Bài 68 trang 35 SGK Toán 6 tập 2

Đề bài

Tính:

a) $\displaystyle {3 \over 5} - {-7 \over {10}} - {{13} \over { - 20}}$

b) $\displaystyle {3 \over 4} + {{ - 1} \over 3} - {5 \over {18}}$

c) $\displaystyle {3 \over {14}} - {5 \over { - 8}} + {{ - 1} \over 2}$

d) $\displaystyle {1 \over 2} + {1 \over { - 3}} + {1 \over 4} - {{ - 1} \over 6}$

Lời giải chi tiết

a) $\displaystyle {3 \over 5} - {{ - 7} \over {10}} - {{13} \over {-20}}$$\displaystyle = {3 \over 5} + {7 \over {10}} + {{13} \over {20}}$$\displaystyle = {{3.4} \over {5.4}} + {{7.2} \over {10.2}} + {{13} \over {20}}$

$\displaystyle = {{12 + 14 + 13} \over {20}} = {{39} \over {20}}$

b) $\displaystyle \frac{3}{4} + \frac{{ - 1}}{3} - \frac{5}{{18}}$$\displaystyle = \frac{{3.9}}{{36}} + \frac{{ - 1.12}}{{36}} + \frac{{ - 5.2}}{{36}}$

$\displaystyle = \frac{{27 - 12 - 10}}{{36}} = \frac{5}{{36}}$

c) $\displaystyle \frac{3}{{14}} - \frac{5}{{ - 8}} + \frac{{ - 1}}{2} = \frac{3}{{14}} + \frac{5}{8} + \frac{{ - 1}}{2}$$\displaystyle = \frac{{12}}{{56}} + \frac{{35}}{{56}} + \frac{{ - 28}}{{56}}$

$\displaystyle = \frac{{12 + 35 - 28}}{{56}} = \frac{{19}}{{56}}$ 

d)  $\displaystyle \frac{1}{2} + \frac{1}{{ - 3}} + \frac{1}{4} - \frac{{ - 1}}{6}$$\displaystyle = \frac{1}{2} + \frac{{ - 1}}{3} + \frac{1}{4} + \frac{1}{6}$ 

$\displaystyle = \frac{6}{{12}} + \frac{{ - 4}}{{12}} + \frac{3}{{12}} + \frac{2}{{12}}$$\displaystyle = \frac{{6 - 4 + 3 + 2}}{{12}} = \frac{7}{{12}}$

Đáp số: $\displaystyle a){{39} \over {20}};b){5 \over {36}};c){{19} \over {56}};d){7 \over {12}}$

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