Bài 1: Tính:
\(C = {{ – 1} \over 5} – {\left( {{1 \over 2} + {3 \over 4}} \right)^2}:{5 \over 8};\)
\(D = {{{5^3} + {{3.5}^2}} \over { – 8}}\).
Bài 2: So sánh: \(A = {3^{222}}\) và \(B = {2^{333}}\).
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Bài 1: \(\eqalign{ C &= {{ – 1} \over 5} – {\left( {{1 \over 2} + {3 \over 4}} \right)^2}:{5 \over 8}\cr&= – {1 \over 5} + {\left( {{{2 + 3} \over 4}} \right)^2}.{8 \over 5} \cr&= {{ – 1} \over 5} – {\left( {{5 \over 4}} \right)^2}.{8 \over 5}\cr& = {{ – 1} \over 5} – {{25} \over {16}}.{8 \over 5} \cr&= – {1 \over 5} – {5 \over 2} = {{ – 2 – 25} \over {10}} = {{ – 27} \over {10}}\cr}\)
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\(D = {{{5^3} + {{3.5}^2}} \over { – 8}} = {{{5^2}.\left( {5 + 3} \right)} \over { – 8}} = {{{5^2}.8} \over { – 8}} = – 25.\)
Bài 2: Ta có
\(A = {3^{222}} = {\left( {{3^2}} \right)^{111}} = {9^{111}};\)
\(B = {2^{333}} = {\left( {{2^3}} \right)^{111}} = {8^{111}};\)
Vì \({9^{111}} > {8^{111}}\). Do đó \({3^{222}} > {2^{333}}\,\;hay\;\,A > B.\)
