Urgent! Wait! Math problems, math problems! Hurry, thank you!

Hello landlord! Factoring is the use of cross multiplication. They all take the first and third numbers apart and match them into the second number.

a⁴+a²b²-56b⁴=(a²-7b²)(a²+8b²) 1×1 -56=-7×8 1=-7+8

x²y²+7xy-44=(xy-4)(xy+11) 1×1 -44=-4×11 7=-4+11

a²-16a+60=(a-6)(a-10) 1×1 60=(-6)×(10) -16=-6-10

a²-7a-60=(a-12)(a+5) 1×1 -60=-12×5 -7=-12+5

a²+32a+60=(a+30)(a+2) 1×1 60=30×2 32=30+2

a²+11a-60=(a+15)(a-4) 1×1 -60=-4×15 11=-4+15

x²-20xy+96y²=(x-12y)(x-8y) 1×1 96=(-8)×(12) -20=-8-12

x²-4xy-96y²=(x-12y)(x+8y) 1×1 -96=-12×8 -4=-12+8

x²+10xy-96y²=(x+16y)(x-6y) 1×1 -96=-6×16 10=-6+16

x²+28xy+96y²=(x+24y)(x+4y) 1×1 96=24×4 28=24+4

I hope it helps you, if you don't understand, you can still ask, or ask me for help directly!

What is this? Didn't you even have an equal sign.

There is no such thing as an equation, how can you solve it?!

3.Solution: It turns out that there are x people in each middle school participating in the competition.

From the first assumption: x is divisible by 15.

From the second hypothesis, x is divisible by 13 and the remainder is 12

x=12 (15-13)*15=90 (person) (90+2)*2=184 (person).

A: In the end, a total of 184 students from the two universities participated in the competition.

The diagonal of the square is the diameter of the circle, and the side length of the square is determined by the Pythagorean theorem: d root number 2, so the area of the square is the square of (d root number 2) = d square 2 is square.

According to the Pythagorean theorem, d'2 is equal to 2 times the square of the side length a (i.e., d'2=2a'2), while the area of the quadrilateral is equal to the square of the length of the side (a'2)。Therefore, the quadrilateral area in this question can also be expressed as ('2).I hope I can understand it, so I don't have to give it points.

1.Let event A be at most two hits by A and event B at least two hits by B

then p(a)=1-4*(1 2) 4-(1 2) 4=11 16, p(b)=1-(1 3) 4-4*(2 3)*(1 3) 3=8 9

A is independent of b, then p(ab) = p(a) * p(b) = 11 16 * 8 9 = 11 18.

2.Hits: 0 1 2 3 4

Score: -4 0 4 8 12

Probability p: 1 81 8 81 24 81 32 81 16 81 Mathematical expectation e=-4 81+0+96 81+256 81+192 81=.

(1)[1-4*1 16-1 16][1-1 81-4*2 81]=11 18 (2)Missed -4 points 1 81 into 1 0 points 8 81 into 2 4 points 24 81 into 3 points 8 points 32 81 All in 12 points 16 81 So B's mathematical expectation is -4*1 81+0*8 81+4*24 81+8*32 81+12*16 81=20 3

(1) According to the knowledge of probability, the probability of A hitting two at most p(A)=11 16;

The probability that B hits at least two of them p(B) = 8 9;

p=p(a)* p(b)=11 18;

2) Let the probability of A hitting a time be p(a), then.

p(0)=(1/3)^4=1/81;

p(1)=c41*(1/3)^3*(2/3)^1=8/81;

p(2)=c42*(1/3)^2*(2/3)^2=24/81;

p(3)=c43*(1/3)^1*(2/3)^3=32/81;

p(4)=c44*(2/3)^4=16/81;

The mathematical expectation is e=p(0)*(4)+p(1)*(0)+p(2)*4+p(3)*8+p(4)*12=20 3

The average of the two numbers must be between these two numbers.

Apparently the third number is 540, the fourth number is 520, the fifth number is 530, the sixth number is 525, the seventh number is, the eighth number is, the ninth number is, and from then on, the average only changes in the decimal part and does not affect the whole number. So the integer part of the 100th number is 526.

2an=an-1 + an-2 ,2(an-an-1)=-(an-1-an-2), let bn=an-an-1, then bn is a proportional series, calculate bn, and then use the superposition method to calculate an, then we get a100

500, because the number is 500. So the nth one is also greater than 500!

It's 500, and you don't need to count them one by one, because the two numbers given are both above 500, and the subsequent numbers will average more than 500 no matter how they are and gradually approach 500.

4x to the eighth power y -5x -xy + x -12a to the third power x seventh power.

x=0, with a minimum value of 10.

The second solution is not unique. For example, when a b takes 0 2, it is -3; When 1 0 is taken, the rolling sensitivity of large dead branches is 5.