bonjour s'il vous plait montrer que √(2+√3)√(3+√(7+√3))√(3−√(7+√3))=1 ⅓√(99+(9÷13))=⅕√(275+(25÷13))

Bonjour Abdellaouikari

[tex]\sqrt{2+\sqrt{3}}\times\sqrt{3+\sqrt{7+\sqrt{3}}}\times\sqrt{3-\sqrt{7+\sqrt{3}}}\\\\=\sqrt{2+\sqrt{3}}\times\sqrt{(3+\sqrt{7+\sqrt{3}})(3-\sqrt{7+\sqrt{3}})}[/tex]

[tex]=\sqrt{2+\sqrt{3}}\times\sqrt{3^2-(\sqrt{7+\sqrt{3}})^2}\\\\=\sqrt{2+\sqrt{3}}\times\sqrt{9-(7+\sqrt{3}})}[/tex]

[tex]=\sqrt{2+\sqrt{3}}\times\sqrt{9-7-\sqrt{3}}}\\\\=\sqrt{2+\sqrt{3}}\times\sqrt{2-\sqrt{3}}}[/tex]

[tex]=\sqrt{(2+\sqrt{3})(2-\sqrt{3})}\\\\=\sqrt{2^2-(\sqrt{3})^2}[/tex]

[tex]=\sqrt{4-3}\\\\=\sqrt{1}\\\\=\boxed{1}[/tex]

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[tex]\dfrac{1}{3}\sqrt{99+(9:13)}=\dfrac{1}{3}\sqrt{99+\dfrac{9}{13}}=\dfrac{1}{3}\sqrt{\dfrac{1287}{13}+\dfrac{9}{13}}=\dfrac{1}{3}\sqrt{\dfrac{1296}{13}[/tex]

[tex]=\dfrac{1}{3}\times\dfrac{\sqrt{1296}}{\sqrt{13}}=\dfrac{1}{3}\times\dfrac{36}{\sqrt{13}}=\boxed{\dfrac{12}{\sqrt{13}}}[/tex]


[tex]\dfrac{1}{5}\sqrt{275+(25:13)}=\dfrac{1}{5}\sqrt{275+\dfrac{25}{13}}=\dfrac{1}{5}\sqrt{\dfrac{3575}{13}+\dfrac{25}{13}}=\dfrac{1}{5}\sqrt{\dfrac{3600}{13}}\\\\[/tex]

[tex]=\dfrac{1}{5}\times\dfrac{\sqrt{3600}}{\sqrt{13}}=\dfrac{1}{5}\times\dfrac{60}{\sqrt{13}}=\boxed{\dfrac{12}{\sqrt{13}}}[/tex]

Par conséquent,

[tex]\boxed{\dfrac{1}{3}\sqrt{99+(9:13)}=\dfrac{1}{5}\sqrt{275+(25:13)}=\dfrac{12}{\sqrt{13}}}[/tex]