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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) \(\sqrt{5-2\sqrt{6}}-\sqrt{5+2\sqrt{6}}=\sqrt{3}-\sqrt{2}-\sqrt{3}-\sqrt{2}=-2\sqrt{2}\)
a) \(21-8\sqrt{5}=16-2\times4\times\sqrt{5}+5=\left(4-\sqrt{5}\right)^2\)
b) \(47-12\sqrt{11}=36-2\times6\times\sqrt{11}+11=\left(6-\sqrt{11}\right)^2\)
c) \(13-4\sqrt{3}=12-2\times1\times\sqrt{3}+1=\left(2\sqrt{3}-1\right)^2\)
d) \(43+30\sqrt{2}=25+2\times5\times3\sqrt{2}+18=\left(5+3\sqrt{2}\right)^2\)
e) \(41+24\sqrt{2}=9+2\times3\times4\sqrt{2}+32=\left(3+4\sqrt{2}\right)^2\)
g) \(29-12\sqrt{5}=9+2\times3\times2\sqrt{5}+20=\left(3+2\sqrt{5}\right)^2\)
h) \(49-8\sqrt{3}=48-2\times4\sqrt{3}\times1+1=\left(4\sqrt{3}-1\right)^2\)
i) \(37-12\sqrt{7}=28-2\times3\times2\sqrt{7}+9=\left(2\sqrt{7}-3\right)^2\)
\(A=\sqrt{5-2\sqrt{5}+1}-\sqrt{5+2\sqrt{5}+1}=\sqrt{\left(\sqrt{5}-1\right)^2}-\sqrt{\left(\sqrt{5}+1\right)^2}\)
\(=\sqrt{5}-1-\sqrt{5}-1=-2\)
Vậy \(A\in Z\)
Làm tương tự với B.
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) \(98-16\sqrt{3}=\left(\sqrt{96}\right)^2-2\sqrt{96}.\sqrt{2}+\left(\sqrt{2}\right)^2=\left(\sqrt{96}-\sqrt{2}\right)^2\)
4. \(\sqrt{x}+\sqrt{y}=6\sqrt{55}\)
\(6\sqrt{55}\) là số vô tỉ, suy ra vế trái phải là các căn thức đồng dạng chứa \(\sqrt{55}\)
Đặt \(\sqrt{x}=a\sqrt{55};\sqrt{y}=b\sqrt{55}\) với \(a,b\in N\)
\(\Rightarrow a+b=6\)
Xét các TH:
a = 0 => b = 6
a = 1 => b = 5
a = 2 => b = 4
a = 3 => b = 3
a = 4 => b = 2
a = 5 => b = 1
a = 6 => b = 0
Từ đó dễ dàng tìm đc x, y
\(1a.\left(\sqrt{72}-3\sqrt{5}+2\sqrt{8}\right).\sqrt{2}+\sqrt{90}=\sqrt{144}-3\sqrt{10}+2.\sqrt{16}+3\sqrt{10}=12+8=20\) \(b.\left(\sqrt{\dfrac{1}{5}}-10\sqrt{\dfrac{27}{5}}+2\sqrt{5}\right):\sqrt{5}+6\sqrt{3}=\left(\sqrt{\dfrac{1}{5}}-30\sqrt{\dfrac{3}{5}}+2\sqrt{5}\right).\dfrac{1}{\sqrt{5}}+6\sqrt{3}=\dfrac{1}{5}-6\sqrt{3}+2+6\sqrt{3}=\dfrac{11}{5}\) \(2.\sqrt{\left(3-\sqrt{10}\right)^2}=\sqrt{10}-3\)
\(b.\sqrt{7+4\sqrt{3}}+\sqrt{7-4\sqrt{3}}=\sqrt{4+2.2\sqrt{3}+3}+\sqrt{4-2.2.\sqrt{3}+3}=2+\sqrt{3}+2-\sqrt{3}=4\) \(c.\dfrac{2+\sqrt{2}}{1+\sqrt{2}}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{1+\sqrt{2}}=\sqrt{2}\)