x-y is not always less than x x+x y +y y. Let x=1 and y=-1. Then, x-y=2 but x x+x y +y y=1-1+1=1

At start you have to Say, Solving on integer numbers

@christianlopez1148

5 ай бұрын

exactly, there are infinite solutions

What had to be solved in whole numbers?

a cubic with 4 solutions. cool. (a double cubic i suppose)

x^3 = 91 - y^3 x = the cube root of (91 - y^3)

THANKS PROFESOR !!!, VERY INTERESTING !!!!

@learncommunolizer

4 ай бұрын

You are welcome! Thank you very much!!

Observe that 91 is sum of cubes of 4 and 3 I.e 91= 64 + 27. Hence, answer is x = 4 and y = -3

How you assume that product of x and y is positive in the beginning?

I just made a list of cubes of 1 to 10, ie, 1 to 1000. Cubed x must be greater than cubed y, so x>=5 (125 for x=5 vs 64 for x=4). The difference between adjacent cubes (ie, smallest distance) must also be less than 91, so y

@user-ye6go4ft7q

4 ай бұрын

Очень уж длинное решение,так явно видно,что х=6 а у=5; Отрицательных ответов не надо для куба-неверно !!! 6³-5³= 91 216-125=91 91=91

@user-im2qo6hh5e

4 ай бұрын

100-9=91

At no time did you indicate that x and y are integers. Your solution is incomplete until you check for other real, and indeed, complex solutions.

@ccudmore

5 ай бұрын

45.5 x 2 also = 91 There’s actually an infinite number of solutions. This probably would map as an interesting 3d object

@francescorusso7730

4 ай бұрын

​@@ccudmore Two dimensional instead.

NONSENSE: one equation with two unknowns, you need as many equations as degrees of freedom.

@MrArcan10

4 ай бұрын

as I understand the only integer solutions required

Sir, because x.y=30 it is not posible for -4,3 and -3,4?

X^3 - Y^3 = 91

X=5&y=4

x=6, y=5

x-y

@user-tetris

4 ай бұрын

多分誤りですね...反例も容易に見つけられてしまいますし...

@user-tetris

4 ай бұрын

(x,y) ∈ R² x-y ⇔x²+xy+y²-x+y > 0 ···① だから f(x,y):=x²+xy+y²-x+y=0 のグラフを書いてみる。 すると明らかに①を満たさない（x,y）∈Z² が存在することがわかる。 またf(x,y)は楕円の方程式3x²+y²=2を反時計回りに45度だけ回転させた後、x軸方向に1,y軸方向に-1だけ平行移動したもの。

@windyyw

3 ай бұрын

wrong. true is 4（x^2+xy+y^2）≥（x-y）^2，that is （x+y）^2≥0。so only （1，91）and （7，13）can be possible solution。

@user-tetris

3 ай бұрын

@@windyyw なぜx-y>0といえるの？

x=6 and y=5.

1x91=91

(6;5) (5;6)

Die Japanner snap niet niet veel van. Oneindig veel oplossingen. Niet de eerste keer dat hij in de fout gaat.

6³-5³=91

Không có điều kiện x,y là số nguyên nên suy ra cách giải này sai. Đáp án bài này là vô số nghiệm

Use soroban

You based your solution on assumptions and trial and error That isn’t logical nor acceptable

Почти 19 минут расписывать то,что решается за 4 минуты? Вы издеваетесь над нами?

@johndavidmyself8039

5 ай бұрын

It's called "teaching." "Teaching" takes longer than the learned doing. There's no indication you've solved this. But feel free to be snarky and irrelevant.

@heinrich.hitzinger

5 ай бұрын

Everything is trivial if you know the solution. However, that's not the point.

废话太多了