Question

$a=\frac{2}{\sqrt{3}+1}+(\sqrt{3}+1)^{2} -\sqrt{108}- \sqrt{7-4\sqrt{3}}$
⇒ a<$\frac{1}{2}$ ???????

Answer 1

Răspuns

Explicație pas cu pas:

a=2/(√3 +1)+(√3 +1)²-√108-√(7-4√3)=

=2(√3 -1)/(√3 +1)(√3- 1)+3 +2√3+1-6√3-√(4-4√3+3)=

=2(√3 -1)/(3- 1)+3 +1-4√3-√(2-√3)²=

=2(√3 -1)/2+4-4√3-(2-√3)=

=√3 -1+4-4√3-2+√3=

=1-2√3 =-2,464  ⇒ a<1/2

Answer 2

 

$\displaystyle\\a=\frac{2}{\sqrt{3}+1}+(\sqrt{3}+1)^2-\sqrt{108}-\sqrt{7-4\sqrt{3}}\\\\a=\frac{2(\sqrt{3}-1)}{3-1}+(3+2\sqrt{3}+1)-\sqrt{36\cdot3}-\sqrt{4+3-2\cdot2\cdot\sqrt{3}}\\\\a=\frac{2(\sqrt{3}-1)}{2}+(4+2\sqrt{3})-6\sqrt{3}-\sqrt{4-2\cdot2\cdot\sqrt{3}+3}\\\\a=(\sqrt{3}-1)+(4+2\sqrt{3})-6\sqrt{3}-\sqrt{2^2-2\cdot2\cdot\sqrt{3}+\Big(\sqrt{3}\Big)^2}\\\\a=\sqrt{3}-1+4+2\sqrt{3}-6\sqrt{3}-\sqrt{\Big(2-\sqrt{3}\Big)^2}$

$\displaystyle\\a=\sqrt{3}-1+4+2\sqrt{3}-6\sqrt{3}-\Big(2-\sqrt{3}\Big)\\\\a=\sqrt{3}-1+4+2\sqrt{3}-6\sqrt{3}-2+\sqrt{3}\\\\a=-1+4-2+\sqrt{3}+2\sqrt{3}-6\sqrt{3}+\sqrt{3}\\\\a=1+4\sqrt{3}-6\sqrt{3}\\\\a=1-2\sqrt{3}~~~\text{a este numar negativ.}\\\\a<0<\frac{1}{2} \\\\\boxed{a<\frac{1}{2}}$