Bài 7. Chứng minh \(\root 3 \of {7 + 5\sqrt 2 } + \root 3 \of {7 - 5\sqrt 2 } = 2\)
Đặt \(x = \root 3 \of {7 + 5\sqrt 2 } + \root 3 \of {7 - 5\sqrt 2 } \) Ta có:
\({x^3} = \left( {\root 3 \of {7 + 5\sqrt 2 } + \root 3 \of {7 - 5\sqrt 2 } } \right)^3\)
\( = 7 + 5\sqrt 2 + 7 - 5\sqrt 2 + 3\root 3 \of {{{\left( {7 + 5\sqrt 2 } \right)}^2}} .\root 3 \of {7 - 5\sqrt 2 } + 3\root 3 \of {7 + 5\sqrt 2 } .\root 3 \of {{{\left( {7 - 5\sqrt 2 } \right)}^2}} \)
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\( = 14 - 3\left( {\root 3 \of {7 + 5\sqrt 2 } + \root 3 \of {7 - 5\sqrt 2 } } \right) = 14 - 3x\).
Từ đó suy ra: \({x^3} + 3x - 14 = 0\,\,\,\,\left( 1 \right)\)
\(\left( 1 \right) \Leftrightarrow \left( {x - 2} \right)\left( {{x^2} + 2x + 7} \right) = 0 \Leftrightarrow x - 2 = 0 \Leftrightarrow x = 2\) ( vì \({x^2} + 2x + 7 > 0\))
Vậy \(\root 3 \of {7 + 5\sqrt 2 } + \root 3 \of {7 - 5\sqrt 2 } = 2\)