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a) Ta có: \(A^3=\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)^3\)
\(=2+\sqrt{5}+2-\sqrt{5}+3\cdot\sqrt[3]{\left(2+\sqrt{5}\right)\left(2-\sqrt{5}\right)}\left(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\right)\)
\(=4-3\cdot A\)
\(\Leftrightarrow A^3+3A-4=0\)
\(\Leftrightarrow A^3-A+4A-4=0\)
\(\Leftrightarrow A\left(A-1\right)\left(A+1\right)+4\left(A-1\right)=0\)
\(\Leftrightarrow\left(A-1\right)\left(A^2+A+4\right)=0\)
\(\Leftrightarrow A=1\)
\(x=\dfrac{3\sqrt[3]{8-3\sqrt{5}}}{\sqrt[3]{57}}.\sqrt[3]{8+3\sqrt{5}}=\dfrac{3\sqrt[3]{\left(8-3\sqrt{5}\right)\left(8+3\sqrt[]{5}\right)}}{\sqrt[3]{57}}=\sqrt[3]{\dfrac{19}{57}}=\dfrac{1}{\sqrt[3]{3}}\)
\(y=\dfrac{\left(\sqrt[3]{3}+\sqrt[4]{2}\right)\left(\sqrt[3]{3}-\sqrt[4]{2}\right)}{\sqrt[3]{3}+\sqrt[4]{2}}+\dfrac{\left(\sqrt[4]{2}-\sqrt[3]{81}\right)\left(\sqrt[4]{2}+\sqrt[3]{81}\right)}{\sqrt[4]{2}-\sqrt[3]{81}}\)
\(=\sqrt[3]{3}-\sqrt[4]{2}+\sqrt[4]{2}+\sqrt[3]{81}=\sqrt[3]{3}+3\sqrt[3]{3}=4\sqrt[3]{3}\)
\(T=xy=\dfrac{4\sqrt[3]{3}}{\sqrt[3]{3}}=4\)
Lời giải:
a.
\(=\sqrt{5+2.2\sqrt{5}+2^2}-\sqrt{5-2.2\sqrt{5}+2^2}\)
$=\sqrt{(\sqrt{5}+2)^2}-\sqrt{(\sqrt{5}-2)^2}$
$=|\sqrt{5}+2|-|\sqrt{5}-2|=(\sqrt{5}+2)-(\sqrt{5}-2)=4$
b.
$=\sqrt{3-2.3\sqrt{3}+3^2}+\sqrt{3+2.3.\sqrt{3}+3^2}$
$=\sqrt{(\sqrt{3}-3)^2}+\sqrt{(\sqrt{3}+3)^2}$
$=|\sqrt{3}-3|+|\sqrt{3}+3|$
$=(3-\sqrt{3})+(\sqrt{3}+3)=6$
c.
$=\sqrt{2+2.3\sqrt{2}+3^2}-\sqrt{2-2.3\sqrt{2}+3^2}$
$=\sqrt{(\sqrt{2}+3)^2}-\sqrt{(\sqrt{2}-3)^2}$
$=|\sqrt{2}+3|-|\sqrt{2}-3|$
$=(\sqrt{2}+3)-(3-\sqrt{2})=2\sqrt{2}$
\(a,=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) \(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{20-2.3\sqrt{20}+9}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(\sqrt{20}-3\right)^2}}}\)\(=\sqrt{\sqrt{5}-\sqrt{6-\sqrt{20}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5-2\sqrt{5}+1}}=\sqrt{\sqrt{5}-\sqrt{\left(\sqrt{5}-1\right)^2}}=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
\(=\sqrt{1}=1\)
\(b,=\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\) \(=\sqrt{3+30\sqrt{2+\sqrt{8+2\sqrt{8}+1}}}\)
\(=\sqrt{3+30\sqrt{2+\sqrt{\left(\sqrt{8}+1\right)^2}}}\)\(=\sqrt{3+30\sqrt{3+\sqrt{8}}}=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{3+30\sqrt{\left(\sqrt{2}+1\right)^2}}=\sqrt{3+30\sqrt{2}+30}=\sqrt{33+30\sqrt{2}}\)
a) Ta có: \(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-\left(2\sqrt{5}-3\right)}}\)
\(=\sqrt{\sqrt{5}-\sqrt{3-2\sqrt{5}+3}}\)
\(=\sqrt{\sqrt{5}-\sqrt{6-2\sqrt{5}}}\)
\(=\sqrt{\sqrt{5}-\sqrt{5}+1}\)
=1
b) Ta có: \(\sqrt{3+30\sqrt{2+\sqrt{9+4\sqrt{2}}}}\)
\(=\sqrt{3+30\sqrt{2+2\sqrt{2}+1}}\)
\(=\sqrt{3+30\left(\sqrt{2}+1\right)}\)
\(=\sqrt{33+30\sqrt{2}}\)
d/ \(x=\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=3+\sqrt{9+\frac{125}{27}}+3-\sqrt{9+\frac{125}{27}}-3\left(\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}-\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\right)\sqrt[3]{3+\sqrt{9+\frac{125}{27}}}.\sqrt[3]{-3+\sqrt{9+\frac{125}{27}}}\)
\(\Leftrightarrow x^3=6-3x\sqrt[3]{9-9-\frac{125}{27}}\)
\(\Leftrightarrow x^3=6-5x\)
\(\Leftrightarrow\left(x-1\right)\left(x^2+x+6\right)=0\)
\(\Leftrightarrow x=1\)
c/
\(\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{\left(4-\sqrt{2}\right)^2}}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{12}+4}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\left(\sqrt{3}+1\right)^2}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{2}\sqrt{2-\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{4-2\sqrt{3}}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{6+2\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{4+2\sqrt{3}}\)
\(=\left(\sqrt{3}-1\right)\sqrt{\left(\sqrt{3}+1\right)^2}\)
\(=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)\)
\(=3-1=2\)
\(A=\sqrt{9-4\sqrt{5}}-\sqrt{9+4\sqrt{5}}\)
\(=\sqrt{\left(\sqrt{5}\right)^2-4\sqrt{5+2^2}}-\sqrt{\left(\sqrt{5}\right)^2+4\sqrt{5}+2^2}\)
\(=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{\left(\sqrt{5}+2\right)^2}\)
\(=\left|\sqrt{5}-2\right|-\left|\sqrt{5}+2\right|\)
\(=\sqrt{5}-2-\left(\sqrt{5}+2\right)\)
\(=-4\)
\(B=\sqrt[3]{9}.\sqrt[3]{-3}+\left(1+\sqrt{2}\right)^2\)
\(=-\sqrt[3]{27}+3+2\sqrt{2}\)
\(=-3+3+2\sqrt{2}\)
\(=2\sqrt{2}\)