Bài 1. Rút gọn : \(a = \root 3 \of {8x} – 2\root 3 \of {27x} + \sqrt {49x} ;\,x \ge 0\)
Bài 2. Tìm x, biết : \(\root 3 \of {3 – x} + 2 = 0\)
Bài 3. Tìm x, biết : \(\root 3 \of {1 – x} < 2\)
Bài 4. Trục căn thức ở mẫu số: \({1 \over {\root 3 \of 3 + \root 3 \of 2 }}\)
Bài 1. Ta có:
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\(\eqalign{ a &= \root 3 \of {{2^3}x} – 2\root 3 \of {{3^3}x} + \sqrt {{7^2}x} \cr & = 2\root 3 \of x – 6\root 3 \of x + 7\sqrt x \cr & = – 4\root 3 \of x + 7\sqrt x \cr} \)
Bài 2. Ta có:
\(\eqalign{ & \root 3 \of {3 – x} + 2 = 0 \Leftrightarrow \root 3 \of {3 – x} = – 2 \cr & \Leftrightarrow 3 – x = – 8 \Leftrightarrow x = 11 \cr} \)
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Bài 3. Ta có: \(\root 3 \of {1 – x} < 2 \Leftrightarrow 1 – x < 8 \Leftrightarrow x > – 7\)
Bài 4. Ta có:
\(\eqalign{ & {1 \over {\root 3 \of 3 + \root 3 \of 2 }} \cr&= {{\root 3 \of 9 – \root 3 \of 6 + \root 3 \of 4 } \over {\left( {\root 3 \of 3 + \root 3 \of 2 } \right)\left( {\root 3 \of 9 – \root 3 \of 6 + \root 3 \of 4 } \right)}} \cr & = {{\root 3 \of 9 – \root 3 \of 6 + \root 3 \of 4 } \over {{{\left( {\root 3 \of 3 } \right)}^3} – {{\left( {\root 3 \of 2 } \right)}^3}}} \cr&= {{\root 3 \of 9 – \root 3 \of 6 + \root 3 \of 4 } \over {3 – 2}} \cr & = \root 3 \of 9 – \root 3 \of 6 + \root 3 \of 4 \cr} \)
