7)
a)
[tex]\sqrt{(\sqrt{2}-\sqrt{3})^2}-\sqrt{(1-\sqrt{3})^2}-\sqrt{(\sqrt{2}-2)^2}[/tex]
În cazul acesta radicalii se simplifică cu puterea:
[tex]|\sqrt{2}-\sqrt{3}|-|1-\sqrt{3}|-|\sqrt{2}-2| \\ \\ \sqrt{2}=1.41 \\ \sqrt{3}=1.71 \\ \\ (\sqrt{3}-\sqrt{2})-(\sqrt{3}-1)-(2-\sqrt{2}) \\ \\ \sqrt{3}-\sqrt{2}-\sqrt{3}+1-2+\sqrt{2} \\ \\ -1[/tex]
Acolo unde modulul conținutul modulului dădea un număr negativ, am inversat termenii astfel încât diferența să dea un număr pozitiv.
b)
[tex]\sqrt{(\sqrt{5}+3)^2}+\sqrt{(2-\sqrt{5})^2}-\sqrt{(2\sqrt{5}+4)^2} \\ \\ |\sqrt{5}+3|+|2-\sqrt{5}|-|2\sqrt{5}+4| \\ \\ (\sqrt{5}+3)+(\sqrt{5}-2)-(2\sqrt{5}+4) \\ \\ \sqrt{5}+3+\sqrt{5}-2-2\sqrt{5}-4 \\ \\ 3-2-4 \\ \\ -3[/tex]
c)
[tex](5\sqrt{3}-3\sqrt{5})^2+4(\sqrt{3}+2\sqrt{5})^2-(2\sqrt{3}-\sqrt{5})(2\sqrt{3}+\sqrt{5})*30 \\ \\ (75-30\sqrt{15}+45)+4(3+4\sqrt{15}+20)-(12-25)*30 \\ \\ 75-30\sqrt{15}+45+12+16\sqrt{15}+80+13*30 \\ \\ 212-14\sqrt{15}+390 \\ \\ 602-14\sqrt{15}[/tex]
d)
[tex]\sqrt{17-12\sqrt{2}}+2\sqrt{3-2\sqrt{2}}+\sqrt{21-12\sqrt{3}}-2\sqrt{4+2\sqrt{3}}[/tex]
[tex]\sqrt{17-12\sqrt{2}}=\sqrt{\frac{A+C}{2}}-\sqrt{\frac{A-C}{2}}=\sqrt{\frac{17+1}{2}}-\sqrt{\frac{17-1}{2}}=3-2\sqrt{2}
\\ \\ \left \{ {{A=289} \atop {B=288}} \right.\rightarrow C^2=A-B=1\rightarrowC=1
\\ \\ \\
\sqrt{3-2\sqrt{2}}=\sqrt{\frac{3+1}{2}}-\sqrt{\frac{3-1}{2}}=\sqrt{2}-1
\\ \\
\left \{ {{A=9} \atop {B=8}} \right. \rightarrow C=1[/tex]
[tex]\sqrt{21-12\sqrt{3}}=\sqrt{\frac{21+3}{2}}-\sqrt{\frac{21-3}{2}}=\sqrt{12}-\sqrt{9}=2\sqrt{3}-3
\\ \\
\left \{ {{A=441} \atop {x=432}} \right. \rightarrow C^2=9 \rightarrow C=3
\\ \\ \\
\sqrt{4+2\sqrt{3}}=\sqrt{\frac{4+2}{2}}+\sqrt{\frac{4-2}{2}}=\sqrt{3}+1
\\ \\
\left \{ {{A=16} \atop {B=12}} \right. \rightarrow C^2=4 \rightarrow C=2[/tex]
[tex]3-2\sqrt{2}+2(\sqrt{2}-1)+(2\sqrt{3}-3)-2(\sqrt{3}+1) \\ \\ 3-2\sqrt{2}+2\sqrt{2}-2+2\sqrt{3}-2-2\sqrt{3}-2 \\ \\ -3[/tex]
8.
a)
[tex]\sqrt{17+12\sqrt{2}}=\sqrt{\frac{17+1}{2}}+\sqrt{\frac{17-1}{2}}=\sqrt{9}+\sqrt{8}=3+2\sqrt{2}
\\ \\
\left \{ {{A=289} \atop {B=288}} \right. \rightarrow C=1
\\ \\ \\
\sqrt{7+4\sqrt{3}}=\sqrt{\frac{7+1}{2}}+\sqrt{\frac{7-1}{2}}=\sqrt{4}+\sqrt{3}=2+\sqrt{3}
\\ \\
\left \{ {{A=49} \atop {B=48}} \right. \rightarrow C=1[/tex]
[tex](3-2\sqrt{2})(3+2\sqrt{2})+(\sqrt{3}+2)(\sqrt{3}-2) \\ \\ (9-8)+(3-4) \\ \\ 0[/tex]
b)
[tex]\frac{1}{\sqrt{2}(\sqrt{2}+1)}+\frac{1}{\sqrt{6}(\sqrt{3}+\sqrt{2})}+\frac{1}{\sqrt{12}(\sqrt{4}+\sqrt{3})} \\ \\ \frac{1}{2+\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+6} \\ \\ \frac{2-\sqrt{2}}{4-2}+\frac{3\sqrt{2}-2\sqrt{3}}{18-12}+\frac{4\sqrt{3}-6}{48-36} \\ \\ \frac{2-\sqrt{2}}{2}+\frac{3\sqrt{2}-2\sqrt{3}}{6}+\frac{4\sqrt{3}-6}{12} \\ \\ \frac{12-6\sqrt{2}+6\sqrt{2}-4\sqrt{3}+4\sqrt{3}-6}{12} \\ \\ \frac{1}{2}[/tex]
Ultimul exercițiu ți-l las ție. Îi 10:30 seara și nu prea am chef de așa ceva...
