What is the value of \(\sqrt{\sqrt{\sqrt[3]{64000}+\sqrt[3]{68921}}}\)
Solution:
we know that \(64000=64 \times {10}^{3}={2}^{6} \times {10}^{3} ={{\left( {{2}^{2}} \right)}^{3}}\times {{10}^{3}}\) so \(\sqrt[3]{64000}=\sqrt[3]{{{\left( {{2}^{2}} \right)}^{3}}\times {{10}^{3}}}={2}^{2} \times 10=40\)
also we have \(68921={41}^{3}\) thus \(\sqrt[3]{68921}=\sqrt[3]{{41}^{3}}=41\)
hence \(\sqrt[2]{40+41}=\sqrt[2]{81}=\sqrt[2]{{9}^{2}}=9\)
therefore \(\sqrt{9}=\sqrt{{3}^{2}}=3\)
Solution:
we know that \(64000=64 \times {10}^{3}={2}^{6} \times {10}^{3} ={{\left( {{2}^{2}} \right)}^{3}}\times {{10}^{3}}\) so \(\sqrt[3]{64000}=\sqrt[3]{{{\left( {{2}^{2}} \right)}^{3}}\times {{10}^{3}}}={2}^{2} \times 10=40\)
also we have \(68921={41}^{3}\) thus \(\sqrt[3]{68921}=\sqrt[3]{{41}^{3}}=41\)
hence \(\sqrt[2]{40+41}=\sqrt[2]{81}=\sqrt[2]{{9}^{2}}=9\)
therefore \(\sqrt{9}=\sqrt{{3}^{2}}=3\)
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