Ask to design a math problem, the answer is 7249, any question type in junior high school can be, sl

Known: and are the two real roots of the equation x -170x+7200=0, and > without solving the equation, try to find the value of the algebraic formula + 10-1)( 10-2)? (Cue: Vedic theorem).

Answer: According to Veda's theorem, +=170, =7200, let :

a=αβ+/10-1)(β/10-2)=αβ+20α-10β+200)/100 ①

b=αβ+/10-1)(α/10-2)=αβ+20β-10α+200)/100 ②

：a+b=2αβ+2αβ-30(α+400]/100=2×7200+(2×7200-30×170+400)/100=14497 ③

:a-b=(10 -10) 100= [(4 ] 10= (170 -4 7200) 10=1 take positive).

：2a=14497+1=14498

a=7249, i.e. the value of the algebraic formula is 7249

Key points: Without solving the equation, when using Veda to understand the value of the asymmetric formula, it is generally necessary to set up the symmetry of the original formula for superposition or subtraction, so that the conditions of Veda theorem can be brought into the formula, and then the problem is often turned into a problem of solving a system of binary equations, which is used here, but this formula usually gives the size relationship between the two roots, such as judging the positive and negative values of the square when the formula is used.

If you have questions, you can ask questions, the code word is not easy, hopefully, if you need other aspects of the content to compile, please reply.

There is a four-digit number, let the number on the single digit be x, the ten-digit number is y, the hundred-digit number is z, the thousand-digit number is, x+y+k=20, 2x+y+z=24, 2y-k=1Find this four-digit number.

1. C 3 to the 75th power = 27 to the 25th power; 2 to the power of 100 = 16 to the power of 25; Because 27 to the 25th power.

Greater than. 16 to the power of 25; So 3 to the 75th power is big.

2、b3、c

AB did not bend the letter of the acre mark, but the answer should be C

ab = 3 reason: one millionth of the volume of the water cube = , which is equivalent to c

8.The topic is endless.

1+2+3...To x, you count him out and you are very powerful;

A watermelon business household to 2 yuan of ** to buy a batch of small watermelons, to yuan 3 kilograms of ****, can be sold kilograms per day, in order to **, the operator decided to reduce the price of sales, after investigation found that this small watermelon per price reduction of yuan kilograms, can be sold more than 40 kilograms per day, in addition, daily rent and other fixed costs of a total of 24 yuan, the operator wants to make a profit of 200 yuan per day, should the price of each kilogram of small watermelon be reduced? (To detail the process, good plus points!) ）

Three people went to buy vegetables, which was originally 25 yuan, and the boss charged 2 yuan more. But out of conscience, he refunded one yuan per person. In this way, the three of them only spent 9 yuan. 3x9=27, plus the boss binary, only spent 29 yuan. What did the other dollar do?

（1）y=-x+2

Problem solving: Substituting x=0 into y=2x+2 to obtain point C coordinates (0,2), and then combining point B coordinates (2,0) to obtain the BC expression: y=-x+2

2) (0, -8 + 4 root number 5) or (0, -2 + 2 root number 2).

Problem solving: Assume the coordinates of the p point (xp, 0).

When xp < 0, (note that xp >-1 at this point).

P is located in the AO segment, connected to PE, according to the triangle similarity, there is OE OC=OP OA, i.e. OE 2= -XP 1, so OE = - 2XP, E point coordinates are (0, - 2XP).

Substituting x=xp into the equation of the straight line ac gives the q point coordinates (xp, 2xp+2), because of the rhombus, there is pq=cq, that is, (2xp+2) 2=(xp) 2+(2xp) 2, and the solution gives xp=4+2 root number 5 (greater than 0, excluded), or xp=4-2 root number 5

So the coordinates of point e are (0, -8+4, root number 5).

When xp > 0, (note that xp < 2 at this point).

P is located in the OB segment, connecting PE, according to the triangle similarity, there is OE OC=OP OB, that is, OE 2= XP 2, so OE= XP, E point coordinates are (0, XP).

Substituting x=xp into the equation of the straight line bc gives the q-point coordinates (xp, -xp+2), due to the rhombus, there is pq=cq, i.e., (-xp+2) 2=(xp) 2+(xp) 2, and the solution gives xp= - 2 + 2 root number 2, or xp = -2 - 2 root number 2 (less than 0, excluded).

In the same way, when p is between ob, op 2 + oe 2 = ep 2 is solved to e(0,2 2-2).

To sum up, there are two (0,2 2-2) (0,4,5-8) that meet the conditions, and you can chase them as soon as possible if you don't understand.

It's time to sleep. The first question is too simple, and the analytic formula is y=-x+2

Solution: (1) Straight line ac: y=2x+2

Let y=0, get x=-1, then a(-1,0).

Let x=0, get y=2, then c(0,2).

Let the expression of the straight line bc be y=kx+b, then.

2=b0=2k+b

So b=2, k=-1

Therefore, the expression for the straight line bc is: y=-x+2

The expression of the straight line bc is y=-2x+4, use your own brain for the rest, don't be lazy, and tell you the rest of the answer tomorrow morning.

Sorry, it's y=—x+2 wrong

The coordinates of e are (0,4 5,8).

If you don't understand the solution, you can contact me.

The area of the first picture can be divided into triangle ABC + triangle BCDS = 1 2m * AE + 1 2m * DE

Mention the common factor:

s=1/2m（ae+de)

1 2mn The second ** method is similar to the above figure, just add to subtract.

(For detailed answers, see Jingyou.com).

Because ae bc the area of the quadrilateral is multiplied diagonally and multiplied by 1 2 s ＝1/2mn

Proof: Because ABCD is a parallelogram.

So ad=bc

ab = dc angle bad = angular bcd

AB parallel DC

AF parallel be

So the angle ead = the angle aeb

Because an bisect angle bad

So the angle ead = angle bae = 1 2 horn bad

So the angle bae = the angle aeb

So ab=be

Because cm bisects the angle bcd

So angular bcf = 1 2 angular bcd

So the angle AEB = the angle BCF

So AE parallel MC

So ab bm = be bc

So BM=BC

So the triangle BMC is an isosceles triangle.

Because of the BF vertical MC

So BF is the midline of the triangle BMC.

So mf=cf

Because AB is parallel to DC

So the angle m = the angle fcd

Angle maf = angle d

So the triangle AMF and the triangle DCF congruence (AAS) so af=df=1 2ad=1 2bc

So af=be

Because AF parallel BE (verified).

So the quadrilateral abef is a parallelogram.

Because ab=be (verified).

So the quadrilateral abef is a diamond.

Proof that AN bisects BAT to E, CM bisects BCD, and AD bisects F fae= bae, abf= EBF

Another quadrilateral ABCD is a parallelogram.

fab+∠abe=180º, af//be∴∠fba+∠bae=90º ，abf=∠ebf=∠bfa∴ae⊥bf, ab＝ae

Let AE cross BF at the point O, then ASA can be known as ABO Ebo BE=BA=AF

The quadrilateral ABEF is a parallelogram.

The quadrilateral abef is a diamond.

Hello, let's talk about this question with you.

Let ae and bf intersect at the point g, 1, bisect by an bad---bae= eaf, by af be---bae= aeb, so ab=be;

2. Divide BAD by AN, divide BCD by cm, and AE FC is easy to prove; BF MC, hence BF AE; That is, if BG is the height of the isosceles triangle ABE, we can know that FBE= FBA.

3. By af be---fbe= bfa, and fbe= fba, so bfa= fba---ab=af;

4. From the conclusion of (1) and (3), be=af, and af be--- abef is a parallelogram, and ab=be, so it is a rhombus.

The key condition for this problem is that BF is perpendicular to MC

The solution is as follows: because BF MC, then FBC+ FCB=90 degrees, because AN bisects BAD, CM bisects BCD, and because ABCD is a parallelogram, AN is parallel to CM, so AEB= FCB, so FBC+ AEB=90 degrees.

So AE BF

So ABEF is diamond-shaped.

ABCD is a parallelogram.

AD = BC, AB = DC, Angle BAD = Angle BCD, AB Parallel DC, AF Parallel BE

So the angle ead = the angle aeb

Because an bisect angle bad

So the angle ead = angle bae = 1 2 horn bad

So the angle bae = the angle aeb

So ab=be

Because cm bisects the angle bcd

So angular bcf = 1 2 angular bcd

So the angle AEB = the angle BCF

So AE parallel MC

So ab bm = be bc

So BM=BC

So the triangle BMC is an isosceles triangle.

Because of the BF vertical MC

So BF is the midline of the triangle BMC.

So mf=cf

Because AB is parallel to DC

So the angle m = the angle fcd

Angle maf = angle d

So the triangle AMF and the triangle DCF congruence (AAS) so af=df=1 2ad=1 2bc

So af=be

Because AF parallel be

So the quadrilateral abef is a parallelogram.

Because ab=be

So the quadrilateral abef is a diamond.

Solution: Because bf mc, fbc+ fcb=90 degrees, because an bisects bad, cm bisects bcd, and because abcd is a parallelogram, an is parallel to cm, so aeb= fcb, so fbc+ aeb=90 degrees.

So AE BF

So ABEF is diamond-shaped.

The sourcing FMA, FCD congruence, the angle bisector is an, cm is parallel, the parallel angle AEB = ban = EAF, AB=be is a parallelogram, and finally ABEF is a diamond.

I don't know if you can understand it, it's almost enough to list what you know.

Dan+ CEA=180, so ECF+ CEA=180 so CF AE

So AECF is a parallelogram.

So ce=af

Then ASA certify FMA, FCD congruence.

I know later.