O número real x, tal que |x - 1 x + 2| = 5, é
| -3 x |
a) -2
b) -1
c) 0
d) 1
Anônimo23:
| -3 x | é embaixo de x-1 e x+2
Soluções para a tarefa
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EAE mano,
aplique a mesma regra usada para determinantes de 2a ordem..
![\left|\begin{array}{ccc}x-1&x+2\\-3&x\\\end{array}\right|=5\\\\\\
x(x-1)-(-3)\cdot(x+2)=5\\
x^2-x+3\cdot(x+2)=5\\
x^2-x+3x+6=5\\
x^2+2x+6-5=0\\
x^2+2x+1=0\\\\
\Delta=2^2-4\cdot1\cdot1\\
\Delta=4-4\\
\Delta=0\\\\
x= \dfrac{-2\pm \sqrt{0} }{2\cdot1}= \dfrac{-2\pm0}{2}\begin{cases}x'=x''=-1\end{cases} \left|\begin{array}{ccc}x-1&x+2\\-3&x\\\end{array}\right|=5\\\\\\
x(x-1)-(-3)\cdot(x+2)=5\\
x^2-x+3\cdot(x+2)=5\\
x^2-x+3x+6=5\\
x^2+2x+6-5=0\\
x^2+2x+1=0\\\\
\Delta=2^2-4\cdot1\cdot1\\
\Delta=4-4\\
\Delta=0\\\\
x= \dfrac{-2\pm \sqrt{0} }{2\cdot1}= \dfrac{-2\pm0}{2}\begin{cases}x'=x''=-1\end{cases}](https://tex.z-dn.net/?f=++%5Cleft%7C%5Cbegin%7Barray%7D%7Bccc%7Dx-1%26amp%3Bx%2B2%5C%5C-3%26amp%3Bx%5C%5C%5Cend%7Barray%7D%5Cright%7C%3D5%5C%5C%5C%5C%5C%5C%0Ax%28x-1%29-%28-3%29%5Ccdot%28x%2B2%29%3D5%5C%5C%0Ax%5E2-x%2B3%5Ccdot%28x%2B2%29%3D5%5C%5C%0Ax%5E2-x%2B3x%2B6%3D5%5C%5C%0Ax%5E2%2B2x%2B6-5%3D0%5C%5C%0Ax%5E2%2B2x%2B1%3D0%5C%5C%5C%5C%0A%5CDelta%3D2%5E2-4%5Ccdot1%5Ccdot1%5C%5C%0A%5CDelta%3D4-4%5C%5C%0A%5CDelta%3D0%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B-2%5Cpm+%5Csqrt%7B0%7D+%7D%7B2%5Ccdot1%7D%3D+%5Cdfrac%7B-2%5Cpm0%7D%7B2%7D%5Cbegin%7Bcases%7Dx%27%3Dx%27%27%3D-1%5Cend%7Bcases%7D+++)
E portanto alternativa B, -1
Tenha ótimos estudos ;D
aplique a mesma regra usada para determinantes de 2a ordem..
E portanto alternativa B, -1
Tenha ótimos estudos ;D
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