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a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
b; 0,5 + \(\dfrac{1}{3}\) + 0,4 + \(\dfrac{5}{7}\) + \(\dfrac{1}{6}\) - \(\dfrac{4}{35}\)
= (\(\dfrac{1}{3}\)+ \(\dfrac{1}{6}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{5}{7}\)- \(\dfrac{4}{35}\)+ \(\dfrac{2}{5}\))
= ( \(\dfrac{1}{2}\) + \(\dfrac{1}{2}\)) + (\(\dfrac{3}{5}\) + \(\dfrac{2}{5}\))
= 1 + 1
= 2
a) \(-1\frac{5}{7}.15+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{2}{3}-\frac{4}{5}+\frac{1}{7}\right)\)
\(=1\frac{5}{7}.\left(-15\right)+\frac{2}{7}.\left(-15\right)+\left(-105\right).\left(\frac{70}{105}-\frac{84}{105}+\frac{15}{105}\right)\)
\(=\left(-15\right)\left(1+\frac{5}{7}+\frac{2}{7}\right)+\left(-105\right).\frac{1}{105}\)
\(=-30-1=-31\)
b) \(\frac{2}{3}+\frac{3}{4}.\left(-\frac{4}{9}\right)\)
= \(=\frac{2}{3}+\frac{3.\left(-4\right)}{4.9}=\frac{2}{3}+\frac{-1}{3}=\frac{1}{3}\)
c) \(\left(\frac{3}{4}-0,2\right).\left(0,4-\frac{4}{5}\right)\)
\(=\left(\frac{3}{4}-\frac{1}{5}\right).\left(\frac{2}{5}-\frac{4}{5}\right)=\frac{11}{20}.\left(\frac{-2}{5}\right)=\frac{11.\left(-1\right).2}{2.10.5}=\frac{-11}{50}\)
a: \(0.4\cdot\sqrt{0.25-\sqrt{\dfrac{1}{4}}}=0.4\cdot\sqrt{0.25-0.5}\)(đề này sai rồi bạn)
b: \(\dfrac{3}{2}+2\left(x-1\right)=-5\dfrac{1}{2}\)
\(\Leftrightarrow2\left(x-1\right)=\dfrac{-11}{2}-\dfrac{3}{2}=-7\)
\(\Leftrightarrow x-1=\dfrac{-7}{2}\)
hay \(x=-\dfrac{5}{2}\)
\(a,0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\dfrac{5}{6}+\dfrac{39}{35}-\dfrac{1}{6}-\dfrac{4}{35}\\ =\left(\dfrac{5}{6}-\dfrac{1}{6}\right)+\left(\dfrac{39}{35}-\dfrac{4}{35}\right)\\ =\dfrac{2}{3}+1\\ =\dfrac{4}{3}.\)
\(b,\left(3-\dfrac{1}{4}+\dfrac{2}{3}\right)-\left(5+\dfrac{1}{3}-\dfrac{6}{5}\right)-\left(-6-\dfrac{7}{4}+\dfrac{3}{2}\right)\\ =3-\dfrac{1}{4}+\dfrac{2}{3}-5-\dfrac{1}{3}+\dfrac{6}{5}+6+\dfrac{7}{4}-\dfrac{3}{2}\\ =\left(3-5+6\right)+\left(-\dfrac{1}{4}+\dfrac{7}{4}\right)+\left(\dfrac{2}{3}-\dfrac{1}{3}\right)+\left(\dfrac{6}{5}+\dfrac{7}{4}\right)\\ =4-\dfrac{3}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{5}{2}+\dfrac{1}{3}+\dfrac{59}{20}\\ =\dfrac{17}{6}+\dfrac{59}{20}\\ =\dfrac{347}{60}.\)
\(c,\dfrac{1}{3}-\dfrac{3}{4}-\left(-\dfrac{3}{5}\right)+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\dfrac{1}{3}+\dfrac{3}{4}+\dfrac{3}{5}+\dfrac{1}{64}-\dfrac{2}{9}-\dfrac{1}{36}+\dfrac{1}{15}\\ =\left(\dfrac{1}{3}-\dfrac{2}{9}\right)+\left(\dfrac{3}{4}-\dfrac{1}{36}\right)+\left(\dfrac{3}{5}+\dfrac{1}{15}\right)+\dfrac{1}{64}\\ =\dfrac{1}{9}+\dfrac{13}{18}+\dfrac{2}{3}+\dfrac{1}{64}\\ =\dfrac{3}{2}+\dfrac{1}{64}\\ =\dfrac{65}{64}.\)
a) \(\frac{4}{5}\)✖\(\left(\frac{7}{2}+\frac{1}{4}\right)^2\)
= \(\frac{4}{5}\)✖ \((\frac{15}{4})^2\)
= \(\frac{4}{5}\)✖ \(\frac{225}{16}\)
= \(\frac{1}{1}\times\frac{45}{4}\)
= \(\frac{45}{4}\)
a)\(3-\left(\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
=\(3-\frac{1}{4}-\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
=\(\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{3}-\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)
=\(-8+\frac{3}{2}-1-\frac{3}{10}\)
=\(\left(-8-1\right)+\left(\frac{3}{2}-\frac{3}{10}\right)\)
=-9+\(\frac{6}{5}\)
=\(\frac{-39}{5}\)
(7 + 3 1/4 - 3/5) + (0,4 - 5) - (4 1/4 - 1)
(7 + 13/4 - 3/5) + (2/5 - 5) - (17/4 - 1)
= 7 + 13/4 - 3/5 + 2/5 - 5 - 17/4 + 1
= (7 - 5 + 1) + (13/4 - 17/4) + (-3/5 + 2/5)
= 3 - 1 - 1/5
= 2 - 1/5
= 9/5
(7 + 3\(\dfrac{1}{4}\) - \(\dfrac{3}{5}\)) + (0,4 - 5) - ( 4\(\dfrac{1}{4}\) - 1)
= 7 + 3 + \(\dfrac{1}{4}\) - 0,6 - 4,6 - 4 - \(\dfrac{1}{4}\) + 1
= ( 7 + 3 - 4 + 1) + (\(\dfrac{1}{4}\) - \(\dfrac{1}{4}\)) - ( 0,6 + 4,6)
= 7 - 5,2
= 1,8