k 取何值时方程组 \begincases x + y + z = 1 }, {x + 2y + kz = 2 }, {x + 4y + k^2 z = 4 \endcases 有唯一解？k 取何值时方程组 \begincases x + y + z = 1 }, {x + 2y + kz = 2 }, {x + 4y + k^2 z = 4 \endcases 有唯一解？

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