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2 tháng 7 2018

\(\left(-0,6\cdot x-\frac{1}{2}\right)\cdot\frac{3}{4}-9-\left(-1\right)=\frac{1}{3}\)

\(\left(-0,6\cdot x-\frac{1}{2}\right)\cdot\frac{3}{4}-10=\frac{1}{3}\)

\(\left(-0,6\cdot x-\frac{1}{2}\right)\cdot\frac{3}{4}=\frac{31}{3}\)

\(-0,6\cdot x-\frac{1}{2}=\frac{124}{9}\)

\(-0,6\cdot x=\frac{257}{18}\)

\(x=\frac{1285}{54}\)

2 tháng 7 2018

\(\left(-0,6.x-\frac{1}{2}\right).\frac{3}{4}-9-\left(-1\right)=\frac{1}{3}\)

\(\frac{-9}{20}x-\frac{3}{8}-\left(9-1\right)=\frac{1}{3}\)

\(\frac{-9}{20}x-\frac{3}{8}-8=\frac{1}{3}\)

\(\frac{-9}{20}x-\left(\frac{3}{8}+8\right)=\frac{1}{3}\)

\(\frac{-9}{20}x-\frac{67}{8}=\frac{1}{3}\)

\(\frac{-9}{20}x=\frac{1}{3}+\frac{67}{8}\)

\(\frac{-9}{20}x=\frac{8}{24}+\frac{201}{24}\)

\(\frac{-9}{20}x=\frac{209}{24}\)

\(x=\frac{209}{24}.\frac{-20}{9}\)

\(x=\frac{-1045}{54}\)

15 tháng 12 2022

a: \(=\dfrac{\left(\dfrac{1}{2}:\dfrac{1}{2}-\dfrac{1}{4}:\dfrac{1}{4}+\dfrac{1}{8}:\dfrac{1}{8}-\dfrac{1}{10}:\dfrac{1}{10}\right)}{1+2+3+...+2008}\)

=0

c: =8,1*5/3*1875+13,5*625

=13,5(1875+625)

=13,5*2500

=33750

21 tháng 9 2017

\(A=\dfrac{\sqrt{\dfrac{9}{4}-3^{-1}+2018^0}}{25\%+1\dfrac{1}{4}-1,3}-\dfrac{\left(-\dfrac{1}{2}\right)^2-\sqrt{\dfrac{4}{9}}+0,4}{0,6-\dfrac{2}{3}.\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)

\(A=\dfrac{\sqrt{\dfrac{9}{4}-\dfrac{1}{3}+1}}{\dfrac{1}{4}+\dfrac{5}{4}-\dfrac{13}{10}}-\dfrac{\dfrac{1}{4}-\dfrac{2}{3}+\dfrac{2}{5}}{\dfrac{3}{5}-\dfrac{2}{3}\left(-\dfrac{1}{4}-\dfrac{1}{2}\right)}\)

\(A=\dfrac{\sqrt{\dfrac{35}{12}}}{\dfrac{1}{5}}-\dfrac{-\dfrac{1}{60}}{\dfrac{11}{10}}\)

\(A=\dfrac{5\sqrt{105}}{6}+\dfrac{11}{66}\)

\(A=\dfrac{55\sqrt{105}}{66}+\dfrac{11}{66}\)

\(A=\dfrac{55\sqrt{105}+11}{66}\)

7 tháng 8 2017

mọi người giải giúp mình với ạ

30 tháng 4 2017

Bài 1:

a)

\(A=\dfrac{-5}{6}\cdot\dfrac{3}{10}\\ =\dfrac{\left(-5\right)\cdot3}{6\cdot10}\\ =\dfrac{-1}{4}\)

b)

\(B=\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{12}\\ =\dfrac{4}{12}-\dfrac{3}{12}+\dfrac{1}{12}\\ =\dfrac{4-3+1}{12}\\ =\dfrac{1}{6}\)

30 tháng 4 2017

Bài 2:

\(A=\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{4}{5}-\dfrac{14}{5}\right|:\dfrac{8}{3}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+\left|\dfrac{-10}{5}\right|\cdot\dfrac{3}{8}\\ =\left(\dfrac{-1}{5}\right)\cdot\dfrac{15}{4}+2\cdot\dfrac{3}{8}\\ =\dfrac{-3}{4}+\dfrac{3}{4}\\ =0\)

\(B=\left(\dfrac{-1}{2}\right)^2:1\dfrac{3}{8}+25\%\cdot\dfrac{3}{11}\\ =\left(\dfrac{-1}{2}\right)^2:\dfrac{11}{8}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{1}{4}\cdot\dfrac{8}{11}+\dfrac{3}{4}\cdot\dfrac{3}{11}\\ =\dfrac{8}{44}+\dfrac{9}{44}\\ =\dfrac{17}{44}\)

\(C=\dfrac{-8}{5}+0,6+\left|\dfrac{-1}{2}\right|+\dfrac{1}{2}\\ =\dfrac{-8}{5}+\dfrac{3}{5}+\dfrac{1}{2}+\dfrac{1}{2}\\ =\left(\dfrac{-8}{5}+\dfrac{3}{8}\right)+\left(\dfrac{1}{2}+\dfrac{1}{2}\right)\\ =\left(-1\right)+1\\ =0\)

\(D=\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}:\dfrac{13}{11}+1\dfrac{5}{9}\\ =\dfrac{-5}{9}\cdot\dfrac{2}{13}+\dfrac{-5}{9}\cdot\dfrac{11}{13}+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot\left(\dfrac{2}{13}+\dfrac{11}{13}\right)+\dfrac{14}{9}\\ =\dfrac{-5}{9}\cdot1+\dfrac{14}{9}\\ =\dfrac{-5}{9}+\dfrac{14}{9}\\ =1\)

\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)

\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)

\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)

\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)

\(\Rightarrow B=-\frac{113}{960}\)

\(C=0\)

\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)

\(\Rightarrow D=1\)

11 tháng 8 2019

D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)

=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)

=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)

=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)

=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)

=\(\frac{1}{99}-1-\frac{1}{99}\)

=1