Question

Calcule X nas questões de Pitágoras abaixo:
a) 3x²=2x²+(5√3)²
b) 20²=4x²+3x²
c) 2x²=x²+(3√3)
d) (x+2)²=x²+(x+1)²

Answer 1

a) Resolvendo:
$3x^2=2x^2+(5\sqrt{3})^2\\ 3x^2=2x^2+25\times3\\ 3x^2=2x^2+75\\ 2x^2-3x^2+75 = 0\\ -x^2+75 = 0\\ -x^2 = -75\\ x = \pm\sqrt{75}\\ \boxed{x = \pm5\sqrt{3}}$

b) Resolvendo:
$20^2=4x^2+3x^2\\ 400=7x^2\\ x^2 = \dfrac{400}{7}\\ \boxed{x = \pm\sqrt{\dfrac{400}{7}}}$

c) Resolvendo:
$2x^2=x^2+(3\sqrt{3})\\ 2x^2-x^2=(3\sqrt{3})\\ x^2=3\sqrt{3}\\ x^2=\pm\sqrt{27}\\ x=\pm\sqrt{\sqrt{27}}\\ \boxed{x=\pm\sqrt[4]{27}}$

d) Resolvendo:
$x^2+(x+1)^2-(x+2)^2 = 0\\ x^2+x^2+2x+1-x^2-4x-4 = 0\\ x^2-2x-3 = 0\\\\ Resolvendo\ trin\^omio\ do\ quadrado\ perfeito:\\ (x-1)^2=3+1\\ (x-1)^2=4\\ x=\pm\sqrt{4}+1\\\\ x'=2+1\\ \boxed{x'=3}\\ x''=-2+1\\ \boxed{x''=-1}$