O conjunto solução da equação irracional
raiz x -3=x-5 é?
(A)-7
(B)7
(C)-6
(D)6
Me ajudem por favor,eu imploro!!
Respostas
respondido por:
5
Boa noite Emanuelle!
Solução!
Primeiro temos que elevar os termos da equação ao quadrado.
![\sqrt{x-3}=x-5\\\\\
(\sqrt{x-3})^{2} =(x-5)^{2} \\\\\
x-3= x^{2} -10x+25\\\\\\
x^{2} -11x+28=0\\\\\\
Uma~~equac\~ao ~~do~~segundo~~grau!\\\\\\\
x= \dfrac{11\pm \sqrt{(-11)^{2}-4.1.-28 } }{2.1}\\\\\\\
x= \dfrac{11\pm \sqrt{121-112 } }{2}\\\\\\\
x= \dfrac{11\pm \sqrt{9} }{2}\\\\\\\
x= \dfrac{11\pm 3 }{2}\\\\\\\\\\\\
{ x_{1}= \dfrac{11+3}{2} }= \dfrac{14}{2}=7 \\\\\\
x_{2}= \dfrac{11-3}{2} }= \dfrac{8}{2}=4 \\\\\\\\\\\\\
\boxed{Resposta:~~S=\{7\}}](https://tex.z-dn.net/?f=+%5Csqrt%7Bx-3%7D%3Dx-5%5C%5C%5C%5C%5C%0A+%28%5Csqrt%7Bx-3%7D%29%5E%7B2%7D+%3D%28x-5%29%5E%7B2%7D+%5C%5C%5C%5C%5C%0Ax-3%3D+x%5E%7B2%7D+-10x%2B25%5C%5C%5C%5C%5C%5C%0A+x%5E%7B2%7D+-11x%2B28%3D0%5C%5C%5C%5C%5C%5C%0AUma%7E%7Eequac%5C%7Eao+%7E%7Edo%7E%7Esegundo%7E%7Egrau%21%5C%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B11%5Cpm+%5Csqrt%7B%28-11%29%5E%7B2%7D-4.1.-28+%7D+%7D%7B2.1%7D%5C%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B11%5Cpm+%5Csqrt%7B121-112+%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B11%5Cpm+%5Csqrt%7B9%7D+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%0Ax%3D+%5Cdfrac%7B11%5Cpm+3+%7D%7B2%7D%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%7B+x_%7B1%7D%3D+%5Cdfrac%7B11%2B3%7D%7B2%7D+%7D%3D+%5Cdfrac%7B14%7D%7B2%7D%3D7+%5C%5C%5C%5C%5C%5C%0A+x_%7B2%7D%3D+%5Cdfrac%7B11-3%7D%7B2%7D+%7D%3D+%5Cdfrac%7B8%7D%7B2%7D%3D4+%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%5C%0A%5Cboxed%7BResposta%3A%7E%7ES%3D%5C%7B7%5C%7D%7D%0A%0A%0A++ "\sqrt{x-3}=x-5\\\\\
(\sqrt{x-3})^{2} =(x-5)^{2} \\\\\
x-3= x^{2} -10x+25\\\\\\
x^{2} -11x+28=0\\\\\\
Uma~~equac\~ao ~~do~~segundo~~grau!\\\\\\\
x= \dfrac{11\pm \sqrt{(-11)^{2}-4.1.-28 } }{2.1}\\\\\\\
x= \dfrac{11\pm \sqrt{121-112 } }{2}\\\\\\\
x= \dfrac{11\pm \sqrt{9} }{2}\\\\\\\
x= \dfrac{11\pm 3 }{2}\\\\\\\\\\\\
{ x_{1}= \dfrac{11+3}{2} }= \dfrac{14}{2}=7 \\\\\\
x_{2}= \dfrac{11-3}{2} }= \dfrac{8}{2}=4 \\\\\\\\\\\\\
\boxed{Resposta:~~S=\{7\}}")
Boa noite!
Bons estudos!
Solução!
Primeiro temos que elevar os termos da equação ao quadrado.
Boa noite!
Bons estudos!
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