Question

ultilizando o teorema de pitagoras determine o valor de x

Answer 1

$a ^{2} = b ^{2} + c ^{2}$
$(x+1) ^{2} = ( \sqrt{7} ) ^{2} + x^{2}$
$x^{2} +2.1.x+ 1^{2} = 7+ x^{2}$
$x^{2} +2x+1=7+ x^{2}$
$x^{2} - x^{2}+2x-7+1=0$
$2x-6=0$
$2x=6$
$x= \frac{6}{2}$
x=3

Logo x é igual a 3.