Question

Помогите,пожалуйста! С решением,если можно)

Answer 1

$sqrt{21-2sqrt{21+2sqrt{19-6sqrt2}}}cdot (3sqrt2+1)=17;$ 

$1)19-6sqrt2=19-2cdot 3sqrt2=18+1-2cdot 3sqrt2=(9cdot 2)-2cdot 3sqrt2+1=\\=(3sqrt2)^2-2cdot 3sqrt2cdot 1+1=(3sqrt2-1)^2\\\2)sqrt{21+2sqrt{19-6sqrt2}}=sqrt{21+2sqrt{(3sqrt2-1)^2}}=sqrt{21+2(3sqrt2-1)}=\\=sqrt{21+2cdot 3sqrt3-2}=sqrt{19+2cdot 3sqrt2}=sqrt{(3sqrt2+1)^2}=3sqrt2+1\\\3)sqrt{21-2sqrt{(3sqrt2+1)^2}}*(3sqrt2+1)=sqrt{21-2(3sqrt2+1)}*(3sqrt2+1)=\\=sqrt{21-2cdot 3sqrt2-2}*(3sqrt2+1)=sqrt{19-2cdot 3sqrt2}*(3sqrt2+1)=sqrt{(3sqrt2-1)^2}*(3sqrt2+1)=\\=(3sqrt2-1)*(3sqrt2+1)=9*2-1=17;$


$P.S.sqrt{(3sqrt2-1)^2}=|3sqrt2-1|=3sqrt2-1,t.k.3sqrt2-1>0\\sqrt{(3sqrt2+1)^2}=|3sqrt2+1|=3sqrt2+1,t.k.3sqrt2+1>0$