Math Algebra and Geometry Problems in the Second Year of Junior High School!!

Solution: The coordinates of point C are (0,-3), right? I see that you are a junior high school student, so you don't need to talk about high school knowledge.

The method in high school is not difficult to do (the distance from the point to the straight line), but if you are a junior high school student, let's talk about a stupid method. But it is practical and tests the ability to calculate. The calculations are slightly cumbersome.

Since the point m is on a hyperbola, let the coordinates of m be (x,-8 x). The crossing point A is parallel to the y-axis to make a straight line L, and the crossing point M is perpendicular to the y-axis to make a straight line, so that it intersects with the straight line L at a point, assuming that it is the point D.

Obviously, the coordinates of point d are (-1, -8 x). This is understandable, right?

Connect point D to point C.

then, Area of ACM = ADM - ACD - CDM.

a(-1,0) ,c(0,-3) ,d(-1,-8/x)，m（x，-8/x）,x >0 。

Area of ACM = 5 2 , Area of ADM = 1 2 * 8 x| *x-(-1)|, area of ACD = 1 2 * 8 x| *1|, area of cdm = 1 2 * x-(-1)| 3-（-8/x）|So, the area of ACM = adm - ACD - CDM , the solution is, x1 = 2, x2 = -3 4 (rounded).

So the coordinates of the point m are (2,-4).

The crossing point m is the perpendicular line of the x-axis, and the intersection of the x-axis is at the point b

The coordinates of the setting point m are (x, y).

Then the area of ACM is the area of ACO + the area of the trapezoidal OCMB is the area of AMB.

So we get the equation: 5 2 1 2 ·3 + 1 2· (3+y)·x -1 2(1+x)·y

Simplify to 3x - y =2

Solve the system of equations 3x - y =2

y = -8/x

We get x =2 y =4 or x =-4 3 y = -6, so the coordinates of the point m are (2, -4) or (-4, 3, 6) (this point is in the second quadrant).

m is a little hyperbola?

It's not hyperbolic in the picture.,Isn't it hyperbolic in junior high school?

On the diagram, the ordinate of c should be <0, but the problem is c(0,3), which is not solved

Proof: The length of the two right-angled sides is a, b, and the length of the hypotenuse is c, so that c=b+kc squared=asquared+bsquared.

C square = b square + 2 * BK + K squared.

So a squared = 2*bk + k squared = k (2b + k) and because a is prime.

So k = 1 so a square = k (2b + k) = 2b + 1

So 2 (a + b + 1) = a square + 2a + 1 = (a + 1).

We know a = (c+b)(c-b).

Because a is a prime number, c+b and c-b are not equal.

So c+b=a c-b=1

i.e. 2b = a -1

So left-right = 2b + 1 - a = 0

Proven hope it helps you.

Hee-hee! It's been a long time since I've done a math problem!

Here's how I did it :

Because 2a+2b+2=a+2a+1

2b+2=a²+1

And because a = c -b

So 2b+1=c -b

c²=(b+1)²

a²=(b+1)²-b²=2b+1

So left equals right!

(1) Extend CB to L, so that BL=DN, then RT ABL RT And, so AL=AN, and then verify AMN AML, we can find MAN= MAL=45°;

2) Let cm=x, cn=y, mn=z, according to x2+y2=z2 and x+y+z=2, the problem can be solved according to =4(z-2)2-32(1-z) 0 :(1) as shown in the figure, extend cb to l, so that bl=dn, then rt abl rt and, so al=an, 1= 2, nal= dab=90°

and mn=2-cn-om=dn+bm

bl+bm=ml

amn≌△aml

man=∠mal=45°

2) Let cm=x, cn=y, mn=z

x2+y2=z2

x+y+z=2, then x=2-y-z

So (2-y-z)2+y2=z2

2y2+(2z-4)y+(4-4z)=0 =4(z-2)2-32(1-z) 0

i.e. (z+2+2 root number 2) (z+2- 2 root number 2) 0 and z 0

z 2 root number 2-2 if and only if x=y=2- root number 2 when the equal sign holds, at this time s amn=s aml= 1 2ml ab= 1 2z, therefore, when z = 2, root number 2-2, x=y=2- root number 2, s amn takes the minimum value as root number 2-1 This question examines the application of the Pythagorean theorem in right triangles, examines the properties that all sides of the square are equal in length, and each inner angle is a right angle!!

Analysis: Because the four sides of the square are equal and the four corners are 90°, i.e. ab ad, b adc 90°, move bam to dam', i.e., rotate the BAM around point A by 90° counterclockwise to dam'location.

Solution: Extend CD to M'to make DM'＝bm，ad＝ab，∠b＝∠adc＝90°

Then bam dam'

bam＝∠dam' am＝am'

mam'＝90°

The circumference of the MCN BC CD

mn＝bm＋dn＝m'n

amn≌△am'n（sss）

man＝∠mam'＝∠bad＝45°

Solution: Extend CD to M'to make DM'＝bm，ad＝ab，∠b＝∠adc＝90°

Then bam dam'

bam＝∠dam' am＝am'

mam'＝90°

The circumference of the MCN BC CD

mn＝bm＋dn＝m'n

amn≌△am'n（sss）

man＝∠mam'＝∠bad＝45°

P is a spike in a right-angled trapezoidal ABCD, where AD is parallel to BC, ABC=90°, and PA=1, PB=2, PC=3, AB=BC=2AD, and the area of the trapezoidal ABCD is found [Figure, above].

is the point inside the isosceles right triangle abc, abc=90°, and pa=1, pb=2, pc=3, find the area of the triangle abc [Figure, middle].

is the midpoint of abc, put the right-angled vertex of the right-angled triangle guess pin ruler at d, and its two right-angled edges intersect ab and ab at the points e, f, try to compare the size of be+cf and ef, and state the reason [with figure, below].

It doesn't seem to be complete.

If you make a graph accurately, the bie is not an equilateral triangle.