Question

3) Se x=0,222... e y=0,2444... a soma x+y será? * 44/90 44/9 68/99 34/9

Answer 1

$\boxed{\begin{array}{l}\sf x=0,222...\cdot10\\\sf 10x=2,222....\\-\underline{\begin{cases}\sf 10x=2,222...\\\sf x=0,222...\end{cases}}\\\sf9x=2\\\sf x=\dfrac{2}{9}\\\sf y=0,2444...\cdot10\\\sf 10y=2,444... \cdot10\\\sf 100y=24,444...\\-\underline{\begin{cases}\sf 100y=24,444...\\\sf10y=2,444...\end{cases}}\\\sf 90y=22\\\sf y=\dfrac{22}{90}\\\sf x+y=\dfrac{2}{9}+\dfrac{22}{90}\\\sf x+y=\dfrac{20+22}{90}\\\huge\boxed{\boxed{\boxed{\boxed{\sf x+y=\dfrac{42}{90}}}}}\end{array}}$