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

Toán Lớp 9: $\sqrt[]{6- √11}$ -$\sqrt[]{6 +√11}$

Toán Lớp 9: $\sqrt[]{6- √11}$ -$\sqrt[]{6 +√11}$

Comments ( 2 )

  1. Giải đáp+Lời giải và giải thích chi tiết:
    \sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}
    =\sqrt{\frac{12-2\sqrt{11}}{2}}-\sqrt{\frac{12+2\sqrt{11}}{2}}
    =\sqrt{\frac{11-2.\sqrt{11}.1+1}{2}}-\sqrt{\frac{11+2.\sqrt{11}.1+1}{2}}
    =\sqrt{\frac{(\sqrt{11}-1)^2}{2}}-\sqrt{\frac{(\sqrt{11}+1)^2}{2}}
    =\frac{|\sqrt{11}-1|}{\sqrt{2}}-\frac{|\sqrt{11}+1|}{\sqrt{2}}
    =\frac{|\sqrt{11}-1|-|\sqrt{11}+1|}{\sqrt{2}}
    =\frac{\sqrt{11}-1-\sqrt{11}-1}{\sqrt{2}}
    =\frac{-2}{\sqrt{2}}
    =-\sqrt{2}
     

  2. #tnvt
    \sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}}
    =\frac{\sqrt{2}.(\sqrt{6-\sqrt{11}}-\sqrt{6+\sqrt{11}})}{\sqrt{2}}
    =\frac{\sqrt{12-2\sqrt{11}}-\sqrt{12+2\sqrt{11}}}{\sqrt{2}}
    =\frac{\sqrt{11-2\sqrt{11}+1}-\sqrt{11+2\sqrt{11}+1}}{\sqrt{2}}
    =\frac{\sqrt{(\sqrt{11}-1)^2}-\sqrt{(\sqrt{11}+1)^2}}{\sqrt{2}}
    =\frac{|\sqrt{11}-1|-|\sqrt{11}+1|}{\sqrt{2}}
    =\frac{\sqrt{11}-1-\sqrt{11}-1}{\sqrt{2}}
    =-2/(\sqrt{2})
    =-\sqrt{2}
     

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

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