Hard Math!!

1.)If x^3-xyz=2 y^3-xyz=6 z^3-xyz=20 Find the maximun of x^3+y^3+z^3

2.)If x^2+y=12=y^2+x Find x^2+y^2

*3.) x=(1/sqrt2)+(1/sqrt3)+(1/sqrt4)+...+(1/sqrt1000) find the most Integer that small than or equal x.

#MathProblem

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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

Comments

The second math...is it have a condition for x and y? If they are both interger, you can just put the table and take your answer there!

Kawaii Kuma - 7 years, 6 months ago

Sr, This is how I solve the second problem.

You have : x^2 +y = 12 x+y^2 = 12

Minus those 2, you will have : (x-y)(x+y-1) = 0 x=y=3 will be the answer so I won't do anything with it here.

So x+y = 1. Add those 2 equation up there, you will have : (x+y)^2 -2xy + x + y = 24

Replace x+y. Find xy. Use xy and x+y to know x and y. I think you won't find it...I don't really know!

Thanks for reading this :)

Kawaii Kuma - 7 years, 6 months ago

how do you know that its are =3?

Areefil Raksanit - 7 years, 6 months ago

@Areefil Raksanit – Do some math. Try to replace and find out. This works everytime if the number you are finding is not big (like from 1->10). When you have the answer, it is easier for you to figure how to do the math :)

Kawaii Kuma - 7 years, 6 months ago

areefil baeng

ashwin chawla - 6 years, 10 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$