Đề bài - bài tập 7 trang 56 tài liệu dạy – học toán 7 tập 1
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\(\eqalign{ & a){{27} \over {{3^x}}} = 3 \cr & {3.3^x} = 27 \cr & {3^{x + 1}} = {3^3} \cr & x + 1 = 3 \cr & x = 2 \cr & b){9^{x - 3}} = {1 \over {81}} \cr & {9^{x - 3}}.81 = 1 \cr & {9^{x - 3}}{.9^2} = 1 \cr & {9^{x - 3 + 2}} = 1 \cr & {9^{x - 1}} = {9^0} \cr & x - 1 = 0 \cr & x = 1 \cr & c){{{2^{7x}}} \over {{2^{3x}}}} = 64 \cr & {2^{7x - 3x}} = {2^6} \cr & {2^{4x}} = {2^6} \cr & 4x = 6 \cr & x = 1{1 \over 2} \cr & d){\left( {{2 \over 3}} \right)^x} = {\left( {{8 \over {27}}} \right)^2} \cr & {\left( {{2 \over 3}} \right)^x} = {\left[ {{{\left( {{2 \over 3}} \right)}^3}} \right]^2} \cr & {\left( {{2 \over 3}} \right)^x} = {\left( {{2 \over 3}} \right)^6} \cr & x = 6 \cr} \) 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 \(\eqalign{ & a){{27} \over {{3^x}}} = 3 \cr & {3.3^x} = 27 \cr & {3^{x + 1}} = {3^3} \cr & x + 1 = 3 \cr & x = 2 \cr & b){9^{x - 3}} = {1 \over {81}} \cr & {9^{x - 3}}.81 = 1 \cr & {9^{x - 3}}{.9^2} = 1 \cr & {9^{x - 3 + 2}} = 1 \cr & {9^{x - 1}} = {9^0} \cr & x - 1 = 0 \cr & x = 1 \cr & c){{{2^{7x}}} \over {{2^{3x}}}} = 64 \cr & {2^{7x - 3x}} = {2^6} \cr & {2^{4x}} = {2^6} \cr & 4x = 6 \cr & x = 1{1 \over 2} \cr & d){\left( {{2 \over 3}} \right)^x} = {\left( {{8 \over {27}}} \right)^2} \cr & {\left( {{2 \over 3}} \right)^x} = {\left[ {{{\left( {{2 \over 3}} \right)}^3}} \right]^2} \cr & {\left( {{2 \over 3}} \right)^x} = {\left( {{2 \over 3}} \right)^6} \cr & x = 6 \cr} \)
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