Respuesta:
a)
[tex]{2}^{2} = x - 8[/tex]
4+8=x
12=x
b)
[tex] 5 = \sqrt{3 x+ 1} [/tex]
[tex] {5}^{2} = 3x + 1[/tex]
25= 3x-1
24=3x
x=8
c)
[tex] \sqrt{5x - 19} - \sqrt{5x} = 1 [/tex]
5x-19 +5x =1
10x=20
x=2
Radicales
Quitamos el radical elevando al cuadrado toda la ecuación
[tex](\sqrt{x-8} =2)^{2} \\\\x-8=4\\\\x=4+8\\\\x=12[/tex]
[tex]\sqrt{12-8} =2\\\\\sqrt{4} =2\\\\2=2[/tex]
[tex](5-\sqrt{3x+1} =0)^{2} } \\\\25-3x-1=0\\\\24-3x=0\\\\-3x=-24\\\\x=\frac{-24}{-3} \\\\x=8[/tex]
[tex]5-\sqrt{3(8)+1}=0\\\\5-\sqrt{24+1} =0\\\\5-\sqrt{25} =0\\\\5-5=0\\\\0=0[/tex]
[tex](\sqrt{5x-19} -\sqrt{5x} =1)^{2} \\\\5x-19-5x=1\\-19\neq 1[/tex]
a) x=12
b) x=8
c) No se puede resolver porque no es una igualdad
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Respuesta:
a)
[tex]{2}^{2} = x - 8[/tex]
4+8=x
12=x
b)
[tex] 5 = \sqrt{3 x+ 1} [/tex]
[tex] {5}^{2} = 3x + 1[/tex]
25= 3x-1
24=3x
x=8
c)
[tex] \sqrt{5x - 19} - \sqrt{5x} = 1 [/tex]
5x-19 +5x =1
10x=20
x=2
Radicales
Quitamos el radical elevando al cuadrado toda la ecuación
a)
[tex](\sqrt{x-8} =2)^{2} \\\\x-8=4\\\\x=4+8\\\\x=12[/tex]
[tex]\sqrt{12-8} =2\\\\\sqrt{4} =2\\\\2=2[/tex]
b)
[tex](5-\sqrt{3x+1} =0)^{2} } \\\\25-3x-1=0\\\\24-3x=0\\\\-3x=-24\\\\x=\frac{-24}{-3} \\\\x=8[/tex]
[tex]5-\sqrt{3(8)+1}=0\\\\5-\sqrt{24+1} =0\\\\5-\sqrt{25} =0\\\\5-5=0\\\\0=0[/tex]
c)
[tex](\sqrt{5x-19} -\sqrt{5x} =1)^{2} \\\\5x-19-5x=1\\-19\neq 1[/tex]
Respuesta:
a) x=12
b) x=8
c) No se puede resolver porque no es una igualdad