quando alla fine si trova con i quadrati, fa un errore, 3 al quadrato + radice di 6 al quadrato - due x 3 x radice di 6 HA 2 SOLUZIONI : (3 - radice di 6) al quadrato, ma anche (radice di 6 - 3) al quadrato ... e idem sotto: (radice di 5 - 2) al quadrato, ma anche (2 - radice di 5) al quadrato ... quindi posso avere 4 soluzioni : 1) num 1/den 1 2) num 1/den 2 3) num 2/den 1 4) num 2/den 2
English is NOT my first language. There's is no letter P in my local dialect making it difficult to pronounce properly. But I'll do something about it. There are English learning materials online to enable me pronounce perfectly. I'm sure I'll achieve all the objectives.
Sqrt[Sqrt[675]-Sqrt[648]]/Sqrt[Sqrt[243]-Sqrt[240]]=6+3Sqrt[5]-2Sqrt[6]-Sqrt[30] It’s in my head.
Rationalizing the denomination didn't make the answer any simpler...so why do it?
You have to simply the radicals.
Merci baucoup
You're welcome 🙏I’m glad you found it helpful! 😎🔥💯
quando alla fine si trova con i quadrati, fa un errore, 3 al quadrato + radice di 6 al quadrato - due x 3 x radice di 6 HA 2 SOLUZIONI : (3 - radice di 6) al quadrato, ma anche (radice di 6 - 3) al quadrato ... e idem sotto: (radice di 5 - 2) al quadrato, ma anche (2 - radice di 5) al quadrato ... quindi posso avere 4 soluzioni :
1) num 1/den 1
2) num 1/den 2
3) num 2/den 1
4) num 2/den 2
Just like √((-1)^2) =1 not -1, so is √((√6-3)^2)= 3-√6.
solution looks like a complication rather than simplification 🙂
I appreciate your feedback! 🤩💯Thanks for sharing your perspective. 🙏💕🥰✅
Good
Thanks🙏
That is B, not P , ok?
English is NOT my first language. There's is no letter P in my local dialect making it difficult to pronounce properly. But I'll do something about it. There are English learning materials online to enable me pronounce perfectly. I'm sure I'll achieve all the objectives.
Lol just pulling your legs. 😅😅
Решил в уме.
Don't all square roots have two solutions?
Only when solving for a variable
@@markyukhno9419
He did just that half way through when he said sqrt 81=9