It can be done this simple way. We have (x+1/x) =1 or x^2_x+1=0 multiplying both sides by (x+1) it is x^3+1=0 , or x3=--1.again x^2+1/x^2=1--2=--1. Now x^32={x^3) ^10.x^2 I. e x^2 since x^3=--1, it's power 10 is 1.hence, the problem is reduced to (x^2+1/x^2) which is =--1.ans.
Since the exponent of x is not so large, I think that this problem can be solved without much effort, without using De Moivre's theorem. That is, (x+1/x)^2=x^2+1/(x^2)+2=1⇒ x^2+1/(x^2)=1^2-2=1-2=-1, Therefore, {x^2+1/(x^2)}^2=x^4+1/(x^4)+2⇒ x^4+1/(x^4)=(-1)^2-2=1-2=-1, Therefore, {x^4+1/(x^4)}^2=x^8+1/(x^8)+2⇒ x^8+1/(x^8)=(-1)^2-2=1-2=-1, Therefore, {x^8+1/(x^8)}^2=x^16+1/(x^16)+2⇒ x^16+1/(x^16)=(-1)^2-2=1-2=-1, Therefore, {x^16+1/(x^16)}^2 =x^32+1/(x^32)+2⇒ x^32+1/(x^32)=(-1)^2-2=1-2=-1
Solo tengo una duda, si x toma cualquier valor real eso siempre dara una suma como sale el negativo. A menos que involucren numeros imaginario. Aunque no dicen en que grupo estan trabajando.
Great one
I agree with Mohammed Ali just square both sides and move constant to left 4 times and you get same result much easier
Yes you right ✅️
Difficult solution there another Easy solution by square the left side 4 times
It can be done this simple way. We have (x+1/x) =1 or x^2_x+1=0 multiplying both sides by (x+1) it is x^3+1=0 , or x3=--1.again x^2+1/x^2=1--2=--1. Now x^32={x^3) ^10.x^2 I. e x^2 since x^3=--1, it's power 10 is 1.hence, the problem is reduced to (x^2+1/x^2) which is =--1.ans.
Nice
Thank you so much 😊🙏
Please how do I solve 1/(a-b)+1/(a+b)
Since the exponent of x is not so large, I think that this problem can be solved without much effort, without using De Moivre's theorem.
That is, (x+1/x)^2=x^2+1/(x^2)+2=1⇒ x^2+1/(x^2)=1^2-2=1-2=-1, Therefore, {x^2+1/(x^2)}^2=x^4+1/(x^4)+2⇒ x^4+1/(x^4)=(-1)^2-2=1-2=-1,
Therefore, {x^4+1/(x^4)}^2=x^8+1/(x^8)+2⇒ x^8+1/(x^8)=(-1)^2-2=1-2=-1, Therefore, {x^8+1/(x^8)}^2=x^16+1/(x^16)+2⇒ x^16+1/(x^16)=(-1)^2-2=1-2=-1,
Therefore, {x^16+1/(x^16)}^2 =x^32+1/(x^32)+2⇒ x^32+1/(x^32)=(-1)^2-2=1-2=-1
I agree with Mohammed Ali
What is the convergent and divergent of sequence
Good morning sir, I’m one of your subscriber, I have a problem in mathematics I don’t know if you can help me with it. Thanks a lot
Sorry. Your exercise is 1/X^2 + X^2 = 1 not X + 1/X
Vc complicou muito num problema simples.
sir which software do you use for writing and also please mention the pen tablet you use...
I used Samsung Galaxy tab s6
Solo tengo una duda, si x toma cualquier valor real eso siempre dara una suma como sale el negativo. A menos que involucren numeros imaginario. Aunque no dicen en que grupo estan trabajando.
Hello, master.
Please send me the lessons of Mathematics from the beginning to the end in the order of the second lesson in the same way, thank you.
It's not easy
Assalamu Alaikum
That means there doesn’t exist a real value for x to satisfy the solution.
Is this for grade 7
Giải sai
Why do you have to multiply with X²
to remove the denomenator to be easy for us
I don't really understand
I thought 4^2=16 sir?
Yes it's
@@tambuwalmathsclass but in the equation you said X^4^2=X^8
I didn't say x^4^2 =x^8
But (x⁴)² = x⁸
Or you can write it as:
(x^4)^2
Too long unnecessarily.
(X)^4=-(X^2+1)???????
Tomorrow is exam 😭😭😭
Thực hiện sai
Is this for grade 7