MacaronLatte Blab

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The answers of these questions :

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$3-7+5=1$

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$\mathrm{(3}+\mathrm{7)}÷5=2$
$\sqrt{\mathrm{(3!}-7+\mathrm{5)}}=\sqrt{4}=2$

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$\sqrt{3^{\mathrm{(7\; -\; 5)}}}=\sqrt{3^2}=3$
$\mathrm{3!}÷\mathrm{(7}-\mathrm{5)}=3$
$-[\sqrt{(-3+7)}]+5=-\sqrt{4}+5=-2+5=3$
$C(\mathrm{3,}\mathrm{ (}7-5))=C_{\mathrm{(7\; -\; 5)}}^3=C_2^3=\frac{\mathrm{3!}}{\mathrm{2! (3\; -\; 2)!}}=3$

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$\mathrm{3!}-7+5=4$

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$3+7-5=5$
$[-(\mathrm{3!})+7]\times 5=1\times 5=5$

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$3\times \mathrm{(7}-\mathrm{5)}=6$
$-(\mathrm{3!})+7+5=6$
$P(\mathrm{3,}\mathrm{ (}7-5))=P_{\mathrm{(7\; -\; 5)}}^3=P_2^3=\frac{\mathrm{3!}}{\mathrm{(3\; -\; 2)!}}=6$
$H(\mathrm{3,}\mathrm{ (}7-5))=H_{\mathrm{(7\; -\; 5)}}^3=H_2^3=C_2^{\mathrm{3\; +\; 2\; -\; 1}}=C_2^4=\frac{\mathrm{4!}}{\mathrm{2! (4\; -\; 2)!}}=6$

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$\sqrt{\mathrm{(-\; 3}+\mathrm{7)}}+5=7$

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$\mathrm{3!}+7-5=8$

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$3^{\mathrm{(7\; -\; 5)}}=3^2=9$
$-3+7+5=9$

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$\sqrt{\mathrm{(-\; 3}+\mathrm{7)}}\times 5=\mathrm{10}$

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$(0!+0!+0!)!=(1+1+1)!=6$

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$(1+1+1)!=6$

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$2+2+2=6$
$2\times 2+2=6$
$2+2\times 2=6$
$2^2+2=6$
$2+2^2=6$
$(2÷2+2)!=(1+2)!=3!=6$
$(2+2÷2)!=(2+1)!=3!=6$

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$3\times 3-3=6$
$3!+3-3=6$
$3!-3+3=6$
$3+3!-3=6$
$3-3+3!=6$
$3!\times (3÷3)=6$
$(3÷3)\times 3!=6$
$\sqrt{(3\times 3)}+3=6$
$3+\sqrt{(3\times 3)}=6$

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$\sqrt{4}+\sqrt{4}+\sqrt{4}=6$
$(\sqrt{4}\times 4)-\sqrt{4}=(2\times 4)-2=8-2=6$
$(4\times \sqrt{4})-\sqrt{4}=(4\times 2)-2=8-2=6$
$[4-(4÷4)]!=(4-1)!=3!=6$
$[\sqrt{4}+(4÷4)]!=(2+1)!=3!=6$
$[(4÷4)+\sqrt{4}]!=(1+2)!=3!=6$
$(4÷\sqrt{4})+4=(4÷2)+4=2+4=6$
$4+(4÷\sqrt{4})=4+(4÷2)=4+2=6$

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$5+(5÷5)=5+1=6$
$(5÷5)+5=1+5=6$

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$6+6-6=6$
$6-6+6=6$
$6\times (6÷6)=6\times 1=6$
$(6÷6)\times 6=1\times 6=6$
$6÷(6÷6)=6÷1=6$
$6\times (6-6)!=6\times 1=6$
$(6-6)!\times 6=1\times 6=6$
$6÷(6-6)!=6÷1=6$

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$7-(7÷7)=7-1=6$

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$8-\sqrt{\sqrt{(8+8)}}=8-\sqrt{\sqrt{\mathrm{16}}}=8-\sqrt{4}=8-2=6$
$\{\sqrt{[8+(8÷8)]}\}!=[\sqrt{(8+1)}]!=(\sqrt{9})!=3!=6$

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$\sqrt{9}\times \sqrt{9}-\sqrt{9}=3\times 3-3=9-3=6$
$(\sqrt{9})!\times (9÷9)=3!\times 1=6$
$(\sqrt{9})!÷(9÷9)=3!÷1=6$
$(\sqrt{9})!+9-9=3!+0=6$
$(\sqrt{9})!-9+9=3!-0=6$
$(\sqrt{9\times 9÷9})!=(\sqrt{9})!=3!=6$
$(\sqrt{9÷9\times 9})!=(\sqrt{9})!=3!=6$

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$\{\sqrt{[\mathrm{10}-(\mathrm{10}÷\mathrm{10})]}\}!=[\sqrt{(\mathrm{10}-1)}]!=(\sqrt{9})!=3!=6$

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Samples :

$\underset{n=1}{\overset{+\mathrm{\infty }}{\Sigma }}\frac{1}{n^2}=\frac{{\pi }^2}{6}$

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$\left|\begin{array}{cc}x& \left(\genfrac{}{}{0ex}{}{5}{3}\right)\\ \sqrt{7+2}& y\end{array}\right|\ne 0$

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$\frac{A}{2}=\left(\begin{array}{ccc}1& 2& 3\\ 4& 5& 6\\ 7& 8& 9\end{array}\right)$

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${\left[123456789\right]}^{\alpha }$

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The infinite sum $\sum _{n=1}^{+\mathrm{\infty }}\frac{1}{n^2}$ is equal to the real number $\frac{{\pi }^2}{6}$ .

$a^2+b^2=c^2$ .

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