Bài 15 trang 9 SBT toán 7 tập 1

Đề bài

Tìm tập hợp các số nguyên \(x\), biết rằng:

\(\displaystyle 4{5 \over 9}:2{5 \over {18}} - 7 < x\displaystyle < \left( {3{1 \over 5}:3,2 + 4,5.1{{31} \over {45}}} \right):\left( { - 21{1 \over 2}} \right)\)

Lời giải

\(\displaystyle 4{5 \over 9}:2{5 \over {18}} - 7 < x < \left( {3{1 \over 5}:3,2 + 4,5.1{{31} \over {45}}} \right):\left( { - 21{1 \over 2}} \right)\) 

\(\displaystyle \frac{{41}}{9}:\frac{{41}}{{18}} - 7 < x < \left( {\frac{{16}}{5}:\frac{{16}}{5} + \frac{9}{2}.\frac{{76}}{{45}}} \right):\frac{{ - 43}}{2}\)

\(\displaystyle {{41} \over 9}.{{18} \over {41}} - 7 < x < \left( {1 + {{38} \over 5}} \right).{{ - 2} \over 43}\)

\(\displaystyle 2 - 7 < x < \left( {\frac{5}{5} + \frac{{38}}{5}} \right).\frac{{ - 2}}{{43}}\)

\(\displaystyle -5 < x < {{43} \over 5}.{{ - 2} \over {43}}\)

\(\displaystyle  - 5 < x < {{ - 2} \over 5}\)

Vì \(x ∈\mathbb Z\) nên \({\rm{x}} \in \left\{ { - 4; - 3; - 2; - 1} \right\}\).


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