Bài 52 trang 29 SGK Toán 6 tập 2

Đề bài

Điền số thích hợp vào ô trống: 

a	\(\displaystyle {6 \over {27}}\)		\(\displaystyle {3 \over 5}\)	\(\displaystyle {5 \over {14}}\)	\(\displaystyle {4 \over 3}\)	\(\displaystyle {2 \over 5}\)
b	\(\displaystyle {5 \over 27}\)	\(\displaystyle {4 \over {23}}\)	\(\displaystyle {7 \over {10}}\)	\(\displaystyle {2 \over 7}\)	\(\displaystyle {2 \over 3}\)
a + b		\(\displaystyle {{11} \over {23}}\)				\(\displaystyle {8 \over 5}\)

 

Lời giải chi tiết

 

a	\(\displaystyle {6 \over {27}}\)	\(\displaystyle {7 \over {23}}\)	\(\displaystyle {3 \over 5}\)	\(\displaystyle {5 \over {14}}\)	\(\displaystyle {4 \over 3}\)	\(\displaystyle {2 \over 5}\)
b	\(\displaystyle {5 \over 27}\)	\(\displaystyle {4 \over {23}}\)	\(\displaystyle {7 \over {10}}\)	\(\displaystyle {2 \over 7}\)	\(\displaystyle {2 \over 3}\)	\(\displaystyle {6 \over 5}\)
a + b	\(\displaystyle {{11} \over {27}}\)	\(\displaystyle {{11} \over {23}}\)	\(\displaystyle {{13} \over {10}}\)	\(\displaystyle {9 \over {14}}\)	2	\(\displaystyle {8 \over 5}\)

Giải thích: 

• \(\displaystyle a + b = \dfrac{6}{{27}} + \dfrac{5}{{27}} = \dfrac{{11}}{{27}}\)

• \(\displaystyle a + \dfrac{4}{{23}} = \dfrac{{11}}{{23}}\) nên \(\displaystyle a = \dfrac{7}{{23}}\) để \(\displaystyle \dfrac{7}{{23}} + \dfrac{4}{{23}} = \dfrac{{11}}{{23}}\)

• \(\displaystyle a + b = \dfrac{3}{5} + \dfrac{7}{{10}} = \dfrac{6}{{10}} + \dfrac{7}{{10}} = \dfrac{{13}}{{10}}\)

• \(\displaystyle a + b = \dfrac{5}{{14}} + \dfrac{2}{7} = \dfrac{5}{{14}} + \dfrac{4}{{14}} = \dfrac{9}{{14}}\)

• \(\displaystyle a + b = \dfrac{4}{3} + \dfrac{2}{3} = \dfrac{6}{3} = 2\)

• \(\displaystyle \dfrac{2}{5} + b = \dfrac{8}{5}\) nên \(\displaystyle b = \dfrac{6}{5}\) để \(\displaystyle \dfrac{2}{5} + \dfrac{6}{5} = \dfrac{8}{5}\) 

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