Toán Lớp 9: Khử mẫu: a) 3√3 / (√2 + √3 + √5) b) (√2 + √3 + √4) / (√2 + √3 + √6 + √8 + 4) c) [căn ( √5 + √2 ) ] / [căn (3 √5 – 3 √2) ]

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1. Giải đáp:

   a) {3\sqrt{3}}/{\sqrt{2}+\sqrt{3}+\sqrt{5}}={3.(2+\sqrt{6}-\sqrt{10})}/4

   b) {\sqrt{2}+\sqrt{3}+\sqrt{4}}/{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}=\sqrt{2}-1

   c) {\sqrt{\sqrt{5}+\sqrt{2}}}/{\sqrt{3\sqrt{5}-3\sqrt{2}}}={\sqrt{5}+\sqrt{2}}/3

   Lời giải và giải thích chi tiết:

   a) {3\sqrt{3}}/{\sqrt{2}+\sqrt{3}+\sqrt{5}}

   ={3\sqrt{3}.(\sqrt{2}+\sqrt{3}-\sqrt{5})}/{(\sqrt{2}+\sqrt{3}+\sqrt{5}).(\sqrt{2}+\sqrt{3}-\sqrt{5})}

   ={3\sqrt{3}.(\sqrt{2}+\sqrt{3}-\sqrt{5})}/{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}

   ={3\sqrt{3}.(\sqrt{2}+\sqrt{3}-\sqrt{5})}/{(\sqrt{2}+\sqrt{3})^2-(\sqrt{5})^2}

   ={3\sqrt{3}.(\sqrt{2}+\sqrt{3}-\sqrt{5})}/{2+2\sqrt{2}.\sqrt{3}+3-5}

   ={3\sqrt{3}.(\sqrt{2}+\sqrt{3}-\sqrt{5})}/{2\sqrt{2}.\sqrt{3}}

   ={3.(\sqrt{2}+\sqrt{3}-\sqrt{5})}/{2\sqrt{2}}

   ={3.(\sqrt{2}+\sqrt{3}-\sqrt{5}).\sqrt{2}}/{2\sqrt{2}.\sqrt{2}}

   ={3.(2+\sqrt{6}-\sqrt{10})}/4

   Vậy: {3\sqrt{3}}/{\sqrt{2}+\sqrt{3}+\sqrt{5}}={3.(2+\sqrt{6}-\sqrt{10})}/4

   $\\$

   b) {\sqrt{2}+\sqrt{3}+\sqrt{4}}/{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}

   ={\sqrt{2}+\sqrt{3}+\sqrt{4}}/{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+\sqrt{4}+\sqrt{4}}

   ={\sqrt{2}+\sqrt{3}+\sqrt{4}}/{(\sqrt{2}+\sqrt{3}+\sqrt{4})+(\sqrt{4}+\sqrt{6}+\sqrt{8})}

   ={\sqrt{2}+\sqrt{3}+\sqrt{4}}/{(\sqrt{2}+\sqrt{3}+\sqrt{4})+\sqrt{2}.(\sqrt{2}+\sqrt{3}+\sqrt{4})}

   ={\sqrt{2}+\sqrt{3}+\sqrt{4}}/{(\sqrt{2}+\sqrt{3}+\sqrt{4}).(1+\sqrt{2})}

   =1/{1+\sqrt{2}}

   ={\sqrt{2}-1}/{(\sqrt{2}-1).(\sqrt{2}+1)}

   ={\sqrt{2}-1}/{(\sqrt{2})^2-1^2}=\sqrt{2}-1

   Vậy: {\sqrt{2}+\sqrt{3}+\sqrt{4}}/{\sqrt{2}+\sqrt{3}+\sqrt{6}+\sqrt{8}+4}=\sqrt{2}-1

   $\\$

   c) {\sqrt{\sqrt{5}+\sqrt{2}}}/{\sqrt{3\sqrt{5}-3\sqrt{2}}}

   ={\sqrt{\sqrt{5}+\sqrt{2}}.\sqrt{\sqrt{5}+\sqrt{2}}}/{\sqrt{3\sqrt{5}-3\sqrt{2}}.\sqrt{\sqrt{5}+\sqrt{2}}}

   ={\sqrt{5}+\sqrt{2}}/{\sqrt{3.(\sqrt{5}-\sqrt{2}).(\sqrt{5}+\sqrt{2})}}

   ={\sqrt{5}+\sqrt{2}}/{\sqrt{3.(5-2)}}

   ={\sqrt{5}+\sqrt{2}}/3

   Vậy: {\sqrt{\sqrt{5}+\sqrt{2}}}/{\sqrt{3\sqrt{5}-3\sqrt{2}}}={\sqrt{5}+\sqrt{2}}/3