Question

teorema de Pitágoras: calcule o valor de x nos ∆ retângulos: x, x+2, x+1, x+5, 6, (x+3), x, 3√5, 2x

Answer 1

(x+2)²=x²+(x+1)²
x²+4x+4=x²+x²+2x+1
x²-2x²+4x-2x+4-1=0
-x²+2x+3=0
x²-2x-3=0
(x+1)(x-3)=0
x'=-1 e x''=3
logo x=3

(x+5)²=6²+(x+3)²
x²+10x+25=36+x²+6x+9
x²-x²+10x-6x=36+9-25
4x=20
x=20/4
x=5

(3V5)²=(2x)²+x²
9.5=4x²+x²
45=5x²
x²=9
x=v9
x=3