Question

AYUDENME PORFA LO NECESITO :(((((

Answer 1

Respuesta:

te ayudo con las 4 primeras... de las demás no estoy seguro por eso nomas no te mando

A)

$\sqrt[]{x-5} = 10 \\\\(\sqrt{x-5} )^{2} = (10)^{2}\\\\x - 5 = 100\\x = 100+5\\x = 105\\\\comprobacion\\\\ \sqrt[]{x-5} = 10\\\sqrt{105-5} = 10\\\sqrt{100} = 10\\ 10 = 10$

B)

$\sqrt{x-8} = 0\\\\(\sqrt{x-8} )^{2} =(0)^{2}\\\\x - 8 = 0\\x = 8\\\\comprobacion\\ \sqrt{x-8} = 0\\\sqrt{8-8} = 0\\\sqrt{0} = 0\\0 = 0$

C)

$x+5 = \sqrt{(x^{2}) +4}\\\\(x+5)^{2} = (\sqrt{x^{2} +4} )^{2} \\\\x^{2} +2(x)(5) + 5^{2} = x^{2} +4\\ x^{2} + 10x +25 = x^{2} +4\\\\x^{2} -x^{2} +10x = 4-25\\10x = -21\\x = -21/10\\\\x= -2.1\\\\comprobacion\\\\x+5 = \sqrt{(x^{2}) +4}\\-2.1 + 5 = \sqrt{(-2.1^{2) } +4} \\2.9 = \sqrt{(-2.1)^{2}+4 }\\2.9 = \sqrt{4.41+4}\\2.9 = \sqrt{8.41}\\2.9 = 2.9$

D)

$x+7 = \sqrt{(x^{2} )+6} \\\\(x+7)^{2} = (\sqrt{x^{2}+6 })^{2} \\\\x^{2} +2(x)(7)+7^{2} = x^{2} +6\\\\x^{2} +14x +49 = x^{2} +6\\x^{2} -x^{2} +14x = 6 -49\\14x = -43\\x = -43/14\\x = -3.071428571\\x = -3.1\\\\comprobacion\\\\x+7 = \sqrt{(x^{2} )+6}\\\\-3.1 + 7 = \sqrt{(-3.1^{2}) +6}\\3.9 = \sqrt{9.61 + 6}\\3.9 = \sqrt{15.61}\\3.9 = 3.9$

Explicación paso a paso: