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\(13-4\sqrt{3}=\left(2\sqrt{3}\right)^2-2.2\sqrt{2}.1+1^2=\left(2\sqrt{3}-1\right)^2\)
a) \(\left(\sqrt{5}+\sqrt{3}\right)\sqrt{8-2\sqrt{15}}=\left(\sqrt{5}+\sqrt{3}\right)\left(\sqrt{5}-\sqrt{3}\right)=5-3=2\)
câu này \(\sqrt{15}\)đúng hơn \(\sqrt{5}\)
b) \(\sqrt{3-\sqrt{5}}-\sqrt{3+\sqrt{5}}=\frac{\sqrt{6-2\sqrt{5}}-\sqrt{6+2\sqrt{5}}}{\sqrt{2}}=\frac{\sqrt{5}-1-\sqrt{5}-1}{\sqrt{2}}=\frac{-2}{\sqrt{2}}=-\sqrt{2}\)c) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)
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) \(98-16\sqrt{3}=\left(\sqrt{96}\right)^2-2\sqrt{96}.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(\sqrt{96}-\sqrt{2}\right)^2\)
Bài 1 bạn nhóm , trục như thường nhé :D
Bài 2. \(a.A=\sqrt{5+2\sqrt{6}}-\sqrt{5-2\sqrt{6}}=\sqrt{3+2\sqrt{3}.\sqrt{2}+2}-\sqrt{3-2\sqrt{3}.\sqrt{2}+2}=\sqrt{3}+\sqrt{2}-\sqrt{3}+\sqrt{2}=2\sqrt{2}\)
\(b.B=\sqrt{17-12\sqrt{2}}-\sqrt{9+4\sqrt{2}}=\sqrt{9-2.2\sqrt{2}.3+8}-\sqrt{8+2.2\sqrt{2}+1}=3-2\sqrt{2}-2\sqrt{2}-1=2-4\sqrt{2}\)
\(c.C=\sqrt{13+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}=\sqrt{13+30\sqrt{2+\sqrt{8+2.2.\sqrt{2}+1}}}=\sqrt{13+30\sqrt{2+2\sqrt{2}+1}}=\sqrt{43+30\sqrt{2}}=\sqrt{25+2.3\sqrt{2}.5+18}=5+3\sqrt{2}\)
\(d.D=\sqrt{12-3\sqrt{7}}-\sqrt{12+3\sqrt{7}}\)
\(D^2=24-2\sqrt{\left(12-3\sqrt{7}\right)\left(12+3\sqrt{7}\right)}=24-2\sqrt{81}=24-18=6\)
\(D=-\sqrt{6}\left(do:D< 0\right)\)
1)
a. \(\sqrt{\dfrac{25}{7}}.\sqrt{\dfrac{7}{9}}=\sqrt{\dfrac{25.7}{7.9}}=\sqrt{\dfrac{25}{9}}=\dfrac{5}{3}\)
b. \(\left(\sqrt{\dfrac{9}{2}}+\sqrt{\dfrac{1}{2}}-\sqrt{2}\right).\sqrt{2}=3+1-2=2\)
c. \(\left(\sqrt{\dfrac{8}{3}}-\sqrt{24}+\sqrt{\dfrac{50}{3}}\right).\sqrt{6}=4-12+10=2\)
d. \(\left(\sqrt{\dfrac{2}{3}}-\sqrt{\dfrac{3}{2}}\right)^2=\dfrac{2}{3}+\dfrac{3}{2}-2\sqrt{\dfrac{2}{3}.\dfrac{3}{2}}=\dfrac{1}{6}\)
2)
a. \(\sqrt{4+2\sqrt{3}}=\sqrt{3+2\sqrt{3}+1}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)
b. \(\sqrt{8-2\sqrt{7}}=\sqrt{7-2\sqrt{7}+1}=\sqrt{\left(\sqrt{7}-1\right)^2}=\sqrt{7}-1\)
c. \(1+\sqrt{6-2\sqrt{5}}=1+\sqrt{5-2\sqrt{5}+1}=1-\sqrt{\left(\sqrt{5}-1\right)^2}=1-\sqrt{5}+1=2-\sqrt{5}\)
d. \(\sqrt{7-2\sqrt{10}}+\sqrt{2}=\sqrt{5-2.\sqrt{5}.\sqrt{2}+2}+\sqrt{2}=\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}+\sqrt{2}=\sqrt{5}-\sqrt{2}+\sqrt{2}=\sqrt{5}\)
3. \(a.A=x^2+2x+16=\left(\sqrt{2}-1\right)^2+2.\left(\sqrt{2}-1\right)+16=2-2\sqrt{2}+1+2\sqrt{2}-2+16=17\)
\(b.B=x^2+12x-14=\left(5\sqrt{2}-6\right)^2+12.\left(5\sqrt{2}-6\right)-14=50+36-60\sqrt{2}+60\sqrt{2}-72-14=0\)
Help me nha 
@Phùng Khánh Linh@Nhã Doanh@Liana@Yukru Cảm ơn trước nhé 
a/ \(A=\frac{30\left(\sqrt{6}-1\right)}{5}+\frac{2\left(\sqrt{6}+2\right)}{2}-\frac{6\left(3+\sqrt{6}\right)}{3}=6\sqrt{6}-6+\sqrt{6}+2-6-2\sqrt{6}\)
\(A=5\sqrt{6}-10\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}\)
\(B=\sqrt{17-6\sqrt{2}+\sqrt{\left(2\sqrt{2}+1\right)^2}}=\sqrt{18-4\sqrt{2}}\)
Đến đây ko rút gọn được nữa, nhưng nếu đề là:
\(B=\sqrt{17+6\sqrt{2}+\sqrt{8+4\sqrt{2}+1}}=\sqrt{18+8\sqrt{2}}=4+\sqrt{2}\)
c/
\(C=\sqrt{8-2\sqrt{7}}+\sqrt{8+2\sqrt{7}}=\sqrt{\left(\sqrt{7}-1\right)^2}+\sqrt{\left(\sqrt{7}+1\right)^2}\)
\(C=\sqrt{7}-1+\sqrt{7}+1=2\sqrt{7}\)
\(D=\sqrt{a-2\sqrt{a}+1}-\sqrt{a-8\sqrt{a}+16}\)
\(D=\sqrt{\left(\sqrt{a}-1\right)^2}-\sqrt{\left(4-\sqrt{a}\right)^2}=\sqrt{a}-1-\left(4-\sqrt{a}\right)=2\sqrt{a}-5\)
\(E=\sqrt{a-2+2\sqrt{a-2}+1}+\sqrt{a-2-2\sqrt{a-2}+1}\) (\(a\ge2\))
\(E=\sqrt{\left(\sqrt{a-2}+1\right)^2}+\sqrt{\left(\sqrt{a-2}-1\right)^2}\)
\(E=\sqrt{a-2}+1+\left|\sqrt{a-2}-1\right|\)
\(\Rightarrow\left[{}\begin{matrix}E=2\sqrt{a-2}\left(a\ge3\right)\\E=2\left(2\le a\le3\right)\end{matrix}\right.\)
\(F=\sqrt[3]{10+6\sqrt{3}}-\sqrt{3}=\sqrt[3]{1+3.1.\sqrt{3}+3.1.\sqrt{3}^2+\sqrt{3}^3}-\sqrt{3}\)
\(F=\sqrt[3]{\left(1+\sqrt{3}\right)^3}-\sqrt{3}=1+\sqrt{3}-\sqrt{3}=1\)
\(G=\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\Rightarrow G^3=\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)^3\)
\(\Rightarrow G^3=14+3\left(\sqrt[3]{7+5\sqrt{2}}+\sqrt[3]{7-5\sqrt{2}}\right)\left(\sqrt[3]{49-50}\right)\)
\(\Rightarrow G^3=14-3G\Rightarrow G^3+3G-14=0\)
\(\Rightarrow G=2\)
\(1a.\left(\sqrt{72}-3\sqrt{5}+2\sqrt{8}\right).\sqrt{2}+\sqrt{90}=\sqrt{144}-3\sqrt{10}+2.\sqrt{16}+3\sqrt{10}=12+8=20\) \(b.\left(\sqrt{\dfrac{1}{5}}-10\sqrt{\dfrac{27}{5}}+2\sqrt{5}\right):\sqrt{5}+6\sqrt{3}=\left(\sqrt{\dfrac{1}{5}}-30\sqrt{\dfrac{3}{5}}+2\sqrt{5}\right).\dfrac{1}{\sqrt{5}}+6\sqrt{3}=\dfrac{1}{5}-6\sqrt{3}+2+6\sqrt{3}=\dfrac{11}{5}\) \(2.\sqrt{\left(3-\sqrt{10}\right)^2}=\sqrt{10}-3\)
\(b.\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}=2+\sqrt{3}+2-\sqrt{3}=4\) \(c.\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}=\sqrt{2}\)
\(a,\)
\(3+2\sqrt{2}=2+2\sqrt{2}+1=\sqrt{2}^2+2\sqrt{2}+1=\left(\sqrt{2}+1\right)^2\)
\(3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}\right)^2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)
\(b,\)
\(6+2\sqrt{5}=5+2\sqrt{5}+1=\left(\sqrt{5}\right)^2+2\sqrt{5}+1=\left(\sqrt{5}+1\right)^2\)
\(6-2\sqrt{5}=5-2\sqrt{5}+1=\left(\sqrt{5}\right)^2-2\sqrt{5}+1=\left(\sqrt{5}-1\right)^2\)
\(c,\)
\(7+4\sqrt{3}=4+2.2\sqrt{3}+3=2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(2+\sqrt{3}\right)^2\)
\(7-4\sqrt{3}=2^2-2.2.\sqrt{3}+\left(\sqrt{3}\right)^2=\left(2-\sqrt{3}\right)^2\)
`a)3+-2sqrt2`
`=2+-2sqrt2+1`
`=(sqrt2+-1)^2`
`b)6+-2sqrt5`
`=5+-2sqrt5+1`
`=(sqrt5+-1)^2`
`7)7+-4sqrt3`
`=4+-2.2.sqrt3+3`
`=(2+-sqrt3)^2`