WOW. This is wow approach. Well explained

Great job. The explanation is so cleared sir, thanks for this video.

Thank you very much. I enjoy learning on this channel

Superb thanks from Bihar India

Mind Blowing Solution..Thanks from India

@onlineMathsTV

Ай бұрын

Smile....😂😂 Its a pleasure to serve you in this capacity and we are most joyous because you derive some positive impacts from this video tutorial sir. We are humble to receive this wonderful, unique, soul touching and encouraging comment from you sir...🙏🙏🙏 We all here @ Onlinemaths TV love you dearly sir...❤️💖💖💖❤️

x=6, y=5. из предположения, что числа последовательные и x-y=1.

@onlineMathsTV

Ай бұрын

Thanks sir. You the boss here because you are good at what you do sir. Much respect sir 🙏🙏🙏

Thanks for the video snd your teachings

@onlineMathsTV

Ай бұрын

Our pleasure! And thank you for watching our contents and commenting sir. Much love sir...❤️💖💖💖

The solution such as mind blowing. But there is a wrong line of solution is in the 2nd vertical line. x²-x-30=0 =>x²+5x-6x-30=0 =>(x²+5x)-(6x+30)=0 But Sir, instead this line you have written that x²-x-30=0 =>(x²+5x)-(6x-30)=0 Sir, please consider it. Thanks of a million Sir.🎉❤🎉

Salut Professeur, I am ivorian, I come From Ivory Coast in Africa. Dieu vous bénisse, mais je n'ai pas bien compris vers la fin. C'est à dire (x-y)=1. Pouvez vous apporter plus d'explication ?

The substitution (x-y) = 1 was the missing step that I was missing, nice trick.

@onlineMathsTV

Ай бұрын

Thanks a million sir. Much love.

@ElishyahIsrael

28 күн бұрын

I wish you would explain that substitution slowly because I don't understand that at all.😢

Thank you very much.

@onlineMathsTV

Ай бұрын

You are most welcome sir! We love you dearly sir.

Aussi pouvez vous expliquer la procédure vous permettant de trouver :x^2+5x-6x-30=0. Je ne comprend aussi pas.

The game actually began at 5:35... 👌🏻👌🏻👌🏻.. By cofficient comparison...rest was just simplification..gr8.. From india 👍🏻👍🏻👍🏻

@onlineMathsTV

Ай бұрын

Hahahahaha...🤣😂😂 You are very much on point sir. Thanks so much for the support given to this channel so far by watching our videos and at the same time commenting. It has really encouraged us a great deal. A million thanks to you sir. You are great and you shall remain great in life boss. Much love from all of us @ Onlinemaths TV to you boss ❤️❤️💖💖💖

wow best

@onlineMathsTV

Ай бұрын

Thanks for the acknowledgment sir.

It is difficult to solve when a condition /substitute is missing.

Muy buen ejercicio

@onlineMathsTV

Ай бұрын

Thanks a million sir and thanks for watching our video. We love you dearly sir.

My comment is not a criticism, it's just another way Nice solutions, but... At moment 6:44 of the video you could use. (x - y)(x² + y²) = 61... ( x - y) = 1, Than (x² + y²) = 61, and y y² + (1 + y)² = 61 2y² + 2y - 60 = y² + y - 30 = 0 (y + 6) (y - 5) = 0 y = 5 is a positive solution. (x² + y²) = 61 (x² + 5²) = 61 x² = 61 - 25 = 36 x = 6 is a positive solution. ( x , y ) = ( 7, 6) = 49 + 36 > 61 ( x , y ) = ( 7, 5 ) = 49 + 25 > 61 ( x , y ) = ( 7, 4) = 49 + 16 > 61 ( x , y ) = ( 7, 3) = 49 + 9 Than x is not seven. ( x , y ) = ( 6 , 5) = 36 + 25 = 61 x = 6 and y = 5 Thanks for your nice work,

Ну, хорошо. Как это надо излагать. Равенство имеет правую и левую части. В правой части неизвестные х и у содержатся только в виде произведения ху. Далее следует разложение (х - у)(х квадрат + ху + у квадрат) = ху + 61. Тогда для того, чтобы левая часть данного равенства могла быть равна правой, необходимо допустить существование системы двух уравнений с двумя неизвестными, из которых первое -- линейное и такое, что х -- у = 1, а второе -- квадратное и такое, что х квадрат + у квадрат = 61. Решение данной системы не представляет сложности и даёт два ответа.

@onlineMathsTV

Ай бұрын

Thanks so much for the exposition sir, Thanks for the eye-opening approach sir. Much respect and love sir... 💖💖❤️❤️🙏🙏

No me aclaro con este paso: (x-y)(x^2+y^2)+xy(x-y)=xy+61; (x-y)=1?????????. Ruego aclaración

@JoseMiguelBuisan

Ай бұрын

Hace lo siguiente: (x-y)(x^2+xy+y^2)=1(xy+61) Entonces: x-y=1 x^2+xy+y^2=xy+61

@elgone8524

Ай бұрын

Muchas gracias Jose Miguel Buisan

@JoseMiguelBuisan

Ай бұрын

Se ha complicado un poco la vida. Yo la había resuelto de la manera que te he explicado. Son dos soluciones las que cumplen la ecuación original (6,5) y (-5,-6) pero en el ejercicio solo pedía solucione de enteros positivos.

@michaelhuppertz6738

Ай бұрын

Simplemente ha sustituido x-y = 1. Este paso obtiene (x^2+y^2) + x*y = x*y + 61, lo que significa que puede utilizar x^2 + y ^2 = 61 , sólo tiene que utilizar la ecuación de sustitución x = 1 + y.

@onlineMathsTV

Ай бұрын

thanks sir