Question

Упростите плииз, а то с корнями проблемы! Заранее спасибо

Answer 1

$(frac{sqrt{3}}{2-sqrt{7-sqrt{48}}}-frac{2}{sqrt{7+sqrt{48}}-1})(2+sqrt3)=\ \ =(frac{sqrt{3}}{2-sqrt{4-4sqrt{3}+3}}-frac{2}{sqrt{4+4sqrt{3}+3}-1}})(2+sqrt3)=\ \ =(frac{sqrt{3}}{2-sqrt{2^2-2*2sqrt{3}+(sqrt{3})^2}}-frac{2}{sqrt{2^2+2*2sqrt{3}+(sqrt{3})^2}-1}})(2+sqrt3)=\ \ =(frac{sqrt{3}}{2-sqrt{(2-sqrt{3})^2}}-frac{2}{sqrt{(2+sqrt{3})^2}-1}})(2+sqrt3)=\ \ = (frac{sqrt{3}}{2-|2-sqrt{3}|}}-frac{2}{|2+sqrt{3}|-1}})(2+sqrt3)$

так как $2-sqrt{3}>0$ то

$(frac{sqrt{3}}{2-(2-sqrt{3})}}-frac{2}{2+sqrt{3}-1}})(2+sqrt3)=(frac{sqrt{3}}{2-2+sqrt{3}}}-frac{2}{1+sqrt{3}}})(2+sqrt3)=\ \ =(frac{sqrt{3}}{sqrt{3}}-frac{2}{1+sqrt{3}}})(2+sqrt3)=(1-frac{2}{1+sqrt{3}}})(2+sqrt3)=\ \ =frac{1+sqrt{3}-2}{1+sqrt{3}}(2+sqrt3)=frac{sqrt{3}-1}{1+sqrt{3}}(2+sqrt3)=frac{(sqrt{3}-1)(2+sqrt3)}{1+sqrt{3}}=\ \ =frac{2sqrt{3}+3-2-sqrt3}{1+sqrt{3}}=frac{sqrt{3}+1}{1+sqrt{3}}=1$