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222-9+11+12:2*14+14 = ? ( )

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) ]

Home/ 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) ]

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) ]

Tháng Hai 9, 2023 Toán học 1 Comment 0 views

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) ]

Comments ( 1 )

  1. trucoanh55
    trucoanh55
    9 Tháng Hai, 2023 at 5:56 chiều
    Reply
    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

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222-9+11+12:2*14+14 = ? ( )

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