Respuesta:
Es 1
Explicación paso a paso:
[tex]\sqrt{\left(x+3\right)^2-\left(x+2\right)\left(x+4\right)}[/tex]
[tex]{\left(x+3\right)^2-\left(x+2\right)\left(x+4\right)}:1[/tex]
[tex]\sqrt{1}[/tex]
1
Espero que te ayude :)
[tex] \sqrt{ {x}^{2} } + 6x \: + 9 - ( {x}^{2} + 4x + 2x + 8) \: \\ \sqrt{ {x}^{2} } + 6x + 9 - ( {x}^{2} + 6x + 8 \\ \sqrt{ {x}^{2} } + 6x + 9 - {x}^{2} 6x - 8 \\ \sqrt{1} [/tex]
todo eso es elevado a la raíz jajzka
" Life is not a problem to be solved but a reality to be experienced! "
© Copyright 2013 - 2025 KUDO.TIPS - All rights reserved.
Verified answer
Respuesta:
Es 1
Explicación paso a paso:
[tex]\sqrt{\left(x+3\right)^2-\left(x+2\right)\left(x+4\right)}[/tex]
[tex]{\left(x+3\right)^2-\left(x+2\right)\left(x+4\right)}:1[/tex]
[tex]\sqrt{1}[/tex]
1
Espero que te ayude :)
Respuesta:
[tex] \sqrt{ {x}^{2} } + 6x \: + 9 - ( {x}^{2} + 4x + 2x + 8) \: \\ \sqrt{ {x}^{2} } + 6x + 9 - ( {x}^{2} + 6x + 8 \\ \sqrt{ {x}^{2} } + 6x + 9 - {x}^{2} 6x - 8 \\ \sqrt{1} [/tex]
todo eso es elevado a la raíz jajzka