Bài 1:
a) \(\sqrt{13-2\sqrt{42}}\)
b) \(\sqrt{46+6\sqrt{5}}\)
c) \(\sqrt{12-3\sqrt{15}}\)
d) \(\sqrt{11+\sqrt{96}}\)
Bài 2:
a) \(A=\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{4+2\sqrt{3}}}}\)
b) \(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}\)
c) \(C=\sqrt{3-\sqrt{5}}\left(\sqrt{10}+\sqrt{2}\right)\left(3+\sqrt{5}\right)\)
d) \(D=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
e) \(E=\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)
g) \(G=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}\)
h) \(H=4x-\sqrt{9x^2-12x+4}\)
i) \(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}+\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 1:
a)
\(\sqrt{13-2\sqrt{42}}=\sqrt{6+7-2\sqrt{6.7}}=\sqrt{(\sqrt{7}-\sqrt{6})^2}=|\sqrt{7}-\sqrt{6}|=\sqrt{7}-\sqrt{6}\)
b)
\(\sqrt{46+6\sqrt{5}}=\sqrt{46+2\sqrt{45}}=\sqrt{45+1+2\sqrt{45.1}}=\sqrt{(\sqrt{45}+1)^2}=\sqrt{45}+1\)
\(=3\sqrt{5}+1\)
c)
\(\sqrt{12-3\sqrt{15}}=\sqrt{\frac{24-6\sqrt{15}}{2}}=\sqrt{\frac{24-2\sqrt{135}}{2}}=\sqrt{\frac{15+9-2\sqrt{15.9}}{2}}\)
\(=\sqrt{\frac{(\sqrt{15}-\sqrt{9})^2}{2}}=\frac{\sqrt{15}-\sqrt{9}}{\sqrt{2}}=\frac{\sqrt{15}-3}{\sqrt{2}}\)
d)
\(\sqrt{11+\sqrt{96}}=\sqrt{11+2\sqrt{24}}=\sqrt{8+3+2\sqrt{8.3}}\)
\(=\sqrt{(\sqrt{8}+\sqrt{3})^2}=\sqrt{8}+\sqrt{3}\)
Bài 2:
a)
\(A=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{4+2\sqrt{3}}}}=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{3+1+2\sqrt{3}}}}\)
\(=\sqrt{6+2\sqrt{2}.\sqrt{3-\sqrt{(\sqrt{3}+1)^2}}}=\sqrt{6+2\sqrt{2}.\sqrt{2-\sqrt{3}}}\)
\(=\sqrt{6+2\sqrt{4-2\sqrt{3}}}=\sqrt{6+2\sqrt{(\sqrt{3}-1)^2}}=\sqrt{6+2(\sqrt{3}-1)}\)
\(=\sqrt{4+2\sqrt{3}}=\sqrt{3+1+2\sqrt{3.1}}=\sqrt{(\sqrt{3}+1)^2}=\sqrt{3}+1\)
b)
\(B=\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}=\sqrt{5}-\sqrt{3-\sqrt{20+9-2\sqrt{20.9}}}\)
\(=\sqrt{5}-\sqrt{3-\sqrt{(\sqrt{20}-\sqrt{9})^2}}=\sqrt{5}-\sqrt{3-(\sqrt{20}-\sqrt{9})}\)
\(=\sqrt{5}-\sqrt{6-2\sqrt{5}}=\sqrt{5}-\sqrt{5+1-2\sqrt{5}}=\sqrt{5}-\sqrt{(\sqrt{5}-1)^2}\)
\(=\sqrt{5}-(\sqrt{5}-1)=1\)
Bài 2:
c)
\(C=\sqrt{3-\sqrt{5}}(\sqrt{10}+\sqrt{2})(3+\sqrt{5})=\sqrt{6-2\sqrt{5}}(\sqrt{5}+1)(3+\sqrt{5})\)
\(=\sqrt{5+1-2\sqrt{5}}(\sqrt{5}+1)(3+\sqrt{5})\)
\(=\sqrt{(\sqrt{5}-1)^2}(\sqrt{5}+1)(3+\sqrt{5})\)
\(=(\sqrt{5}-1)(\sqrt{5}+1)(3+\sqrt{5})=4(3+\sqrt{5})\)
d)
\(D=\sqrt{4+\sqrt{7}}-\sqrt{4-\sqrt{7}}\)
\(D\sqrt{2}=\sqrt{8+2\sqrt{7}}-\sqrt{8-2\sqrt{7}}=\sqrt{7+1+2\sqrt{7.1}}-\sqrt{7+1-2\sqrt{7.1}}\)
\(=\sqrt{(\sqrt{7}+1)^2}-\sqrt{(\sqrt{7}-1)^2}=\sqrt{7}+1-(\sqrt{7}-1)=2\)
\(\Rightarrow D=\sqrt{2}\)
Bài 2:
e)
\(E^2=16+2\sqrt{(8+2\sqrt{10+2\sqrt{5}})(8-2\sqrt{10+2\sqrt{5}})}\)
\(=16+2\sqrt{8^2-(2\sqrt{10+2\sqrt{5}})^2}\)
\(=16+2\sqrt{64-4(10+2\sqrt{5})}=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{20+4-2\sqrt{20.4}}\)
\(=16+2\sqrt{(\sqrt{20}-\sqrt{4})^2}=16+2(\sqrt{20}-\sqrt{4})\)
\(=12+2\sqrt{20}=10+2+2\sqrt{10.2}=(\sqrt{10}+\sqrt{2})^2\)
\(\Rightarrow E=\sqrt{10}+\sqrt{2}\) (do $E$ không âm)
g)
\(G=\sqrt{x+2\sqrt{x-1}}+\sqrt{x-2\sqrt{x-1}}=\sqrt{(x-1)+2\sqrt{x-1}+1}+\sqrt{(x-1)-2\sqrt{x-1}+1}\)
\(=\sqrt{(\sqrt{x-1}+1)^2}+\sqrt{(\sqrt{x-1}-1)^2}\)
\(=|\sqrt{x-1}+1|+|\sqrt{x-1}-1|\)
Bài 2:
h)
\(H=4x-\sqrt{9x^2-12x+4}=4x-\sqrt{(3x)^2-2.3x.2+2^2}\)
\(=4x-\sqrt{(3x-2)^2}=4x-|3x-2|\)
\(=\left\{\begin{matrix} 4x-(3x-2): \text{nếu x}\geq \frac{2}{3}\\ 4x-(2-3x): \text{nếu x}< \frac{2}{3}\end{matrix}\right.\)
\(=\left\{\begin{matrix} x+2:\text{nếu x}\geq \frac{2}{3}\\ 7x-2:\text{nếu x}< \frac{2}{3}\end{matrix}\right.\)
i)
\(\frac{\sqrt{7}-\sqrt{2}}{\sqrt{7}+\sqrt{2}}+\frac{\sqrt{7}+\sqrt{2}}{\sqrt{7}-\sqrt{2}}=\frac{(\sqrt{7}-\sqrt{2})^2+(\sqrt{7}+\sqrt{2})^2}{(\sqrt{7}+\sqrt{2})(\sqrt{7}-\sqrt{2})}\)
\(=\frac{7-2\sqrt{14}+2+7+2\sqrt{14}+2}{7-2}=\frac{18}{5}\)
