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

Toán Lớp 9: $\frac{2\sqrt{3+\sqrt{5-\sqrt{13+ \sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}$

Home/ Toán Lớp 9: $\frac{2\sqrt{3+\sqrt{5-\sqrt{13+ \sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}$

Toán Lớp 9: $\frac{2\sqrt{3+\sqrt{5-\sqrt{13+ \sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}$

Tháng Chín 6, 2022 Toán học 2 Comments 0 views

Toán Lớp 9: $\frac{2\sqrt{3+\sqrt{5-\sqrt{13+ \sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}$

Comments ( 2 )

  1. ngoclankhue
    ngoclankhue
    6 Tháng Chín, 2022 at 4:39 sáng
    Reply
    ~rai~
    \(\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{13+4\sqrt{3}}}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{12+2.2\sqrt{3}+1}}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+\sqrt{5-\sqrt{(2\sqrt{3}+1)^2}}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+\sqrt{5-|2\sqrt{3}+1|}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+\sqrt{5-2\sqrt{3}-1}}}{\sqrt{6}+\sqrt{2}}\quad\text{(vì }2\sqrt{3}+1>0)\\=\dfrac{2\sqrt{3+\sqrt{4-2\sqrt{3}}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+\sqrt{3-2\sqrt{3}+1}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+\sqrt{(\sqrt{3}-1)^2}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+|\sqrt{3}-1|}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{2\sqrt{3+\sqrt{3}-1}}{\sqrt{6}+\sqrt{2}}\quad\text{(vì }\sqrt{3}-1>0)\\=\dfrac{2\sqrt{2+\sqrt{3}}}{\sqrt{6}+\sqrt{2}}\\=\dfrac{\sqrt{2}.\sqrt{2}.\sqrt{2+\sqrt{3}}}{\sqrt{2}(\sqrt{3}+1)}\\=\dfrac{\sqrt{4+2\sqrt{3}}}{\sqrt{3}+1}\\=\dfrac{3+2\sqrt{3}+1}{\sqrt{3}+1}\\=\dfrac{\sqrt{(\sqrt{3}+1)^2}}{\sqrt{3}+1}\\=\dfrac{|\sqrt{3}+1|}{\sqrt{3}+1}\\=\dfrac{\sqrt{3}+1}{\sqrt{3}+1}\quad\text{(vì }\sqrt{3}+1>0)\\=1.\)
     

  2. ngochuong2k2
    ngochuong2k2
    6 Tháng Chín, 2022 at 4:39 sáng
    Reply
    $\displaystyle \begin{array}{{>{\displaystyle}l}} \frac{2\sqrt{3+\sqrt{5-\sqrt{13+\sqrt{48}}}}}{\sqrt{6} +\sqrt{2}} \ \\ =\frac{2\sqrt{3+\sqrt{5-\sqrt{12+2\sqrt{12} +1}}}}{\sqrt{6} +\sqrt{2}}\\ =\frac{2\sqrt{3+\sqrt{5-\sqrt{12} -1}}}{\sqrt{6} +\sqrt{2} \ }\\ =\frac{2\sqrt{3+\sqrt{1-2\sqrt{3} +3}} \ }{\sqrt{2}\left(\sqrt{3} +1\right)}\\ =\frac{2\sqrt{3+\sqrt{\left( 1-\sqrt{3}\right)^{2}}}}{\sqrt{2}\left(\sqrt{3} +1\right)}\\ =\frac{2\sqrt{3+\sqrt{3} -1}}{\sqrt{2}\left(\sqrt{3} +1\right)}\\ =\frac{2\sqrt{2+\sqrt{3}}}{\sqrt{2}\left(\sqrt{3} -1\right)} =\frac{\sqrt{2}\sqrt{4+2\sqrt{3}}}{\sqrt{2}\left(\sqrt{3} +1\right)} \ \\ =\frac{\sqrt{3} +1}{\sqrt{3} +1} =1\ \\ \end{array}$

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

Thúy Mai

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