Simplified Middle School Mathematics Problem 5, Simplified Middle School Mathematics Problem

That doesn't seem right. Isn't it wrong?

I was dizzy when I saw it.

It's different from the one on the first floor.

Write it clearly, okay? No.

Day, I don't understand this! Can you write clearly?

The numerator and denominator are multiplied by the denominator at the same time, using the formula a2-b2=(a+b)(a-b), and the root number in the denominator is often eliminated in this method in the root fraction: =(3+ 3)[ 2+ (2- 3)] [( 2+ (2- 3))(2- (2- 3))]=[(3+ 3)[ 2+ (2- 3)] [2-(2- 3)] =[(3+ 3)[ 2+ (2- 3)] 3 = (1+ 3)[ 2+ (2- 3)] Double radical, which is matched into (a b) 2 form and then squared, becomes a single-layer radical: 2- 3) = (4-2 3) 2 = (3-2 3+1) 2 = [(3)2-2 3+1] 2 = ( 3-1)2 2 =( 3-1) 2 Original = (1 + 3)[ 2+( 3-1) 2] =(1+ 3)[2+( 3-1)] 2 =(1+ 3)(1 + 3) 2 = (1+ 3)2 =(4+2 3) 2 = 2(4+2 3) 2 = 2(2+ 3) =2 2+ 6

kn-2k-2n=0

Add 4 on both sides at the same time

kn-2k-2n+4=4

It can be deformed as: k(n-2)-2(n-2)=4

Re-deform, it's fine.

k-2）×(n-2)=4

n(k-2)=2k

k(n-2)=2n

Multiply two to get nk(n-2)(k-2)=4kn and simplify to get (n-2)(k-2)=4

Hello First of all, you need to know a formula: a 3-b 3=(a-b)(a 2+ab+b 2) so there is (n+1) 3 -n 3=((n+1)-n)((n+1) 2+(n+1)n+n 2)=3n 2+3n+1 I hope it can help you.

According to this formula: x 3-y 3=(x-y)(x 2+xy+y 2) substituting yields (n+1-n)[(n+1) +n+1)n+n]=(n+1) +n+1)n+n

n²+2n+1+n²+n+n²

3n²+3n+1

Original = n to the third power + 1 + three times n square + 3n to the third power.

Three times n squared + 3n + 1

a²a*[（p+q）*1/2%]²

a*[（p+q)*1/200]²

a* [p+q)²/40000]

a*(p²+2pq+q²/40000)

ap²+aq²+2apq)/40000

I don't know if you can read it!

Minus eighths a to the sixth power and b to the third power.

I just got it (covering my face).

To answer this question, we must first know that the form of a quadratic equation can be (x+a)(x+b)=x +bx+ax+ab=x +(a+b)x+ab

The quadratic equation of the general form x +bx+c=0

It is not difficult to find that b=a+b

c=ab After you have done your preparations, go back to the question.

x²-（k+3）x+2k+2=0

i.e. -k-3=a+b

2k+2=ab

Well, just guess.

a=k+1，b=2

No, then a=-k-1, b=-2

Apparently yes. Go back to where you started (x+a)(x+b)=x +bx+ax+ab=x +(a+b)x+ab

Then apparently (x-k-1)(x-2)=0

This is my own method of pushing (covering my face) because I can't understand what the teacher is saying =. = hhhhh above.

Solution: Simplify. a (a-b) + (b+1) (b-a) fractional addition and subtraction, obtain.

a-b-1）/（a-b）

I'll suggest a place for you. Go to Math Know-it-all and ask a professional teacher. I was also introduced by a classmate. It's really good. The teachers were prompt.

Where did the question come from? Weird, right? Just treat it as if it doesn't have a solution

The subject is unknown, but the best one should be a (a-b) +b (b-a) +1, so that the final result is 2

Is it the sum of a part of a to b and a part of a b+1 of a b-a?

Or is it the sum of a part of a -b and a part of b and a of b-a?

Or is it the sum of a-b parts a + b-a parts b +1?