Tìm tập hợp các số nguyên x biết: 1/2-(1/3+1/4)

• Lớp 7
• Toán Học

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Đáp án:

x=0

Giải thích các bước giải: $\begin{array}{l} \frac{1}{2} - (\frac{1}{3} + \frac{1}{4}) < x < \frac{1}{{48}} - (\frac{1}{{16}} - \frac{1}{6})\\ \Leftrightarrow \frac{1}{2} - \frac{{4 + 3}}{{3 \times 4}} < x < \frac{1}{{48}} - (\frac{3}{{48}} - \frac{8}{{48}})\\ \Leftrightarrow \frac{1}{2} - \frac{7}{{12}} < x < \frac{1}{{48}} - ( - \frac{5}{{48}})\\ \Leftrightarrow \frac{6}{{12}} - \frac{7}{{12}} < x < \frac{6}{{48}}\\ \Leftrightarrow \frac{{ - 1}}{{12}} < x < \frac{1}{8}\\ \Rightarrow x = 0 \end{array}$

$\begin{array}{l} \dfrac{1}{2} - \left( {\dfrac{1}{3} + \dfrac{1}{4}} \right) = \dfrac{1}{2} - \left( {\dfrac{4}{{12}} + \dfrac{3}{{12}}} \right) = \dfrac{6}{{12}} - \dfrac{7}{{12}} = \dfrac{{ - 1}}{{12}};\\ \dfrac{1}{{48}} - \left( {\dfrac{1}{{16}} - \dfrac{1}{6}} \right) = \dfrac{1}{{48}} - \left( {\dfrac{3}{{48}} - \dfrac{8}{{48}}} \right) = \dfrac{1}{{48}} - \dfrac{{ - 5}}{{48}}\\ = \dfrac{1}{{48}} + \dfrac{5}{{48}} = \dfrac{6}{{48}} = \dfrac{1}{8}\\ \Rightarrow \dfrac{{ - 1}}{{12}} < x < \dfrac{1}{8}\\ \Rightarrow x = 0 \end{array}$