algebra

[oblique pagoda]

How can i solve it?

[night silo]

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[turbid glen]

@oblique pagoda do you want to rationalise it?

[oblique pagoda]

Yea

[turbid glen]

@oblique pagoda show me the full question

[oblique pagoda]

Ok

[turbid glen]

@oblique pagoda what's the conjugate of the bottom

[oblique pagoda]

Idk

[turbid glen]

wtf

@oblique pagoda I say multiply everything and see what works

[oblique pagoda]

How

[turbid glen]

bro just equate them and see

[oblique pagoda]

What

[slow cairn]

$\frac{a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d}{(a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d) \cdot (\sqrt{6}-\sqrt{3}+\sqrt{2}+1)} \$ The idea is to multiply this : $\ (a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d) \cdot (\sqrt{6}-\sqrt{3}+\sqrt{2}+1)= \ \sqrt{6}(a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d) \ -\sqrt{3}(a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d) \ +\sqrt{2}(a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d) \ +a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d= \ 6a+3b\sqrt{2}+2c\sqrt{3}+d\sqrt{6} \ -3a\sqrt{2}-3b-c\sqrt{6}-d\sqrt{3} \ +2a\sqrt{3}+b\sqrt{6}+2c+d\sqrt{2} \ +a\sqrt{6}+b\sqrt{3}+c\sqrt{2}+d= \ \sqrt{6}(d-c+b+a)+\sqrt{3}(2c-d+2a+b)+\sqrt{2}(3b-3a+d+c)+6a-3b+2c+d$

[gilded oak]

Ludwig

[oblique pagoda]

But ir show variant:A

[turbid glen]

skip the question @oblique pagoda

[slow cairn]

The next step is to solve this : $\ d-c+b+a=0 \ 2c-d+2a+b=0 \ 3b-3a+d+c=0$

[gilded oak]

Ludwig

[bright finch]

How can i solve it

[rough burrow]

is this a test 👀

[bright finch]

Unfortunaly yes

[rough burrow]

whaa

okie one thing ☠️

sadly we can't help in tests

[bright finch]

ii tried to solve it,but i couldnt