I used an other method. y=x-6. You substitute and add the first fraction with the last and add the two others. You keep the -4. In the end, you have (y^2-4)/(y^2+5y+4) + (y^2-6)/(y^2+5y+6) =-2. (We simplify by 2). Let a=y^2+5y+5 b=y^2-5 We substitute, we developp and we have. (ab+1)/(a^2-1)=-1 (We simplify by 2 again) a^2+ab=0 a(a+b)=0 a=0 ou a+b=0 And after it is easy.
Each term in LHS add 1 --> RHS=0 Simplifying yields common numerator 2(x-6) which can be cancell out giving: [1/(x-2)]+[1/(x-3)]+[1/(x-4)]+[1/(x-5)=0 Add the two terms in the middle as well as the remaning two, yielding common numerator 2x-7 which can be cancelled out, yielding [1/(x²-7x+10)+[1//(x²-7x+12)]=0 Let u=x²-7x+11 [1/(u-1)]+[1/(u+1)]=0 --> u=0 x²-7x+11=0 --> x=½[7±sqrt(5)]
@@MyOneFiftiethOfADollar "solving this needs a lot of intuition" - However the presenter came upon the solution we are grateful to him for bringing it to us.
😊 hay que dejar claro que todos estos artificios aplicados solo surten efecto, solo si existen algunos patrones, en caso contrario, no quedará otra que PICAR PIEDRA 😊😊
Esta questão parece ser complicada, mas não é! É linda! Muito linda! Parabéns pela escolha! Brasil - Setembro de 2024. This question seems complicated, but it's not! It's beautiful! Very beautiful! Congratulations on your choice! Brazil - September 2024.
Yo habría usado el mínimo comun multiplo y habría resultado la ecuación como una fraccion normal.....si después me queda una ecuación de 4° grado uso Ruffini y listo....no es lo mas rápido, pero si lo mas efectivo.
Respected Vijay Sir, Good afternoon. Although there is method of solution which follows grammar of mathematics but I tried in a different way. Pls let me know this way is permitted by Mathematics or not -4=(-1)+(-1)+(-1)+(-1) So (X-10)/(x-2)=-1 X-10=2-x X+x=2+10 2x=12 X=6 2nd term X-9=3-x 2x=12 X=6 Similarly all terms produce the same same result.... Addition of t1&t4 will be (2x^2-24x+64)/(x^2-7x+10) Let a=2x^2-24x/60 b=x^2-7x+10 So the expression will be (a+4)/b...... eqn1 Again the addition of t2 &t3 will be (2x^2-24x+60)/(X^2 -7x+12) =a/(b+2).......eqn2 Eqn1 + eqn2 {(a+4)/b}+{a/(b+2)} =ab+2a+4b+8+ab=-4(b^2+2b) =2ab+2a+4b+8=-4b^2-8b 2ab+2a+8b+8=-4b^2-4b 2a(b+1)+8(b+1)=-4b(b+1) (2a+8)(b+1)=-4b(b+1) 2a+8=4b a+4=(-2b) Put the values of a & b 2x^2-24x+64=-2(X^2 -7x+10) 2X^2 -24x+64=-2x^2+14x-20 4x^2-38x+84=0 4x^2-24x-14x+84=0 4x(x-6)-14(x-6)=0 (X-6)(4x-14)=0 X-6=0 or 4x-14=0 X=6 or 4x=14 X=14/4=7/2 Hence x=6 or (7/2)
You husbands, in the same way, continue dwelling with them according to knowledge. Assign them honor as to a weaker vessel, the feminine one, since they are also heirs with you of the undeserved favor of life, in order for your prayers not to be hindered. 1Peter 3:7 see saksi ni Jehova
In the numerator and denominator x-6 missed. Hence asume x = 6
x-10/x-2 + x-9/x-3 +
x-8/x-4 + x-7/x-5
=6-10/6-2 + 6-9/6-3 + 6-8/6-4 + 6-7/6-5
--4/4 + -3/3 + -2/2 + -1/1
-1-1-1-1=-4
I used an other method.
y=x-6.
You substitute and add the first fraction with the last and add the two others. You keep the -4.
In the end, you have
(y^2-4)/(y^2+5y+4) + (y^2-6)/(y^2+5y+6) =-2. (We simplify by 2).
Let a=y^2+5y+5
b=y^2-5
We substitute, we developp and we have.
(ab+1)/(a^2-1)=-1 (We simplify by 2 again)
a^2+ab=0
a(a+b)=0
a=0 ou a+b=0
And after it is easy.
If product two numbers is zero, any one can be equal to zero. 2x-12=0, x=6.
(2х-12)=0, х=6
(1/(х-2)+1/(х-3)+1/(х-4)+1/(х-5))=0, х=3,5
Thank you for the video! Very well
Explained ❤❤❤❤❤
Thank you so much for your appreciation 🙋
Very nice! Thank you.
Each term in LHS add 1 --> RHS=0
Simplifying yields common numerator 2(x-6) which can be cancell out giving:
[1/(x-2)]+[1/(x-3)]+[1/(x-4)]+[1/(x-5)=0
Add the two terms in the middle as well as the remaning two, yielding common numerator 2x-7 which can be cancelled out, yielding
[1/(x²-7x+10)+[1//(x²-7x+12)]=0
Let u=x²-7x+11
[1/(u-1)]+[1/(u+1)]=0 --> u=0
x²-7x+11=0 --> x=½[7±sqrt(5)]
good one. but while resolving we should have compared x with 2,3,4,5 to avoid 0 in denominators.
Thanks! You're right, we should be careful about the denominators otherwise it will becomes zero
You must right,
x is not equal to 2,3,4,5
That's an important rule
Yes, you are right other wise denominator will becomes zero 👍
Excellent problem with four REAL solutions! Solving this needs a lot of intuition. Thank you!
Intuition? , More probable that presenter had to look at the solution first.
Very few could solve this based on intuition.
A nice problem nonetheless.
@@MyOneFiftiethOfADollar "solving this needs a lot of intuition" - However the presenter came upon the solution we are grateful to him for bringing it to us.
15:34 Excellent work
Thank you
15:39 The denominator can't be less than 5, so I try x = 6 and get 4 times - 1 , which is indeed - 4.
Okay, I see, 4 solutions...Nice problem!
Thanks for your valuable feedback 🌺
Greetings , you explained well ,thank you we enjoyed it,.
Thanks for your valuable feedback
X-10/x-2=-1. X=6 or,x-9/x-3=-1so x=6
😊 hay que dejar claro que todos estos artificios aplicados solo surten efecto, solo si existen algunos patrones, en caso contrario, no quedará otra que PICAR PIEDRA 😊😊
x = 6. (= (10+2)/2 = (9+3)/2 = (8+4)/2 = (7+5)/2. Mental work, 20 sec for verifying.
Ok 👍 I agree it takes 20 seconds for verifing but not for solution
Kademe, kademe anlatan, iyi bir öğretmen sınıfına giriyorsunuz. ❤👍
Esta questão parece ser complicada, mas não é! É linda! Muito linda! Parabéns pela escolha! Brasil - Setembro de 2024. This question seems complicated, but it's not! It's beautiful! Very beautiful! Congratulations on your choice! Brazil - September 2024.
Thank you
7777j@@vijaymaths5483
I like the subtitles, he speaks x-5 x-2 then two seconds of gibberish I can't discern, then again speaking math.
SENSATIONAL!
Great explanation
(X-10)/(X-2)+(X-9)/(X-3)+(X-8)/(X-4)+(X-7)/(X-5)=-4 X=7/2=3.5 X=6 X=(7 ± Sqrt[5])/2=3.5±0.5Sqrt[5]
Oh .. i was looking forward to another 15minutes of checking each of the answers by substituting into the equation lol
How to thumbnail like u?
Excellent
Thanks a lot 🎈
Yo habría usado el mínimo comun multiplo y habría resultado la ecuación como una fraccion normal.....si después me queda una ecuación de 4° grado uso Ruffini y listo....no es lo mas rápido, pero si lo mas efectivo.
숫자 3과 7을 저렇게 헷갈리게 쓰는 사람을 처음 보네.. 수학을 다루면서 어떻게 숫자 3과 7을 저렇게 쓸 수 있지?
I think substitution is an easier way
Very good
Thank you for watching!
Thank you sir
Most welcome 👍
Common denominator 60. Combine X's,, combine numerals. Am I missing something?
X=6❤❤❤
Всё мельче и мельче
👏👏👏
И для какого класса в Америке это уравнение ?? 🤔🤔🤔🤔
В России это уровень 7 класса
X=6
Thx a lot
Thank you for your feedback!
0:50
Пишите, пожалуйста не так мелко. На телефоне не видно ❤
OK 👍 I will try my best in next time, thank you for your good suggestion ⚘️
Ouch! Long problem!
Long!!! But best 👌 problem
Is there is any chance that this will came in PMO
6
X=12
Уснуть можно
Fala língua de gente, esse trem já difícil entender em português agora vc. Imagina em grego.
-3y,1
Is that math olympiad 🤣🤣
u = x - 6 => x = u + 6
(u - 4)/(u + 4) + 1 + (u - 3)/(u + 3) + 1 + (u - 2)/(u + 2) + 1 + (u - 1)/(u + 1) + 1 = 0
(2u)[1/(u + 4) + 1/(u + 3) + 1/(u + 2) + 1/(u + 1)] = 0
u = 0 => *x = 6*
1/(u + 4) + 1/(u + 3) + 1/(u + 2) + 1/(u + 1) = 0
(2u + 5){1/[(u + 4)(u + 1)] + 1/[(u + 3)(u + 2)} = 0
2u + 5 = 0 => u = -5/2 => *x = 7/2*
1/[(u + 4)(u + 1)] + 1/[(u + 3)(u + 2) = 0
(u + 4)(u + 1) = -(u + 3)(u + 2)
u² + 5u + 4 = -(u² + 5u + 6)
2u² + 10u + 10 = 0
u² + 5u + 5 = 0
u = (-5 ± √5)/2 => *x = (7 ± √5)/2*
Bu sorunun dört tane cevabı mı var
What if the number is like 3,5..... somethings else and not 4?
beautiful
Thanks for your valuable feedback 👍
2x _12 xuất hiện nếu Cộng thêm 1 cho các Phân số
X=6; x =7/2
Vous n avez pas le droit de mettre des parenthèses
Великолепное решение!!!❤
Thank you
What language is this fellow speaking,in between English ?.
Beautiful Indian language HINDI
@@vijaymaths5483 Okay,thanks.🤗..
(X-12)(1/x-2+1/x-3+1/x-4+1/x-5)=0
(X-12)×k=0
K différentes de0
X-12=0
X=12.
(x ➖ 10)^2/(x ➖ 2)^2 ➖ (x ➖ 9)^2/(x ➖3)= (x^2 ➖ 100)/(x^2 ➖ 4) ➖ (x^2 ➖ 81)/x^2 ➖ 9)= {x^0+x^0 ➖ }/(x^0+x^0 ➖) ➖ (x^0+x^0 ➖)/x^0+x^0 ➖ )={x^1/x^1 +x^1/x^1}= x^2/x^2 +(x ➖ 8)^2/(x ➖ 4)^2=x^2/x^2 +(x ➖ 64)/(x ➖ 16)=x^2/x^2+ {x^0+x^0 ➖}/{x^0+x^0 ➖ }={x^1/x^1+x^1/x^1=x^3/x^3+ (x ➖ 7)^2/x ➖ 5)^2=x^3/x^3+ (x^2 ➖ 49)/(x^2 ➖ 25)=x^3/x^3+(x^0+x^0 ➖)/(x^0+x^0 ➖)={x^3/x^3+x^1/x^1}=x^4/x^4(x ➖ 4x+4) I liked loved video as aways .
Respected Vijay Sir, Good afternoon. Although there is method of solution which follows grammar of mathematics but I tried in a different way. Pls let me know this way is permitted by Mathematics or not
-4=(-1)+(-1)+(-1)+(-1)
So
(X-10)/(x-2)=-1
X-10=2-x
X+x=2+10
2x=12
X=6
2nd term
X-9=3-x
2x=12
X=6
Similarly all terms produce the same same result....
Addition of t1&t4 will be
(2x^2-24x+64)/(x^2-7x+10)
Let a=2x^2-24x/60
b=x^2-7x+10
So the expression will be
(a+4)/b...... eqn1
Again the addition of t2 &t3 will be
(2x^2-24x+60)/(X^2 -7x+12)
=a/(b+2).......eqn2
Eqn1 + eqn2
{(a+4)/b}+{a/(b+2)}
=ab+2a+4b+8+ab=-4(b^2+2b)
=2ab+2a+4b+8=-4b^2-8b
2ab+2a+8b+8=-4b^2-4b
2a(b+1)+8(b+1)=-4b(b+1)
(2a+8)(b+1)=-4b(b+1)
2a+8=4b
a+4=(-2b)
Put the values of a & b
2x^2-24x+64=-2(X^2 -7x+10)
2X^2 -24x+64=-2x^2+14x-20
4x^2-38x+84=0
4x^2-24x-14x+84=0
4x(x-6)-14(x-6)=0
(X-6)(4x-14)=0
X-6=0 or 4x-14=0
X=6 or 4x=14
X=14/4=7/2
Hence x=6 or (7/2)
Excellent,Thank you for sharing Manoj sir but we get 4 solutions at the end🙏☺️☺️
In English, please? Ah fuck it ill use Mathematica
You husbands, in the same way, continue dwelling with them according to knowledge. Assign them honor as to a weaker vessel, the feminine one, since they are also heirs with you of the undeserved favor of life, in order for your prayers not to be hindered. 1Peter 3:7 see saksi ni Jehova
Очень долго и занудно
문제를 위한 문제일 뿐입니다. 저런 문제를 만드는 사람은 할 일이 전혀 없는 무가치한 사람임이 틀림이 없습니다. 창조적인 것에 시간을 쓰세요!
Excellent
x=6
x=6