Toán Lớp 9: chứng minh : A=∛(1+√84/9)+∛(1-√84/9) là số nguyên
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Toán Lớp 9: chứng minh : A=∛(1+√84/9)+∛(1-√84/9) là số nguyên
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A = \sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}} + \sqrt[3]{{1 – \dfrac{{\sqrt {84} }}{9}}}\\
\Leftrightarrow {A^3} = 1 + \dfrac{{\sqrt {84} }}{9} + 3{\left( {\sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}}} \right)^2}.\sqrt[3]{{1 – \dfrac{{\sqrt {84} }}{9}}}\\
+ 3.\sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}}.{\left( {\sqrt[3]{{1 – \dfrac{{\sqrt {84} }}{9}}}} \right)^2} + 1 – \dfrac{{\sqrt {84} }}{9}\\
\Leftrightarrow {A^3} = 2 + 3.\sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}}.\sqrt[3]{{1 – \dfrac{{\sqrt {84} }}{9}}}.\left( {\sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}} – \sqrt[3]{{1 + \dfrac{{\sqrt {84} }}{9}}}} \right)\\
\Leftrightarrow {A^3} = 2 + 3.\sqrt[3]{{1 – \dfrac{{84}}{{81}}}}.A\\
\Leftrightarrow {A^3} = 2 + 3.\sqrt[3]{{\dfrac{{ – 3}}{{81}}}}.A\\
\Leftrightarrow {A^3} = 2 + 3.\sqrt[3]{{ – \dfrac{1}{{27}}}}.A\\
\Leftrightarrow {A^3} = 2 + 3.\dfrac{{ – 1}}{3}.A\\
\Leftrightarrow {A^3} + A – 2 = 0\\
\Leftrightarrow {A^3} – {A^2} + {A^2} – A + 2A – 2 = 0\\
\Leftrightarrow \left( {A – 1} \right)\left( {{A^2} + A + 2} \right) = 0\\
\Leftrightarrow A = 1
\end{array}$