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b: D=(-3/31-28/31)+(-6/17-11/17)+(1/25-5/25)
=-2-4/25=-54/25
\(a.\)
\(\dfrac{27}{13}-\dfrac{106}{111}+-\dfrac{5}{111}=\dfrac{27}{13}-\dfrac{106}{111}-\dfrac{5}{111}=\dfrac{27}{13}-\left(\dfrac{106+6}{111}\right)=\dfrac{27}{13}-1=\dfrac{14}{13}\)
\(b.\)
\(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\dfrac{7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+1=\dfrac{23}{11}\)
\(c.\)
\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+-\dfrac{6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)-\left(\dfrac{25}{31}+\dfrac{6}{31}\right)=1-1=0\)
a) \(\dfrac{27}{13}-\dfrac{106}{111}+\dfrac{-5}{111}=\dfrac{27}{13}+\left(\dfrac{-106}{111}+\dfrac{-5}{111}\right)=\dfrac{27}{13}+-1=\dfrac{14}{13}\)
b) \(\dfrac{12}{11}-\dfrac{-7}{19}+\dfrac{12}{19}=\dfrac{12}{11}+\left(\dfrac{7}{19}+\dfrac{12}{19}\right)=\dfrac{12}{11}+1=\dfrac{23}{11}\)
c)\(\dfrac{5}{17}-\dfrac{25}{31}+\dfrac{12}{17}+\dfrac{-6}{31}=\left(\dfrac{5}{17}+\dfrac{12}{17}\right)+\left(\dfrac{-25}{31}+\dfrac{-6}{31}\right)=1+-1=0\)
\(M=\dfrac{1,6:\left(1\dfrac{3}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(5\dfrac{5}{9}-2\dfrac{1}{4}\right).2\dfrac{1}{4}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{1,6:\left(\dfrac{8}{5}.1,25\right)}{0,64-\dfrac{1}{25}}+\dfrac{\left(1,08-\dfrac{2}{25}\right):\dfrac{4}{7}}{\left(\dfrac{50}{9}-\dfrac{9}{4}\right).\dfrac{35}{17}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{1,6:2}{0,64-\dfrac{1}{25}}+\dfrac{1:\dfrac{4}{7}}{\dfrac{119}{36}.\dfrac{35}{17}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{0,8}{0,6}+\dfrac{1,75}{\dfrac{245}{36}}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{4}{3}+\dfrac{9}{35}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+0,6.0,5:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+\dfrac{3}{10}:\dfrac{2}{5}\)
\(M=\dfrac{167}{105}+\dfrac{3}{4}\)
\(M=\dfrac{983}{420}.\)
a) Ta có: \(\dfrac{-5}{18}+\dfrac{32}{45}-\dfrac{9}{10}\)
\(=\dfrac{-25}{90}+\dfrac{64}{90}-\dfrac{81}{90}\)
\(=\dfrac{-42}{90}=-\dfrac{7}{15}\)
b) Ta có: \(\left(-\dfrac{1}{4}+\dfrac{51}{33}-\dfrac{5}{3}\right)-\left(-\dfrac{15}{12}+\dfrac{6}{11}-\dfrac{42}{29}\right)\)
\(=\dfrac{-1}{4}+\dfrac{17}{11}-\dfrac{5}{3}+\dfrac{5}{4}-\dfrac{6}{11}+\dfrac{42}{29}\)
\(=\dfrac{-5}{3}+\dfrac{42}{29}\)
\(=\dfrac{-145}{87}+\dfrac{126}{87}=\dfrac{-19}{87}\)
c) Ta có: \(1-\dfrac{1}{2}+2-\dfrac{2}{3}+3-\dfrac{3}{4}+4-\dfrac{1}{4}-3-\dfrac{1}{3}-2-\dfrac{1}{2}-1\)
\(=\left(1-1\right)-\left(\dfrac{1}{2}+\dfrac{1}{2}\right)+\left(2-2\right)-\left(\dfrac{2}{3}+\dfrac{1}{3}\right)+\left(3-3\right)-\left(\dfrac{3}{4}+\dfrac{1}{4}\right)+4\)
\(=-1-1-1+4\)
=1
a) Ta có: −518+3245−910−518+3245−910
=−2590+6490−8190=−2590+6490−8190
=−4290=−715=−4290=−715
b) Ta có: (−14+5133−53)−(−1512+611−4229)(−14+5133−53)−(−1512+611−4229)
=−14+1711−53+54−611+4229=−14+1711−53+54−611+4229
=−53+4229=−53+4229
=−14587+12687=−1987=−14587+12687=−1987
c) Ta có: 1−12+2−23+3−34+4−14−3−13−2−12−11−12+2−23+3−34+4−14−3−13−2−12−1
=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4=(1−1)−(12+12)+(2−2)−(23+13)+(3−3)−(34+14)+4
=−1−1−1+4=−1−1−1+4
=1
\(\Leftrightarrow-\dfrac{16}{279}< \dfrac{x}{9}< =\dfrac{2}{3}\)
\(\Leftrightarrow\dfrac{x}{9}=0\)
hay x=0
1) (-6 - 2) . (-6 + 2) = -8 . (-4) = 32
2) 627 + 125 + (-627) + |-46| -21 = 125 + 46 - 21 = 150
3) (-25) + 30 + (-16) + 25 + (-4) = 30 + (-20) = 10
4) 75 - 5.(15 - 40) - (-60) = 75 - 75 + 80 + 60 = 140
5) |31 - 17| - |15 - 52| = |14| - |-37| = 14 - 37 = -23
6) 13 - 18 - (-42) - 15 = 13 - 18 + 42 - 15 = 22
3 bài sau hết kiên nhẫn rồi, thông cảm nhé.
a) Ta có: \(\dfrac{2}{3}x-1=\dfrac{3}{2}\)
\(\Leftrightarrow x\cdot\dfrac{2}{3}=\dfrac{5}{2}\)
hay \(x=\dfrac{5}{2}:\dfrac{2}{3}=\dfrac{5}{2}\cdot\dfrac{3}{2}=\dfrac{15}{4}\)
b) Ta có: \(\left|5x-\dfrac{1}{2}\right|-\dfrac{2}{7}=25\%\)
\(\Leftrightarrow\left|5x-\dfrac{1}{2}\right|=\dfrac{1}{4}+\dfrac{2}{7}=\dfrac{15}{28}\)
\(\Leftrightarrow\left[{}\begin{matrix}5x-\dfrac{1}{2}=\dfrac{15}{28}\\5x-\dfrac{1}{2}=\dfrac{-15}{28}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}5x=\dfrac{29}{28}\\5x=\dfrac{-1}{28}\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{29}{140}\\x=\dfrac{-1}{140}\end{matrix}\right.\)
c) Ta có: \(\dfrac{x-3}{4}=\dfrac{16}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=64\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=8\\x-3=-8\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=11\\x=-5\end{matrix}\right.\)
d) Ta có: \(\dfrac{-8}{13}+\dfrac{7}{17}+\dfrac{21}{31}\le x\le\dfrac{-9}{14}+4-\dfrac{5}{14}\)
\(\Leftrightarrow\dfrac{3246}{6851}\le x\le3\)
\(\Leftrightarrow x\in\left\{1;2;3\right\}\)
\(\left(\dfrac{31}{42}\times\dfrac{17}{25}+\dfrac{31}{42}\right)\times\left(-5\right)^2\)
\(=\dfrac{31}{42}\times\left(\dfrac{17}{25}+1\right)\times25\)
\(=\dfrac{31}{42}\times\dfrac{42}{25}\times25=31\)