Determine a solução de:
2 3 -2
0 1 x = 2
2 x -3
Soluções para a tarefa
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Usando a regra de Sarrus:
![[(2 \times 1 \times -3) + (3 \times x \times 2) +\\ (-2 \times 0 \times x)] + [(-2 \times 1 \times 2) +\\ (x \times x \times 2) + (-3 \times 0 \times 3)] = 2\\\\
[-6 + 6x + 0] - [-4 + 2x^2 + 0] = 2\\
-2x^2 +6x - 4 = 0\\
2x^2 - 6x + 4 = 0\\\\
\Delta = (-6)^2 - 4 \times 2 \times 4\\
\Delta = 36 - 32\\
\Delta = 4\\\\
x = \frac{-(-6) \pm \sqrt 4}{2 \times 2}\\
x = \frac{6 \pm 2}{4}\\\\
\boxed{\boxed{S = \{2; 1\}}}](https://tex.z-dn.net/?f=%5B%282+%5Ctimes+1+%5Ctimes+-3%29+%2B+%283+%5Ctimes+x+%5Ctimes+2%29+%2B%5C%5C+%28-2+%5Ctimes+0+%5Ctimes+x%29%5D+%2B+%5B%28-2+%5Ctimes+1+%5Ctimes+2%29+%2B%5C%5C+%28x+%5Ctimes+x+%5Ctimes+2%29+%2B+%28-3+%5Ctimes+0+%5Ctimes+3%29%5D+%3D+2%5C%5C%5C%5C%0A%0A%5B-6+%2B+6x+%2B+0%5D+-+%5B-4+%2B+2x%5E2+%2B+0%5D+%3D+2%5C%5C%0A-2x%5E2+%2B6x+-+4+%3D+0%5C%5C%0A2x%5E2+-+6x+%2B+4+%3D+0%5C%5C%5C%5C%0A%0A%5CDelta+%3D+%28-6%29%5E2+-+4+%5Ctimes+2+%5Ctimes+4%5C%5C%0A%5CDelta+%3D+36+-+32%5C%5C%0A%5CDelta+%3D+4%5C%5C%5C%5C%0Ax+%3D+%5Cfrac%7B-%28-6%29+%5Cpm+%5Csqrt+4%7D%7B2+%5Ctimes+2%7D%5C%5C%0Ax+%3D+%5Cfrac%7B6+%5Cpm+2%7D%7B4%7D%5C%5C%5C%5C%0A%5Cboxed%7B%5Cboxed%7BS+%3D+%5C%7B2%3B+1%5C%7D%7D%7D "[(2 \times 1 \times -3) + (3 \times x \times 2) +\\ (-2 \times 0 \times x)] + [(-2 \times 1 \times 2) +\\ (x \times x \times 2) + (-3 \times 0 \times 3)] = 2\\\\
[-6 + 6x + 0] - [-4 + 2x^2 + 0] = 2\\
-2x^2 +6x - 4 = 0\\
2x^2 - 6x + 4 = 0\\\\
\Delta = (-6)^2 - 4 \times 2 \times 4\\
\Delta = 36 - 32\\
\Delta = 4\\\\
x = \frac{-(-6) \pm \sqrt 4}{2 \times 2}\\
x = \frac{6 \pm 2}{4}\\\\
\boxed{\boxed{S = \{2; 1\}}}")
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