Đề bài
Tìm x, biết:
\[\eqalign{ & a]\,\,{{27} \over {{3^x}}} = 3\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b]{9^{x - 3}} = {1 \over {81}} \cr & c]\,\,{{{2^{7x}}} \over {{2^{3x}}}} = 64\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,c]\,\,{\left[ {{2 \over 3}} \right]^x} = {\left[ {{8 \over {27}}} \right]^2} \cr} \]
Lời giải chi tiết
\[\eqalign{ & a]{{27} \over {{3^x}}} = 3 \cr & {3.3^x} = 27 \cr & {3^{x + 1}} = {3^3} \cr & x + 1 = 3 \cr & x = 2 \cr & b]{9^{x - 3}} = {1 \over {81}} \cr & {9^{x - 3}}.81 = 1 \cr & {9^{x - 3}}{.9^2} = 1 \cr & {9^{x - 3 + 2}} = 1 \cr & {9^{x - 1}} = {9^0} \cr & x - 1 = 0 \cr & x = 1 \cr & c]{{{2^{7x}}} \over {{2^{3x}}}} = 64 \cr & {2^{7x - 3x}} = {2^6} \cr & {2^{4x}} = {2^6} \cr & 4x = 6 \cr & x = 1{1 \over 2} \cr & d]{\left[ {{2 \over 3}} \right]^x} = {\left[ {{8 \over {27}}} \right]^2} \cr & {\left[ {{2 \over 3}} \right]^x} = {\left[ {{{\left[ {{2 \over 3}} \right]}^3}} \right]^2} \cr & {\left[ {{2 \over 3}} \right]^x} = {\left[ {{2 \over 3}} \right]^6} \cr & x = 6 \cr} \]