The master of solving middle school math problems comes in and has 3 problems... Give 10 points

1.Substitute t=3s= and t=5s= respectively

3=2u

u=put u= in

sa=1 when t=7.

s=1+s=2. a+b=2 ①

c-3=-2 ②

2a+(-6b)=2③

Solution c=1

2- Gotcha. 2a+2b-2a+6b=2

8b=2 b=

Substitute b=.

a+a=a= b= c=1

3.Solution: Set a type x kilogram and B type y kilogram.

16x+12y=720

14(x+y)=720-20

The solution yields x=30 y=20

A: 30,000 grams for type A and 20,000 grams for type B.

Question 1,. The second question, you made a mistake and couldn't do it. The third question, A 30 kg, B 20 kg.

Question 1,.

Question 2, a=, b=, c=1;

The third question, A 30 kg, B 20 kg.

Question 1 s= Question 2. a= b= c=1 Question 3: 30,000 grams for A and 20,000 grams for B.

Let cm=x, then cn=14-x, and amc and bnc are both right-angled triangles, which are according to the title.

6^2+x^2=8^2+(14-x)^2

Solution x = 8 km.

Answer: Station C should be built at a distance of 8 kilometers from point m.

Do you understand?

Hope it helps.

o(∩_o~

Solution: The painting carries this out of its side picture, and the envy is as follows:

Then: abcd must be a parallelogram, and dc = cf = ab = ag is known to be obtained: ac = 2

If a is used as a perpendicular brother of BC, then: ae = 1, and it is easy to know that cae = abc = a

Therefore, we can see that cos a = 1 (2 ).

BPC and APD are equilateral triangular missing mu disfigurement, BC=PC, BCP= CPB=60°, APD=60°

PCD= Vozhi BCD- BCP=90°-60°=30°, and CPD=180°- Stuffy CPB- APD=60°, CDP=180°- PCD- CPD=90°

cd=cp*cos30°=√3/2*bc，tandbc=cd/bc=√3/2<>

Analysis (1) Connecting AG, the circumferential angle of the diameter pair is a right angle and the perpendicular diameter theorem knows AGF= AEF=90°, then the four points A, E, G, F are on the circle with AF as the diameter, and the midpoint of AF is the center of this circle, so the distance from the midpoint of AF to the four points of A, E, G, F is equal, and it is known by the circumferential angle theorem, the circumferential angle of the chord fg is known fag= feg, which is known by the equal co-angle of the same angle, bag= bfe, and the outer angle of the triangle is equal to the sum of the two inner angles that are not adjacent to it, BGN= bfe+ feg, and bam= fag+ bag, there is mab= ngb, which is known by the circumferential angle theorem ngb= nab, so there is mab= nab, that is, ab bisects man;

3) Connecting OC and BM, since it is known that there is OC=5 and CE=3, then in RT OEC we get OE=4 from the Pythagorean theorem, so AE=OA+OE=9, in RT AEF EF=6, from the Pythagorean theorem AF=313, it is easy to obtain RT ABM RT AFE to obtain AME=ABAF, we can find AM=AB AEAF=301313 from (1) AB bisects Man, so AN=AM=301313

Answer: Solution: (1) If AG is connected, then AGF= AEF=90°, and the distance from the midpoint of AF to the four points A, E, G, and F is equal, that is, the four points A, E, G, and F are on the same circle

The circumferential angle of the string fg to which fag = feg

bag+∠abg=∠bfe+∠fbe=90°，∴bag=∠bfe．

BGN= BFE+ FEG, while BAM= FAG+ BAG, MAB= NGB

ngb=∠nab，∠mab=∠nab．

ab divide the man

3) Connect OC, BM, OC=5, CE=3, and get OE=4 in RT OEC

AE=9 at RT AEF, EF=6, AF=313

ab=10, obtained by rt abm rt afe amae=abaf, am=ab aeaf=301313

ab divides man, an=am=301313

2) The question will not be, please choose me, I gave it in detail, and my hands were tired.

(a2+4a+4）+（b2-2b+1）-4-1+5=0（a+2)2+(b-1)2=0

So a=-2, b=1

Substituting the solution, the original formula = 1 3

2.Because x-y=1, x=y+1

Substituting x2+y2=2; (y+1)2+y2=2y2+2y+1+y2=2, i.e. 2y(y+1)=1, so, 2xy=1

xy=1/2

I'm sorry, I don't know how to square it, I use it"2"Instead,

1 It is known that the square of a + the square of b + 4a-2b + 5 = 0 Find a+b in the fraction of a-b

The square of a + the square of b + 4a-2b + 5 = 0

a^2+4a+4)+(b^2-2b+1)=0（a+2)^2+(b-1)^2=0

And because (a+2) 2 0, (b-1) 2 0, a+2=0, b-1=0

A=-2 and b=1

So a+b a-b=[(-2)+1] [(-2)-1]=1 32 Knowing the square of x + the square of y=2 x-y=1 then the value of xy is ( ).

x-y=1x-y)^2=1^2

x^2-2xy+y^2=1

2xy=1-(x^2+y^2)=1-2=-1xy=1/2

1 a squared + b squared + 4a-2b + 5

a^2+4a+4)+(b^2-2b+1)=（a+2）^2+(b-1)^2=0

a=-2 ,b=1

a-b parts a+b = 1 3

Let's talk about the second question first, let's make it simpler, use the perfect square formula, square x-y = 1 overall square, get the square of x plus the square of y minus 2 times xy is equal to 1, from the square of the known condition x plus the square of y = 2, the whole substitution, get -2xy = -1 so xy = 1 2All in all, this question is a test of the use of a perfectly squared formula.

Let's talk about the second question, it's actually very simple, it depends on how you use this 5. You split 5 into 4 and 1, and match a and b respectively in a perfect flat way, that is, the square of (b-1) on the square of (a plus 2) = 0, and the square of a home 2 is greater than or equal to 0, and the same goes for it. So a family 2 is equal to 0, a is solvable, and b is solvable in the same way.

Another generation into, solved put!

These two mentions are the mentions of the full square formula used to make use of the test, and the ground is relatively rusty at one time, but after practicing more and getting familiar with it, there is no problem, these two mentions are very representative mentions, and the exam also loves to test this kind of mention!

I hope my answer will help you, my old level is not high, but only two times in the three-year mathematics test of junior high school did not get a full score, so, at least the foundation is still dropped!

Split 5 into 4 plus 1, match all the terms of A with 4, and all the terms of B with 1 to form a perfect square to solve the values of A and B.

Question 1: Since the total number of people is 50, we can set the number of people in the fourth group is x; Add up the total number of people to fifty! Here's how: (3a+2b)+ do the math and you're done! x= substitution is very simple, do the math yourself!

Question 1: There is no way to do what you described! Question 4: Take a good look at it, and extract one of the last two items - what does it become if you "minus"? 10-(2x-3y)=10-1=9 Note that the negative sign is to change the sign!

Did you find the same question? Take a good look! The answer to the first question is 36, and the second question is of the same type as the question you asked earlier, and you can directly substitute a 2 in the first two items, 2 (2x squared - 2) - 8 = 20 - 8 = 12

1.Group 4: 43-13a 2-4b; A=2, B=3, the fourth group of 18 people 2, original = 3x square - [5xy square - 4xy square + 3 + 2x square y square] 3x square - xy square - 3-2x square y square.

When x=3, y=2, the original formula = 3*3 squared - 3*2 squared - 3-2*3 squared *2 squared.

3, the original = -2a square b + 4ab square + 3a square b-2ab square.

A squared B + 2AB squared.

4, 10-2x+3y=10-(2x-3y)=10-1=95, k=5 4, the algebraic formula does not contain the xy term.

1.The second group has (3a+2b) 2+7=people, so the fourth group has 50-(3a+2b)- (individuals, when a=2, b=3, the fourth group has 18 people.

x^2-[5xy^2-(4xy^2-3)+2x^2y^2]=3x^2-xy^2-3-2x^2y^2=-36

3.-2a^2b-(-4ab^2)-(3a^2b)-2ab^2=ab(-2a+4b+3a-2b)=ab(a+2b)

Because 5-4k=0, k=.

1. The fourth group: 50-[(3a+2b)+(3a+2b) 2+7+(2a+b)]=

When a=2, b=3, =18 people.

2. Original formula = 3x square - xy square - 2x square y square - 3; Original = -363Original formula = a square b + 2ab square;

4. 10-（2x-3y）=10-1=9；

5.（ 4k+ 5）=0 ， k=5/4

182.This question is a bit messy, I didn't understand it.

3.This question is a bit messy, I didn't understand it.

Write the first two questions clearly, and I'll give it to you!

1. 43-13a/2-4b；18；

2.Original = 3x squared - xy square - 2x square y square - 3; -36；

3.A squared B + 2AB squared.

5. k=5/4

Group 4: 43-13a 2-4b; A=2, B=3, the fourth group of 18 people.

A squared B + 2AB squared.

cax（-4x+y-xy)

1.Group 4: 43-13a 2-4b; A=2, B=3, the fourth group of 18 people.

The score is so high. Therefore, in the process of learning, you must pay attention to the details and master all the knowledge points. Xuehai Education provides you with:

pq should be equal to.

To form a rhombus egfh, then one must take the midpoint of the ab, bc, cd, da sides ah=2ae=, then by the Pythagorean theorem eh is equal to pq so pq is equal to.