Bài 44 trang 26 SGK Toán 6 tập 2

Đề bài

Điền dấu thích hợp (<, >, = ) vào ô vuông.

Lời giải chi tiết

a) Ta có \(\dfrac{{ - 4}}{7} + \dfrac{3}{{ - 7}} = -1\)

Vì

\(\dfrac{{ - 4}}{7} + \dfrac{3}{{ - 7}} = \dfrac{{ - 4}}{7} + \dfrac{{ - 3}}{7} \)\(\,= \dfrac{{ - 4 + \left( { - 3} \right)}}{7} = \dfrac{{ - 7}}{7} =  - 1\)

b) Ta có \(\dfrac{{ - 15}}{{22}} + \dfrac{{ - 3}}{{22}} < \dfrac{{ - 8}}{{11}}\)

Vì

\(\dfrac{{ - 15}}{{22}} + \dfrac{{ - 3}}{{22}} = \dfrac{{ - 15 + \left( { - 3} \right)}}{{22}} \)\(\,= \dfrac{{ - 18}}{{22}} = \dfrac{{ - 9}}{{11}} < \dfrac{{ - 8}}{{11}}\)

c) Ta có \(\dfrac{3}{5} > \dfrac{2}{3} + \dfrac{{ - 1}}{5}\)

Vì 

\(\begin{array}{l}
\dfrac{2}{3} + \dfrac{{ - 1}}{5} = \dfrac{{10}}{{15}} + \dfrac{{ - 3}}{{15}} \\= \dfrac{{10 + \left( { - 3} \right)}}{{15}} = \dfrac{7}{{15}}\\
\dfrac{3}{5} = \dfrac{9}{{15}}\\
Do\,\,\dfrac{9}{{15}} > \dfrac{7}{{15}} \\\Rightarrow \dfrac{3}{5} > \dfrac{2}{3} + \dfrac{{ - 1}}{5}
\end{array}\)

d) Ta có \(\dfrac{1}{6} + \dfrac{{ - 3}}{4} < \dfrac{1}{{14}} + \dfrac{{ - 4}}{7}\)

Vì

\(\begin{array}{l}
\dfrac{1}{6} + \dfrac{{ - 3}}{4} = \dfrac{2}{{12}} + \dfrac{{ - 9}}{{12}} = \dfrac{{2 + \left( { - 9} \right)}}{{12}} = \dfrac{{ - 7}}{{12}}\\
\dfrac{1}{{14}} + \dfrac{{ - 4}}{7} = \dfrac{1}{{14}} + \dfrac{{ - 8}}{{14}} = \dfrac{{1 + \left( { - 8} \right)}}{{14}} = \dfrac{{ - 7}}{{14}}\\
\dfrac{{ - 7}}{{12}} = \dfrac{{ - 7.7}}{{12.7}} = \dfrac{{ - 49}}{{84}};\\
\dfrac{{ - 7}}{{14}} = \dfrac{{ - 7.6}}{{14.6}} = \dfrac{{ - 42}}{{84}}\\
Do\,\,\dfrac{{ - 49}}{{84}} < \dfrac{{ - 42}}{{84}}\\ \Rightarrow \dfrac{1}{6} + \dfrac{{ - 3}}{4} < \dfrac{1}{{14}} + \dfrac{{ - 4}}{7}
\end{array}\)

 

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