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a)\(\left|-0.75\right|+\dfrac{1}{4}-2\dfrac{1}{2}\)
=0.75+0.25-2.5
=1-2.5=-1.5
b)\(15.\dfrac{1}{5}:\left(\dfrac{-5}{7}\right)-2\dfrac{1}{5}.\left(\dfrac{-7}{5}\right)\)
=3.(-1.4)+3.08
=-4.2+3.08=-1.12
c)\(\dfrac{5}{17}+\dfrac{2}{3}-\dfrac{20}{12}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{49}{51}-\dfrac{5}{3}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{-12}{17}+\dfrac{7}{9}+\dfrac{12}{17}\)
=\(\dfrac{11}{153}+\dfrac{12}{17}\)
=\(\dfrac{7}{9}\)
d)\(\dfrac{5}{15}+\dfrac{14}{25}-\dfrac{12}{9}+\dfrac{2}{7}+\dfrac{11}{25}\)
=\(\dfrac{67}{75}-\dfrac{4}{3}+\dfrac{2}{7}+\dfrac{11}{25}\)
=-0.44+\(\dfrac{127}{175}\)
=\(\dfrac{2}{7}\)
a,\(\dfrac{5}{6}+\left(-\dfrac{1}{2}\right)+\dfrac{3}{4}\)
\(=\dfrac{10}{12}+\left(-\dfrac{6}{12}\right)+\dfrac{9}{12}\)
\(=\dfrac{10-6+9}{12}=\dfrac{13}{12}\)
b,\(\left(0,75-\dfrac{1}{3}\right):\dfrac{7}{15}\)
\(=\left(\dfrac{3}{4}-\dfrac{1}{3}\right):\dfrac{7}{15}\)
\(=\left(\dfrac{9}{12}-\dfrac{4}{12}\right):\dfrac{7}{15}\)
\(=\dfrac{5}{12}:\dfrac{7}{15}\)
\(=\dfrac{25}{28}\)
c,\(\dfrac{7}{12}-\dfrac{3}{4}.\dfrac{5}{6}\)
\(=\dfrac{7}{12}-\dfrac{5}{8}\)
\(=\dfrac{14}{24}-\dfrac{15}{24}\)
\(=-\dfrac{1}{24}\)
d,\(\left(2\dfrac{1}{3}+1\dfrac{3}{4}\right).\dfrac{12}{13}\)
\(=\left(\dfrac{7}{3}+\dfrac{7}{4}\right).\dfrac{12}{13}\)
\(=\left(\dfrac{28}{12}+\dfrac{21}{12}\right).\dfrac{12}{13}\)
\(=\dfrac{49}{12}.\dfrac{12}{13}\)
\(=\dfrac{49}{13}\)
a) \(\dfrac{5}{6}+\left(\dfrac{-1}{2}\right)+\dfrac{3}{4}\)
\(=\dfrac{10}{12}-\dfrac{6}{12}+\dfrac{9}{12}\)
\(=\dfrac{13}{12}\)
b) \(\left(0,75-\dfrac{1}{3}\right):\dfrac{7}{15}\)
\(=\left(\dfrac{3}{4}-\dfrac{1}{3}\right).\dfrac{15}{7}\)
\(=\left(\dfrac{9}{12}-\dfrac{4}{12}\right).\dfrac{15}{7}\)
\(=\dfrac{5}{12}.\dfrac{15}{7}\)
\(=\dfrac{25}{28}\)
c) \(\dfrac{7}{12}-\dfrac{3}{4}.\dfrac{5}{6}\)
\(=\dfrac{7}{12}-\dfrac{5}{8}\)
\(=\dfrac{14}{24}-\dfrac{15}{24}\)
\(=\dfrac{-1}{24}\)
d) \(\left(2\dfrac{1}{3}+1\dfrac{3}{4}\right).\dfrac{12}{13}\)
\(=\left(\dfrac{7}{3}+\dfrac{7}{4}\right).\dfrac{12}{13}\)
\(=\left(\dfrac{28}{12}+\dfrac{21}{12}\right).\dfrac{12}{13}\)
\(=\dfrac{49}{12}.\dfrac{12}{13}\)
\(=\dfrac{49}{13}\)
a) \(\frac{1}{12}+\frac{3}{15}+\frac{11}{12}+\frac{1}{71}-\frac{12}{10}=\left(\frac{1}{12}+\frac{11}{12}\right)+\left(\frac{1}{5}-\frac{1}{5}\right)+\frac{1}{71}\)
\(=\frac{12}{12}+0+\frac{1}{71}=1+\frac{1}{71}=1\frac{1}{71}=\frac{72}{71}\)
b) \(\frac{2}{3}-4\left(\frac{1}{2}+\frac{3}{4}\right)=\frac{2}{3}-4.\frac{5}{4}=\frac{2}{3}-5=\frac{2}{3}-\frac{15}{3}=-\frac{13}{3}\)
c) \(\frac{-4}{13}.\frac{3}{17}+\frac{-12}{13}.\frac{4}{7}+\frac{4}{13}=\frac{4}{13}.\frac{-3}{17}+\frac{4}{13}.\frac{-12}{17}+\frac{4}{13}.1\)
\(=\frac{4}{13}\left(\frac{-3}{17}+\frac{-12}{17}+1\right)=\frac{4}{13}\left(\frac{-15}{17}+\frac{17}{17}\right)=\frac{4}{13}.\frac{2}{17}=\frac{8}{221}\)
d) \(\frac{10^3+2.5+5^3}{55}=\frac{1000+10+125}{55}=\frac{1135}{55}=\frac{227}{11}\)
Bài 1
\(=-\frac{21}{60}=-\frac{7}{20}\)
\(b,\left(2-\frac{1}{3}\right)^2+|-\frac{5}{6}|+\frac{-7}{12}-\frac{25}{9}\)
\(=\frac{25}{9}+\frac{5}{6}-\frac{7}{12}-\frac{25}{9}\)
\(=\left(\frac{25}{9}-\frac{25}{9}\right)+\left(\frac{5}{6}-\frac{7}{12}\right)\)
\(=0+\frac{1}{4}=\frac{1}{4}\)
Bài 2
\(a,x+\frac{2}{5}=-\frac{3}{10}\)
\(x=-\frac{3}{10}-\frac{2}{5}\)
\(x=-\frac{3}{10}-\frac{4}{10}\)
\(x=-\frac{7}{10}\)
\(b,|\frac{2}{3}+x|=\frac{5}{7}\)
\(\Rightarrow\orbr{\begin{cases}\frac{2}{3}+x=\frac{5}{7}\\\frac{2}{3}+x=-\frac{5}{7}\end{cases}\Rightarrow\orbr{\begin{cases}x=\frac{5}{7}-\frac{2}{3}\\x=-\frac{5}{7}-\frac{2}{3}\end{cases}\Rightarrow}\orbr{\begin{cases}x=\frac{1}{21}\\x=-\frac{29}{21}\end{cases}}}\)
== chắc trog quá trình lm lỡ xóa đó
\(a,-\frac{3}{4}.\frac{7}{15}\)
\(=-\frac{21}{60}=-\frac{7}{20}\)
với lại bài trên mk tính nhẩm ko bấm máy sai == sửa giúp
Ta có:
\(\left(\frac{3}{15}-\frac{12}{16}-\frac{7}{14}\right)\cdot\left(-1\frac{5}{14}\right)\)
\(=\left(\frac{1}{5}-\frac{3}{4}-\frac{1}{2}\right)\cdot\left(-\frac{19}{14}\right)\)
\(=\left(\frac{4}{20}-\frac{15}{20}-\frac{10}{20}\right)\cdot\left(-\frac{19}{14}\right)\)
\(=\left(-\frac{21}{20}\right)\cdot\left(-\frac{19}{14}\right)\)
\(=\frac{57}{40}\)
a) Ta có:
\(3^3+3\cdot\left(-\frac{1}{2}\right)^3-0,75\)
\(=27+3\cdot\left(-\frac{1}{8}\right)-\frac{3}{4}\)
\(=27+\left(-\frac{3}{8}\right)-\frac{6}{8}\)
\(=27+\left[\left(-\frac{3}{8}\right)+\left(-\frac{6}{8}\right)\right]\)
\(=27+\left(-\frac{9}{8}\right)\)
\(=\frac{207}{8}\)