2x]/2sinxcosx=2(cosx)^2/2sinxcosx+8(sin)^2x/2sinxcosx

cotx+4tanx

00,cotx>0

f(x)>=2 root number cotx*4tanx=4cotx=4tanx is an equal sign.

i.e. (tanx) 2=1 4, tanx=2

So you can get an equal sign.

So the minimum value = 4

m^2+2m-sinx=0

m^2+2m+1=1+sinx

m+1)^2=1+sinx

0<=1+sinx<=2

So 0<=(m+1) 2 <=2

2-1<=m<=√2-1

Solution 1 (1): When x belongs to r, there is always f(x) 0 i.e., ) = (a-2) x 2 + 2 (a-2) x-4 0 classification discussion) 1, when a-2 = 0, that is, a=2-4 0, conform.

2. When A-2≠0, there is.

A-2 0, i.e. A2

a^2-4a≥0

A 4 or A 0

In summary: A 4 or A 0

2): Eyes of the x belong to [

Next, let's use Vedic theorem to find the range of numbers for a).

1. 2x-1 times of a When a is less than 1, 2x-1》x-4, x is greater than -3;

2. One-half of x + one-half of the negative half of x, obtained by opening the square, -1 time of x+x + 2=4+2=6, so one-half of x + one-half of the negative half of x is equal to the positive and negative root number 6;

x+x-1 times = 4

Both sides are squared at the same time, the power of x + the negative power of x +2 = 16, then the power of x + the negative power of x = 14

x to the third power -x to the negative power.

x-1/x)^3+3x^2*1/x-3x*1/x^2=（x-1/x)^3+3x-3/x

x-1/x)[(x-1/x)^2+3]=4*(16+3)

763、 (a+b)^3=a^3+3a^2b+3ab^2+b^3(a-b)^3=a^3-3a^2b+3ab^2-b^3

1. When discussing a>=12x-1a>0. 2x-1>x-4 to get x>-32, with 6 The reason is that if one of the numerators of x is equal to a, then the square of a is subtracted by two to get x plus one x to get 4, so the square of a is 6, that is, the value is evaluated as 6!!

The same goes for the latter two!

3. In the second formula, add three times a squared by three times a squared and then triple ab squared and then add b to the third power of the second formula, replace the first two plus signs in the first formula with minus signs!

1.Discuss the case of dividing a, when a > 1, when a < 1, discuss the range 2 of x, (one half of x + one negative half of x) square x + x -1 time 2 so.

Other analogies. 3.(a+b).

(a+b).

Just take it apart and calculate it yourself.

Good luck with your studies!

<0a^(x-4)(a^(x+3)-1)<0

a (x-4) constant "0

a^(x+3)-1<0

a^(x+3)<1

If 00, x>-3

If a>1, then x+3<0, x<-3

If a=1, x is an empty set.

2.（1）（x^(1/2)+x^(-1/2)）^2=x+x^(-1)+2

x^(1/2)+x^(-1/2)=√6

2）x^2+x^(-2)=(x+x^(-1))^2-2=14(3) x^3+x^(-3)=(x+x^(-1))^3-3(x+x^(-1))=52

3.(a+b)^3=a^3+3a^2b+3ab^2+b^3(a-b)^3=a^3-3a^2b+3ab^2-b^3