Root number x 1 x 1 puzzle solved!!

Your teacher's unraveling method is right.

Because it's an equation, the left and right sides are equal, but in different forms, for example, if the numbers on both sides of the 2=2 equal sign are obviously equal, and square them to 4=4, isn't it the same?

Let me explain:

2 means squared.

Root number x+1=x-1

Solution: (root number x+1) 2=(x-1) 2

x+1 =x^2-2x+1

x-x^2+2x=1-1

3x-2^2=0

x(3-x)=0

The solution is x1=0 x2=3

So the solution of the equation is x1=0 x2=3

The two sides are squared to remove the root number.

If they are all root number x+1, then the root number on the right is still not solved, and the solution is still not solved, isn't it?

The teacher is right, you are right, both sides are multiplied by a root number, the left and right sides are still equal, but the root number still can't go, so in order to solve, it is still necessary to square.

You are a very creative thinker, which I admire, and you can see things that others don't notice! There's a future.

Let me illustrate a: root number x+1=x-1

b: root number x+1=x-1

Multiply the two formulas a and b to get it.

x+1）=(x-1)^2 ( b)

Okay to explain your thoughts.

The root number x+1=x-1, you mean multiply the root number x+1 at the same time, then x+1=(x-1)*(root number x+1)= (x-1) 2 ( c) to know that you and the teacher are doing the right thing.

But do you see that the solution of c can solve x;

The teacher's purpose is to remove the root number.

Common practice.

Let a=b then the teacher's approach.

a^2=b^2

Your practice. a^2=a*b=b^2;

Actually, it's the same.

2 sides squared. x+1=x^2-2x+1

x^2-3x=0

x=3 or x=0

x-1>=0 ,x>=1

So, at the end x=3

Explanation: If a=b then aa=bb ab=bb aa=ab aa=bb

Let t=[(1-x) (1+x)] 1 2) then x=(1-t 2) hail (1+t 2).

dx=-4tdt\(1+t^2)^2

Friends are chaotic [(1-x) (1+x)] 1 2)*(1 x)dx 4t 2dt (t 4-1).

2dt\(t^2+1)+∫dt\(t-1)-dt\(t+1)2arctant+in|(t-1)\(t+1)|+c2arctan+in|[1-(1-x^2)^(1\2)]\x|+c

The method of arguing and fighting is as follows, carrying the grind.

Please do the divination test:

It is relatively simple to excite the points of the cherry blossoms, and it can be directly set or formula.

Parse: let t x, then x t, dx dt 2tdt so primitive (arctan x) x(1 x)dx [arctant t(1 t )]2tdt 2 arctant (1 t )dt

2∫arctant d（arctant）＝2*1/2*(arctant)²＋c.

arctan√x)²＋c.

Use the equivalent infinite burn to make a laugh Xiao Hu Yin:

ln(1+x)~x；(1+x 2)-1 (x 2) 2 In this way, the original limit formula becomes.

lim(x*x) Picosis contains [(x 2) 2]=2

Answer: (1-x)+x-1)+x 2-1 algebraic formula makes sense:

1-x>=0

x-1>=0

So: 1<=x<=1

So: x=1

Substitution = (1-1) + 1-1) + 1 2-1 = 0 + 0 + 0 = 0 so: (1-x) + x-1) + x 2-1 = 0

According to the squared difference formula (a+b) (a-b) = a -b

So the original formula = (root number x) - root number (x-1)) = x - (x-1)=1

= square of the root number x - square of the root number x-1.

x-(x-1)=1

According to the squared difference formula.

Original = square of root number x - square of root number x - square of root number x-1.

x-(x-1)=1