Question

Ajutati-ma va rog la exercitiul 10.....URGENT!!!!!

Answer 1

 

$\displaystyle\\\\x=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\\\\x=\sqrt{5+3-2\sqrt{5}\sqrt{3}}-\sqrt{5+3+2\sqrt{5}\sqrt{3}}\\\\x=\sqrt{5-2\sqrt{5}\sqrt{3}+3}-\sqrt{5+2\sqrt{5}\sqrt{3}+3}\\\\x=\sqrt{\Big(\sqrt{5}\Big)^{2} -2\sqrt{5}\sqrt{3}+\Big(\sqrt{3}\Big)^{2}}-\sqrt{\Big(\sqrt{5}\Big)^{2}+2\sqrt{5}\sqrt{3}+\Big(\sqrt{3}\Big)^{2}}\\\\x=\sqrt{\Big(\sqrt{5}-\sqrt{3}\Big)^{2}}-\sqrt{\Big(\sqrt{5}+\sqrt{3}\Big)^{2}}\\\\x=\Big(\sqrt{5}-\sqrt{3}\Big)-\Big(\sqrt{5}+\sqrt{3}\Big)$

$\displaystyle\\x=\underline{\sqrt{5}}-\sqrt{3}-\underline{\sqrt{5}}-\sqrt{3}\\\\x=-\sqrt{3}-\sqrt{3}\\\\\boxed{\bf x=-2\sqrt{3}}\\\\\text{a) Calculam pe x la puterea a 2-a.}\\\\x^2=\Big(-2\sqrt{3}\Big)^{\b2}=2^2\cdot 3=4\cdot 3=12\\\\\boxed{\boxed{\bf x^2=12}}$

$\displaystyle\\\\b)\\\Big(x+2\sqrt{3}\Big)^{1996}=\Big(-2\sqrt{3}+2\sqrt{3}\Big)^{1996}=0^{1996}=0\\\boxed{\boxed{\Big(x+2\sqrt{3}\Big)^{1996}=0}}$

Answer 2

10.a.  x=√8 +√64-60 /2 - √8 -√64-60 /2 -√8 +√64 -60 /2 +√8 -√64-60 /2 =>

x=√5 -√3 -√5 -√3 <=> x=-2√3 => x²=(-2√3)²=12 => x²=12 .

b.  (x +2√3)¹⁹⁹⁶ =(-2√3 +2√3)¹⁹⁹⁶ =0¹⁹⁹⁶ =0 .