Ask a few math questions! A few math problems in the third year of junior high school, urgent!!

y=4/3x^2-4/3

y=4 15x&+16 15x+12 15. y=ax 2+bx+c substituting a,b and expressing b,c with an algebraic expression containing a. Substituting b and c into the preceding equation and using Vedica's theorem to express the relationship between the two roots.

and the arithmetic square root 2 of the absolute value of x1-x2 (the arithmetic square root of ((x1 x2) 2-4x1x2), and substituting the relation into the above equation can find a, so that the analytic formula of the quadratic function can be obtained.

Let y=a(x-1) 2-4

Find the two relations of a(x-1) 2-4=0, which can be found by the method of 1 problem.

Note that you want to go to a negative number, because a is negative and there is no minimum value.

1. The chord centroid distance is the radius of one and a half of the liquid sun 2, and the radius is 4, so half of the chord length is 2 3, so the chord length is 4 3

2. The radius oc 13, the chord center distance is 13-8 5, so half of the chord length is 12, therefore, the chord is buried in the long ab limb chain 24

3. Make OH CD from the center of the circle, and the vertical foot is H. then h is the midpoint of cd, so, he ce ch 7 5 2. The quadrilateral ofeh is rectangular, so of he 2

Solution: Because ABCD is a rectangle (AB=BC), BE is perpendicular to AC and DF is perpendicular to AC, so the right-angled triangle ABE is congruent with the right-angled triangle CDF. So cf=ae=1

And because the angle ACB + angle BAC = 90 degrees, and the angle ABE + angle BAC = 90 degrees, so the angle ABE = the angle BAC, so the right triangle ABE is similar to the right triangle ACB, according to the nature of the similar triangle.

ab ac=ae ab where ac=ae+ef+fc=1+2+1=4, ae=1

So the square of ab = ae times ac=4

So ab=2

As described above, fc=1, ab=2

That's my answer, I hope it helps, my favorite subject in junior high school was math).

Fc is equal to 1 because ab=dc, ab cd, be perpendicular to ac, df perpendicular to ac so aeb is equal to dfc, dcf is equal to aeb so abe dfc, so ae is equal to cf=1

fc=ae=1, you can use congruent triangles to prove it, and the conditions are not enough to find ab.

(1) Let ad=x meters, then ab=(80-2x) meters x * (80-2x)=750

The solution is x=15 or 25

Because when x=15, it doesn't fit the topic and is discarded.

So x=25

A: 25 meters for AD and BC and 30 meters for AB.

2) Same as above: x*(80-2x)=810

There is no solution to the original equation.

So no.

(1) Let ab=x, ad=(80-x) 2

So s=x*(80-x) 2=750

So x = 30 or 50 (rounded).

So x=30

So ad=bc=25, ab=30

2) In the same way, we get x*(80-x) 2=810, so x 2-80x+1620=0

Because =80 2-4*1620=-480 is less than 0, it can't.

The first question is ad=x, then ab=80-2x, s=x(80-2x)=750, then x=15 or 25, (when x=15, ab exceeds 45, rounded off), so ad=25, ab=30

The second question is s=x(80-2x)=810, the equation can be reduced to xx-40x+405=0, that is, (x-20)(x-20)+5=0, and the equation has no solution, so it cannot be enclosed into 810 square meters.

Because: 45 = 5 9 = 15 3;

So: ab=dc=15;ad=bc=50 can meet the requirements of the question;

Answer: Make ab=dc=15m; AD=BC=50m to make the area of the rectangular site 750m2

In the same way, because: 810 = 9 9 10 = 9 90 so: ab=dc=9;ad=bc=90 can meet the requirements of the question;

Answer: Make ab=dc=9m; AD=BC=90m to make the area of the rectangular site 810m2

a^2+ab-b^2=0

Because the value of a b is required, b ≠ 0

So divide the original by b 2 to get it.

a^2/b^2）+a/b-1=0

Think of a b as a whole and let a b = x

So there is x 2 + x - 1 = 0

The solution yields x=( 5-1) 2

or x= ( 5+1) 2

x is the value of a b.

1) Draw the quadrilateral abcd, so that ab cd, and adc=90°, and then draw a little b to the perpendicular line of ad, be, and the vertical foot is e

2) Among the four line segments AE, BE, CD, DE, there is a certain quantitative relationship between some line segments, please write it.

The two equations represent these quantitative relationships (each equation contains 2 or 3 line segments), and choose one of the equations to explain why it is true.

,ab=30;

2.Let ad=x, then ab=80-2x;

x(80-2x)=810;

After simplification: 2x*x-80x+810=0;

Discriminant: 80*80-4*2*810<0;There is no real solution.

So it can't reach 810cm2

1. Let the length of AB be XM and the length of BC be YM, then there is:

xy=750

x+2y=80

Solve the system of equations to obtain: x=30, y=25.

That is, the fence is divided into three sections, two sections are 25 m long, one section is 30 meters long, and two sections of 25 m fence are perpendicular to the wall.

2. According to the title, assuming that there is a satisfying area of 810m, it can be obtained:

xy=810 ①

x+2y=80 ②

According to the above system of equations, we can get by substituting x=80-2y into the equation:

80-2y)y=810

The deformation is: (y-20) +25=0

Since: (y-20) 0, the above equation will not hold.

That is, there is no such a value of y, so that the area of the enclosed rectangular field is 810m.

Because it's a quadratic equation.

The coefficient in front of x 2 cannot be equal to 0

So k+3≠0 k≠-3

And k 2-4 (k+3) 0

k-6)(k+2)≥0

k 6 or k -2

To sum up: k 6 or k -2 and k ≠-3

1. a=x 2-3x+m=-x 2-3x+m, so x 2=-x 2, so x=0, m=0, a=0

2、(2a^2+2b^2)(2a^2+2b^2-2)=4(a^2+b^2)(a^2+b^2-1)=3

Let x=a2+b2

4x(x-1)=3

x^2-x=3/4

x^2-x-3/4=0

x = 3 2 or -1 2

3. x=(k+1 or - under the root number (2k-3)) 2x1 or =0 is |x1|=x1=x2

2k-3=0,k=3/2

2) x1<0 is |x1|=-x1=x2

k+1=-k-1

k=-1 (2k-3) is not a real number and is not true.

So k=3 2

The area of ABC is: 1 2AB*AC*Sina=4ADE, and the area of ADE is: 1 2AD*AE*Sina=1 and AD=AC*COSA

ae=ab*cosa

Substituting into the above two formulas, you get |cosa|=1 2, get a = 60 or 120

The area of the triangle ADE is 1:4 compared to the area of the triangle ABC, then AD*EF AB*CD=1 4, and EF CD=AE AC, substituting the previous formula, (AD AC)*(AE AB)=1 4, that is, the cosine of the angle DAC = 1 2, so the angle DAC = 60 degrees, so the angle A is equal to 60 degrees or 120 degrees.