A Nice Math Olympiad Problem • You should know this Trick!
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- čas přidán 9. 09. 2024
- This is an interesting question with amazing concepts!
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解释得很清楚,受教了。谢谢!
Good solution. I did it by cubing both sides of 2=a-b.
8=(a-b)^3 =a^3-b^3+3 a b (b-a)=98+3 a b (-2)
15=a b
Now, cube both sides to get
15^3=a^3 b^3 = (x+49)(x-49)=x^2-49^2
x^2=15^3+49^2=5776
It's not easy to see x=+-76 without a calculator but if you know trick for cubing numbers ending in 5, you can see 75^2=5625 and from there you could try 76^2=(75+1)^2=5625+150+1=5776.
Good job Man 👏👏👏
Nice work
Set the left hand side to be Y. So Y=2, and Y^3=8. Or x+49-3(x+49)^(2/3)(x-49)^(1/3)+3(x-49)^(2/3)(x+49)^(1/3)-x+49. One has 8=98-3[(x+49)^(1/3)-(x-49)^3](x^2-49^2)^(1/3)=98-3*Y*(x^2-49^2)^(1/3)=98-6*(x^2-49^2)^(1/3). One gets 90=6*(x^2-49^2)^(1/3), or 15=(x^2-49^2)^(1/3). It follows 15^3=x^2-49^2, or x^2=15^3+49^2=3375+2401=5776=76^2. Finally, x=-76 or x=76.
Let's analyse f(x)=cuberoot(x+49)-cuberoot(x-49)
Observe that x is shifted 49 units in negative and positive directions, so the function is symmetric around x=0. If x is a solution, then -x is also a solution.
Also, f(x) is monotonic and continuous with an extremum occurring at the axis of symmetry x=0. f(0)=2cuberoot(49). Let's try some values: 5-3=2 cuberoot(125)-cuberoot(27)=2. An obvious solution is x=76, f(76)=cuberoot(125)-cuberoot(27)=2. So another solution is x=-76. These are only points of intersections of the graph of f(x) with horizontal line y=2. There are no other points of intersection (roots) in real numbers.
Too long you just power. Cube the initial équation...and replace by the 2 value whenever ...you meet it
Nice ❤
Great
Nice job
Finally a new one!
A bit complex for me
Change your dp
My answer:
x=76