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

Toán Lớp 9: BT1: Tính (rút gọn) a) √8-2√7 b) √7+4√3 c) √7-2√6

Toán Lớp 9: BT1: Tính (rút gọn)
a) √8-2√7
b) √7+4√3
c) √7-2√6

Comments ( 2 )

  1. a/ $\sqrt{8-2\sqrt 7}\\=\sqrt{7-2\sqrt 7+1}\\=\sqrt{(\sqrt 7)^2-2.\sqrt 7.1+1^2}\\=\sqrt{(\sqrt 7-1)^2}\\=|\sqrt 7-1|\\=\sqrt 7-1(vì\,\,\sqrt 7-1>0)$
    Vậy $\sqrt{8-2\sqrt 7}=\sqrt 7-1$
    b/ $\sqrt{7+4\sqrt 3}\\=\sqrt{4+4\sqrt 3+3}\\=\sqrt{2^2+2.2.\sqrt 3+(\sqrt 3)^2}\\=\sqrt{(2+\sqrt 3)^2}\\=|2+\sqrt 3|\\=2+\sqrt 3(vì\,\,2+\sqrt 3>0)$
    Vậy $\sqrt{7+4\sqrt 3}=2+\sqrt 3$
    c/ $\sqrt{7-2\sqrt 6}\\=\sqrt{6-2\sqrt 6+1}\\=\sqrt{(\sqrt 6)^2-2.\sqrt 6.1+1^2}\\=\sqrt{(\sqrt 6-1)^2}\\=|\sqrt 6-1|\\=\sqrt 6-1(vì\,\,\sqrt 6-1>0)$
    Vậy $\sqrt{7-2\sqrt 6}=\sqrt 6-1$

  2. Giải đáp:
     
    Lời giải và giải thích chi tiết:
    a) \sqrt{8-2\sqrt{7}}
    =\sqrt{7+1-2\sqrt{7}}
    =\sqrt{(\sqrt{7})^2+(1)^2-2\sqrt{7}}
    =\sqrt{(\sqrt{7}-1)^2}
    =|\sqrt{7}-1|
    =\sqrt{7}-1
    b) \sqrt{7+4\sqrt{3}}
    =\sqrt{4+3+2.2\sqrt{3}}
    =\sqrt{(2)^2+(\sqrt{3})^2+2.2\sqrt{3}}
    =\sqrt{(2+\sqrt{3})^2}
    =|2+\sqrt{3}|
    =2+\sqrt{3}
    c) \sqrt{7-2\sqrt{6}}
    =\sqrt{6+1-2\sqrt{6}}
    =\sqrt{(\sqrt{6})^2+(1)^2-2\sqrt{6}}
    =\sqrt{(1-\sqrt{6})^2}
    =|1-\sqrt{6}|
    =\sqrt{6}-1

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

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