Hướng dẫn giải

Đáp án đúng là: A

A = $1 - \frac{3}{4} + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{3} + {(\frac{3}{4})}^{4} - ... - {(\frac{3}{4})}^{2021} + {(\frac{3}{4})}^{2022}$

$\frac{3}{4} A =$ $\frac{3}{4} - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - {(\frac{3}{4})}^{4} + {(\frac{3}{4})}^{5} - ... - {(\frac{3}{4})}^{2022} + {(\frac{3}{4})}^{2023}$

$\frac{3}{4} A + A = [\frac{3}{4} - {(\frac{3}{4})}^{2} + {(\frac{3}{4})}^{3} - {(\frac{3}{4})}^{4} + {(\frac{3}{4})}^{5} - ... - {(\frac{3}{4})}^{2022} + {(\frac{3}{4})}^{2023}]$ + $[1 - \frac{3}{4} + {(\frac{3}{4})}^{2} - {(\frac{3}{4})}^{3} + {(\frac{3}{4})}^{4} - ... - {(\frac{3}{4})}^{2021} + {(\frac{3}{4})}^{2022}]$

$\frac{7}{4} A =$ $1 + {(\frac{3}{4})}^{2023}$

A = $[1 + {(\frac{3}{4})}^{2023}] : \frac{7}{4} = [1 + {(\frac{3}{4})}^{2023}] . \frac{4}{7} = \frac{4}{7} + \frac{4}{7} . \frac{3}{4} . {(\frac{3}{4})}^{2022} = \frac{4}{7} + \frac{3}{7} . {(\frac{3}{4})}^{2022}$

Vậy đáp án đúng là A.

Tìm x sao cho 182^x: 13^x=196.