Bài 1 : Tính
1) 1/4 – ( 7/10 + -20/25 ) – |1/2 – 2/3|
2) 1/5 – ( 10/8 – 16/4 + 1/2 ) -|3/10 – 7/4|
3) |-3/4 + 5/10 + -8/6| + ( -5/3 – -7/2 )
4) -4/5 – |-7/3 + 3/4| + (14/6 – 12/8 )
5) 1/7 – ( 5/3 – -4/21 ) -|5/6 – 1/2 – 3/4|
6) -|-7/5 – -3/10 + -5/2| + ( 7/-20 + 1/4 )
7) 7/8 – ( 1/4 + -2/3 + 5/6 ) + |-5/3 + 1/6 |
8) 5/7 – 4/21 + | -3/2 – -1/7 | – ( 1/3 – 5/6 |
9) -5/3 – ( 4/5 – 1/2 ) – | 3/4 – 5/2 + 1/3 |
10) 1/2 – ( 5/3 – 7/4 + 1/6 ) + ( 11/12 – 7/4 – 1/2 )
Đáp án:
\(1)\dfrac{{11}}{{60}}\)
Giải thích các bước giải:
\(\begin{array}{l}
1)\dfrac{1}{4} – \dfrac{7}{{10}} + \dfrac{4}{5} – \left| { – \dfrac{1}{6}} \right|\\
= \dfrac{1}{4} – \dfrac{7}{{10}} + \dfrac{4}{5} – \dfrac{1}{6}\\
= \dfrac{{15 – 7.6 + 4.12 – 10}}{{60}} = \dfrac{{11}}{{60}}\\
2)\dfrac{1}{5} – \dfrac{5}{4} + 4 – \dfrac{1}{2} – \left| { – \dfrac{{29}}{{20}}} \right|\\
= \dfrac{{49}}{{20}} – \dfrac{{29}}{{20}} = 1\\
3)\left| { – \dfrac{3}{4} + \dfrac{1}{2} – \dfrac{4}{3}} \right| – \dfrac{5}{3} + \dfrac{7}{2}\\
= \left| { – \dfrac{{19}}{{12}}} \right| + \dfrac{{11}}{6} = \dfrac{{19}}{{12}} + \dfrac{{11}}{6} = \dfrac{{41}}{{12}}\\
4) – \dfrac{4}{5} – \left| { – \dfrac{{19}}{{12}}} \right| + \dfrac{7}{3} – \dfrac{3}{2}\\
= – \dfrac{4}{5} – \dfrac{{19}}{{12}} + \dfrac{7}{3} – \dfrac{3}{2} = – \dfrac{{31}}{{20}}\\
5)\dfrac{1}{7} – \dfrac{5}{3} + \dfrac{4}{{21}} – \left| { – \dfrac{5}{{12}}} \right|\\
= \dfrac{1}{7} – \dfrac{5}{3} + \dfrac{4}{{21}} – \dfrac{5}{{12}} = – \dfrac{7}{4}\\
6) – \left| { – \dfrac{7}{5} + \dfrac{3}{{10}} – \dfrac{5}{2}} \right| – \dfrac{1}{{10}}\\
= – \left| { – \dfrac{{18}}{5}} \right| – \dfrac{1}{{10}} = – \dfrac{{18}}{5} – \dfrac{1}{{10}}\\
= – \dfrac{{37}}{{10}}\\
7)\dfrac{7}{8} – \dfrac{5}{{12}} + \left| { – \dfrac{3}{2}} \right| = \dfrac{{11}}{{24}} + \dfrac{3}{2}\\
= \dfrac{{47}}{{24}}\\
9) – \dfrac{5}{3} – \dfrac{3}{{10}} – \left| { – \dfrac{{17}}{{12}}} \right|\\
= – \dfrac{{59}}{{30}} – \dfrac{{17}}{{12}} = – \dfrac{{203}}{{60}}\\
10)\dfrac{1}{2} – \dfrac{1}{{12}} – \dfrac{4}{3} = \dfrac{{6 – 1 – 4.4}}{{12}} = – \dfrac{{11}}{{12}}
\end{array}\)