Đáp án:

\(\eqalign{ & a)\,x = 51 \cr & b)\,\,x = 76 \cr & c)\,\,x = 405 \cr & d)\,\,x = 17 \cr & e)\,\,x = 9 \cr} \)

Giải thích các bước giải:

\(\eqalign{ & a)\,\,2\left( {x - 1} \right) - 86 = 14 \cr & \Leftrightarrow 2\left( {x - 1} \right) = 14 + 86 \cr & \Leftrightarrow 2\left( {x - 1} \right) = 100 \cr & \Leftrightarrow x - 1 = 50 \cr & \Leftrightarrow x = 51 \cr & b)\,\,9x - 13 = 671 \cr & \Leftrightarrow 9x = 671 + 13 \cr & \Leftrightarrow 9x = 684 \cr & \Leftrightarrow x = 76 \cr & c)\,\,31\left( {x - 405} \right).67 = 0 \cr & \Leftrightarrow x - 405 = 0 \cr & \Leftrightarrow x = 405 \cr & d)\,\,3\left( {x + 7} \right) - 21 = 51 \cr & \Leftrightarrow 3\left( {x + 7} \right) = 51 + 21 \cr & \Leftrightarrow 3\left( {x + 7} \right) = 72 \cr & \Leftrightarrow x + 7 = 24 \cr & \Leftrightarrow x = 17 \cr & e)\,\,4\left( {x + 5} \right) + 44 = 100 \cr & \Leftrightarrow 4\left( {x + 5} \right) = 100 - 44 \cr & \Leftrightarrow 4\left( {x + 5} \right) = 56 \cr & \Leftrightarrow x + 5 = 14 \cr & \Leftrightarrow x = 9 \cr} \)

Đáp án:

Giải thích: