Bài tập 1 trang 13 VBT Toán 5 tập 1
Bài tập 1 trang 13 VBT Toán 5 tập 1
Điền dấu thích hợp vào chỗ chấm \((>\,;\; <\,;\; =)\)
\(\eqalign{ & 5{1 \over 7}\;...\;2{6 \over 7}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad 3{2 \over 7}\;...\;3{5 \over 7} \cr & 8{6 \over {10}}\;...\;8{3 \over 5}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\quad 9{1 \over 2}\;...\;5{1 \over 2} \cr}\)
a) \(5\dfrac{1}{{7}} = \dfrac{{36}}{{7}}\;;\;\;2\dfrac{6}{{7}} = \dfrac{{20}}{{7}}.\)
Mà \(\dfrac{{36}}{{7}} > {\rm{ }}\dfrac{{20}}{{7}}\)
Vậy: \(5\dfrac{1}{{7}} > {\rm{ }}2\dfrac{6}{{7}}\)
b) \(3\dfrac{2}{{7}} = \dfrac{{23}}{{7}}\;;\;\;3\dfrac{5}{{7}} = \dfrac{{26}}{{7}}\)
Mà \(\dfrac{{23}}{{7}} < \dfrac{{26}}{{7}}\)
Vậy : \(3\dfrac{2}{{7}} < {\rm{ }}3\dfrac{5}{{7}}\)
c) \(8\dfrac{6}{{10}} = \dfrac{{86}}{{10}}\;;\;\;8\dfrac{3}{{5}} = \dfrac{{43}}{{5}}\)
Mà \(\dfrac{{43}}{{5}} = \dfrac{{43 \times 2}}{{5 \times}} =\dfrac{{86}}{{10}}\)
Vậy: \(8\dfrac{6}{{10}} = {\rm{ }}6\dfrac{3}{{5}}\)
d) \(9\dfrac{1}{{2}} = \dfrac{{19}}{{2}} \;;\;\;5\dfrac{1}{2} = \dfrac{{11}}{2}\)
Mà \(\dfrac{{19}}{2} > \dfrac{{11}}{2}\)
Vậy : \(9\dfrac{1}{{2}} > {\rm{ }}5\dfrac{1}{2}\)
-- Mod Toán lớp 5
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