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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