Tìm x a) $\sqrt[]{2x+3}$ = 1+$\sqrt[]{2}$ b) $\sqrt[]{10+\sqrt[]{3}x}$ =2+$\sqrt[]{6}$

Đáp án:

a) $\sqrt[]{2x+3}$=1+ $\sqrt[]{2}$ 

⇔ 2x+3 = 3+ 2$\sqrt[]{2}$ 

⇔ 2x = 2$\sqrt[]{2}$ 

⇔ x = $\sqrt[]{2}$ 

b) $\sqrt[]{10+\sqrt[]{3}x}$ = 2 +$\sqrt[]{6}$ 

⇔ 10 +  $\sqrt[]{3}$ x  = 10 +4$\sqrt[]{6}$ 

⇔ x$\sqrt[]{3}$ = 4$\sqrt[]{6}$ 

⇔ x = 4$\sqrt[]{2}$