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

Toán Lớp 9: Thực hiện phép tính sau a) P=√11+6√2 – √11-6√2 b) Q=√17+12√2 – √17-12√2

Toán Lớp 9: Thực hiện phép tính sau
a) P=√11+6√2 – √11-6√2
b) Q=√17+12√2 – √17-12√2

Comments ( 2 )

  1. Giải đáp + Lời giải và giải thích chi tiết:
    a)P=\sqrt{11+6\sqrt{2}}-\sqrt{11-6\sqrt{2}}
    =\sqrt{9+2.3.\sqrt{2}+2}-\sqrt{9-2.3.\sqrt{2}+2}
    =\sqrt{(9+\sqrt{2})^2}-\sqrt{(9-\sqrt{2})^2}
    =|9+\sqrt{2}|-|9-\sqrt{2}|
    =9+\sqrt{2}-9+\sqrt{2}=2\sqrt{2}
    b)Q=\sqrt{17+12\sqrt{2}}-\sqrt{17-12\sqrt{2}}
    =\sqrt{9+2.3.2\sqrt{2}+(2\sqrt{2})^2}-\sqrt{9-2.3.2\sqrt{2}+(2\sqrt{2})^2}
    =\sqrt{(3+2\sqrt{2})^2}-\sqrt{(3-2\sqrt{2})^2}
    =|3+2\sqrt{2}|-|3-2\sqrt{2}|
    =3+2\sqrt{2}-3+2\sqrt{2}=4\sqrt{2}

  2. $a)P=\sqrt[]{11+6\sqrt[]{2}}-\sqrt[]{11-6\sqrt[]{2}}$
    $=\sqrt[]{(3+\sqrt[]{2})^2}-\sqrt[]{(3-\sqrt[]{2})^2}$
    $=|3+\sqrt[]{2}|-|3-\sqrt[]{2}|$
    $=3-3+\sqrt[]{2}+\sqrt[]{2}$
    $=2\sqrt[]{2}$
    $b)Q=\sqrt[]{17+12\sqrt[]{2}}-\sqrt[]{17-12\sqrt[]{2}}$
    $=\sqrt[]{(3+2\sqrt[]{2})^2}-\sqrt[]{(3-2\sqrt[]{2})^2}$
    $=|3+2\sqrt[]{2}|-|3-2\sqrt[]{2}|$
    $=3-3+2\sqrt[]{2}+2\sqrt[]{2}$
    $=4\sqrt[]{2}$

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

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