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Câu hỏi: Tính giá trị của biểu thức A,B,C rồi sắp xếp các kết quả tìm được theo thứ tự từ nhỏ đến lớn:
\(\displaystyle{\rm{A}} = {2 \over 3} + {3 \over 4}.\left( {{{ - 4} \over 9}} \right)\)
\(\displaystyle B = 2{3 \over {11}}.1{1 \over {12}}.\left( { - 2,2} \right)\)
\(\displaystyle C = \left( {{3 \over 4} - 0,2} \right).\left( {0,4 - {4 \over 5}} \right)\)
\(\displaystyle{\rm{A}} = {2 \over 3} + {3 \over 4}.\left( {{{ - 4} \over 9}} \right)\)
\(\displaystyle B = 2{3 \over {11}}.1{1 \over {12}}.\left( { - 2,2} \right)\)
\(\displaystyle C = \left( {{3 \over 4} - 0,2} \right).\left( {0,4 - {4 \over 5}} \right)\)
Phương pháp giải
\(\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{ad}}{{bd}} + \dfrac{{cb}}{{bd}} = \dfrac{{ad + cb}}{{bd}}\)
\(\dfrac{a}{b} - \dfrac{c}{d} =\dfrac{a}{b} + \dfrac{-c}{d}= \dfrac{{ad}}{{bd}} + \dfrac{{-cb}}{{bd}} \)\( = \dfrac{{ad - cb}}{{bd}}\)
\( \dfrac{a}{b} . \dfrac{c}{d} =\dfrac{a.c}{b.d}\)
Lời giải chi tiết
\(\displaystyle {\rm{A}} = {2 \over 3} + {3 \over 4}.\left( {{{ - 4} \over 9}} \right) = {2 \over 3} + {{ - 1} \over 3} \)\( \displaystyle= {1 \over 3}\)
\(\displaystyle B = 2{3 \over {11}}.1{1 \over {12}}.\left( { - 2,2} \right) \)
\(\displaystyle = {{25} \over {11}}.{{13} \over {12}}.{{ - 22} \over {10}} \)
\(\displaystyle = {{5.5.13.(-2).11} \over {11.12.5.2}} = {{ - 65} \over {12}}\)
\(\displaystyle C = \left( {{3 \over 4} - 0,2} \right).\left( {0,4 - {4 \over 5}} \right) \)
\( \displaystyle = \left( {{3 \over 4} - {1 \over 5}} \right).\left( {{2 \over 5} - {4 \over 5}} \right)\)
\( \displaystyle = \left( {{{15} \over {20}} - {4 \over {20}}} \right).\left( {{{ - 2} \over 5}} \right)\)
\(\displaystyle = {{11} \over {20}}.\left( {{{ - 2} \over 5}} \right) = {{ - 11} \over {50}}\)
\(\begin{array}{l}
\dfrac{{ - 65}}{{12}} = \dfrac{{\left( { - 65} \right).25}}{{12.25}} = \dfrac{{ - 1625}}{{300}}\\
\dfrac{{ - 11}}{{50}} = \dfrac{{\left( { - 11} \right).6}}{{50.6}} = \dfrac{{ - 66}}{{300}}\\
\Rightarrow \dfrac{{ - 65}}{{12}} < \dfrac{{ - 11}}{{50}} < \dfrac{1}{3}
\end{array}\)
Do đó: \(B<C<A.\)
\(\dfrac{a}{b} + \dfrac{c}{d} = \dfrac{{ad}}{{bd}} + \dfrac{{cb}}{{bd}} = \dfrac{{ad + cb}}{{bd}}\)
\(\dfrac{a}{b} - \dfrac{c}{d} =\dfrac{a}{b} + \dfrac{-c}{d}= \dfrac{{ad}}{{bd}} + \dfrac{{-cb}}{{bd}} \)\( = \dfrac{{ad - cb}}{{bd}}\)
\( \dfrac{a}{b} . \dfrac{c}{d} =\dfrac{a.c}{b.d}\)
Lời giải chi tiết
\(\displaystyle {\rm{A}} = {2 \over 3} + {3 \over 4}.\left( {{{ - 4} \over 9}} \right) = {2 \over 3} + {{ - 1} \over 3} \)\( \displaystyle= {1 \over 3}\)
\(\displaystyle B = 2{3 \over {11}}.1{1 \over {12}}.\left( { - 2,2} \right) \)
\(\displaystyle = {{25} \over {11}}.{{13} \over {12}}.{{ - 22} \over {10}} \)
\(\displaystyle = {{5.5.13.(-2).11} \over {11.12.5.2}} = {{ - 65} \over {12}}\)
\(\displaystyle C = \left( {{3 \over 4} - 0,2} \right).\left( {0,4 - {4 \over 5}} \right) \)
\( \displaystyle = \left( {{3 \over 4} - {1 \over 5}} \right).\left( {{2 \over 5} - {4 \over 5}} \right)\)
\( \displaystyle = \left( {{{15} \over {20}} - {4 \over {20}}} \right).\left( {{{ - 2} \over 5}} \right)\)
\(\displaystyle = {{11} \over {20}}.\left( {{{ - 2} \over 5}} \right) = {{ - 11} \over {50}}\)
\(\begin{array}{l}
\dfrac{{ - 65}}{{12}} = \dfrac{{\left( { - 65} \right).25}}{{12.25}} = \dfrac{{ - 1625}}{{300}}\\
\dfrac{{ - 11}}{{50}} = \dfrac{{\left( { - 11} \right).6}}{{50.6}} = \dfrac{{ - 66}}{{300}}\\
\Rightarrow \dfrac{{ - 65}}{{12}} < \dfrac{{ - 11}}{{50}} < \dfrac{1}{3}
\end{array}\)
Do đó: \(B<C<A.\)