$u.v = √2$
$=> u = √2/v$
vì $u-v= √3 -√ 2$
$=> √2/v -v= √3 -√ 2$
$<=> √2 -v²= (√3 -√ 2)v$
$<=> v² + (√3 -√ 2)v - √2 = 0$
$\text{ đen ta = (√3 -√ 2)² - 4.(- √2)}$
$\text{ = 3+2-2√6 + 4√2 = 5-2√6 + 4√2 >0 }$
=> PT có 2 nghiệm phân biệt
$v_{1} = \dfrac{√ 2-√3-\sqrt[]{5-2√6 + 4√2}}{2} $
$=> v_{2} = \dfrac{√ 2-√3+\sqrt[]{5-2√6 + 4√2}}{2} $
+) Với $v= \dfrac{√ 2-√3-\sqrt[]{5-2√6 + 4√2}}{2} $
=> $u= √ 2 : \dfrac{√ 2-√3-\sqrt[]{5-2√6 + 4√2}}{2} $
u= $\dfrac{2√ 2}{√ 2-√3-\sqrt[]{5-2√6 + 4√2}}$
.
+) Với $v= \dfrac{√ 2-√3+\sqrt[]{5-2√6 + 4√2}}{2} $
=> $u= √ 2 : \dfrac{√ 2-√3+\sqrt[]{5-2√6 + 4√2}}{2} $
u= $\dfrac{2√ 2}{√ 2-√3+\sqrt[]{5-2√6 + 4√2}}$
