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\(a,8-2\sqrt{7}=\sqrt{7}^2-2\sqrt{7}+1^2=\left(\sqrt{7}-1\right)^2\)
\(b,8-2\sqrt{15}=\sqrt{5}^2-2.\sqrt{3}.\sqrt{5}+\sqrt{3}^2=\left(\sqrt{5}-\sqrt{3}\right)^2\)
\(c,8+4\sqrt{3}=2^2+2.2.\sqrt{3}+\sqrt{3}^2=\left(2+\sqrt{3}\right)^2\)
a,\(5+\sqrt{24}=5+\sqrt{6.4}=5+2\sqrt{6}=\left(\sqrt{2}\right)^2+2\sqrt{2}\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{2}+\sqrt{3}\right)^2\)
b,\(14+6\sqrt{5}=14+2.3.\sqrt{5}=3^2+2.3\sqrt{5}+\left(\sqrt{5}\right)^2=\left(3+\sqrt{5}\right)^2\)
b: \(5+2\sqrt{6}=\left(\sqrt{3}+\sqrt{2}\right)^2\)
c: \(13+\sqrt{48}=13+4\sqrt{3}=\left(2\sqrt{3}+1\right)^2\)
d: \(4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)
1) \(5-2\sqrt{6}=\left(\sqrt{3}\right)^2-2\sqrt{3}.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(\sqrt{3}-\sqrt{2}\right)^2\)
2) \(8+2\sqrt{15}=\left(\sqrt{5}\right)^2+2\sqrt{5}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{5}+\sqrt{3}\right)^2\)
3) \(10-2\sqrt{21}=\left(\sqrt{7}\right)^2-2\sqrt{7}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{7}-\sqrt{3}\right)^2\)
4) \(21+6\sqrt{6}=\left(\sqrt{18}\right)^2+2.\sqrt{18}.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(\sqrt{18}+\sqrt{3}\right)^2\)
5) \(14+8\sqrt{3}=\left(\sqrt{8}\right)^2+2.\sqrt{8}.\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{8}+\sqrt{6}\right)^2\)
6) \(36-12\sqrt{5}=\left(\sqrt{30}\right)^2-2.\sqrt{30}.\sqrt{6}+\left(\sqrt{6}\right)^2=\left(\sqrt{30}-\sqrt{6}\right)^2\)
7) \(25+4\sqrt{6}=\left(\sqrt{24}\right)^2+2\sqrt{24}.1+1^2=\left(\sqrt{24}+1\right)^2\)
8) \(98-16\sqrt{3}=\left(\sqrt{96}\right)^2-2\sqrt{96}.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(\sqrt{96}-\sqrt{2}\right)^2\)
1) \(15-\sqrt{216}=15-\sqrt{4}.\sqrt{54}\)=\(9-2.\sqrt{9}.\sqrt{6}+6\)=\(\left(\sqrt{9}-\sqrt{6}\right)^2=\left(3-\sqrt{6}\right)^2\)
2)\(20-\sqrt{76}=20-\sqrt{4}.\sqrt{19}=19-2\sqrt{19}.1+1=\left(\sqrt{19}-1\right)^2\)
3)\(24-12\sqrt{3}=6\left(4-2\sqrt{3}\right)=6\left(3-2.\sqrt{3}.1+1\right)=6\left(\sqrt{3}-1\right)^2\)
4)\(7-\sqrt{13}=\frac{14-2\sqrt{13}}{2}=\frac{13-2\sqrt{13}.1+1}{2}=\frac{\left(\sqrt{13}-1\right)^2}{2}\)
5)\(16-\sqrt{31}=\frac{32-2\sqrt{31}}{2}=\frac{31-2\sqrt{31}.1+1}{2}=\frac{\left(\sqrt{31}-1\right)^2}{2}\)
1)\(43-30\sqrt{2}=\left(5-3\sqrt{2}\right)^2\)
2)\(21+4\sqrt{5}=\left(1+2\sqrt{5}\right)^2\)
a, \(\sqrt{3+2\sqrt{2}}=\sqrt{\sqrt{2}^2+2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}+1\right)^2}=\left|\sqrt{2}+1\right|=\sqrt{2}+1\)
b, \(\sqrt{3-2\sqrt{2}}=\sqrt{\sqrt{2}^2-2\sqrt{2}+1}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)
c, \(\sqrt{8-2\sqrt{15}}=\sqrt{\sqrt{5}^2-2\sqrt{5.3}+\sqrt{3}^2}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}\)
\(=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\)
\(A=22-4\sqrt{10}\)
\(A=2\left(11-2\sqrt{10}\right)\)
\(A=2\left(\sqrt{10}-1\right)^2\)
\(A=\left(\sqrt{20}-\sqrt{2}\right)^2\)
1 ) \(9+4\sqrt{2}=9+2\sqrt{8}=[\left(\sqrt{8}\right)^2+2.\sqrt{8}.1+1]=\left(\sqrt{8}+1\right)^2\)
2 ) \(31+12\sqrt{3}=31+2\sqrt{108}=\left[\left(\sqrt{27}\right)^2+2.\sqrt{27}.2+2^2\right]=\left(\sqrt{27}+4\right)^2\)
\(8-\sqrt{25}=8-5=3=\left(\sqrt{3}+0\right)^2\)