Question

Aflati x, y € R, unde
√[(1-x)(x-y)] + √[(x+y)(2-x)] = 3/2

Answer 1

 

$\displaystyle\bf\\\sqrt{(1-x)(x-y)} + \sqrt{(x+y)(2-x)} = \frac{3}{2}$

Exercitiul are o singura ecuatie cu 2 necunoscute.

Daca numarul necunoscutelor este mai mare decat numarul ecuatiilor,

atunci avem o infinitate de solutii.

Voi incerca sa gasesc o solutie (una din cele o infinitate) folosind o metoda empirica sau babeasca sau prin incercari  sau toate la un loc.

Trebuie sa scapam de radicali, dar nu putem ridica la putere.

Facem asa:

Daca sub fiecare radical cele 2 paranteze ar fi egale,

atunci vom avea radical din ceva la patrat si ramanem fara radicali.

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$\displaystyle\bf\\\sqrt{(1-x)(x-y)} + \sqrt{(x+y)(2-x)} = \frac{3}{2}\\\\(1-x)=(x-y)\\\\1>x>y\\\\Alegem\!:~~~x=0,75\\\implies~(1-0,75)=(0,75-y)\\0,75-y=0,25\\y=0,75-0,25=0,5\\Avem~o~solutie:\\\boxed{\bf~x=0,75~~si~~y=0,5}Verificam~solutia~la~radicalul~al~doilea.\\\\(x+y)=0,75+0,5=1,25\\2-x=2-0,75=1,25\\Corect$

.

$\displaystyle\bf\\Verificam~ecuatia:\\\\\sqrt{(1-x)(x-y)} + \sqrt{(x+y)(2-x)}=\\\\=\sqrt{(1-0,75)(0,75-0,5)} + \sqrt{(0,75+0,5)(2-0,75)}=\\\\=\sqrt{(0,25)\cdot(0,25)} + \sqrt{(1,25)\cdot(1,25)}=\\\\=\sqrt{(0,25)^2} + \sqrt{(1,25)^2}=\\\\=0,25+1,25=1,5=\frac{3}{2} \\\\Corect!\\\\Rezulta~ca~solutia:\\\\\boxed{\bf x=0,75~~~si~~~y=0,5}\\\\este~corecta.\\\\Sa~nu~uitam:\\\\Solutia~este~una~din~cele~o~infinitate~de~solutii.$