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\(a)=\frac{7}{25}+\frac{4}{13}-\frac{5}{2}+\frac{18}{25}-\frac{17}{13}\)
\(=1-1-\frac{5}{2}\)
\(=-\frac{5}{2}\)
3,
a) (−23+37):45+(−13+47):45
= \(-\frac{5}{21}:\frac{4}{5}+\frac{5}{21}:\frac{4}{5}\)
= \(\left(-\frac{5}{21}+\frac{5}{21}\right):\frac{4}{5}\)
= \(0:\frac{4}{5}=0\)
2,
a) \(\frac{-3}{4}\).\(\frac{12}{-5}\).(\(\frac{-25}{6}\))
= \(\frac{-3.4.3.\left(-5\right).5}{4.\left(-5\right).3.3}\)
= \(-5\)
b) (−2).\(\frac{-38}{21}\).\(\frac{-7}{4}\).(\(\frac{-3}{8}\))
= \(\frac{-2.\left(-38\right)\left(-7\right)\left(-3\right)}{\left(-7\right)\left(-3\right)\left(-2\right)\left(-2\right).8}\)
= \(\frac{19}{8}\)
c) (\(\frac{11}{12}:\frac{33}{16}\)).\(\frac{3}{5}\)
= \(\left(\frac{11}{12}.\frac{16}{33}\right).\frac{3}{5}\)
= \(\frac{4}{9}.\frac{3}{5}\)
= \(\frac{4}{15}\)
d) \(\frac{7}{23}\left[\left(\frac{-8}{6}\right)-\frac{45}{18}\right]\)
= \(\frac{7}{23}.\left(\frac{-41}{10}\right)\)
= \(\frac{-287}{203}\)
3. Tính:
a) (\(\frac{-2}{3}+\frac{3}{7}\)):\(\frac{4}{5}\)+(\(\frac{-1}{3}+\frac{4}{7}\)):\(\frac{4}{5}\)
= (\(\frac{-2}{3}+\frac{3}{7}\)\(+\)\(\frac{-1}{3}+\frac{4}{7}\)) : \(\frac{4}{5}\)
= 0 : \(\frac{4}{5}\)
= 0
b) \(\frac{5}{9}\):(\(\frac{1}{11}-\frac{5}{22}\))+\(\frac{5}{9}\):(\(\frac{1}{15}-\frac{2}{3}\))
= \(\frac{5}{9}\): \(\frac{-3}{22}\)+ \(\frac{5}{9}\): \(\frac{-3}{5}\)
= \(\frac{5}{9}\): \(\frac{-81}{110}\)
= \(\frac{-550}{729}\)
Nhiều thế :( Làm 1,2 câu thôi nhé
a) \(\frac{1}{3}+\frac{1}{4}=\frac{4}{12}+\frac{3}{12}=\frac{7}{12}\) (bị mất nét nhưng vẫn nhìn ra là số 12 nhỉ?)
b) \(\frac{-2}{5}+\frac{7}{21}=\frac{-42}{105}+\frac{35}{105}=\frac{-7}{105}=\frac{-1}{15}\)
a) Ta có: \(\frac{3}{8}-\frac{1}{5}+\frac{3}{40}\)
\(=\frac{15}{40}-\frac{8}{40}+\frac{3}{40}\)
\(=\frac{10}{40}=\frac{1}{4}\)
b) Ta có: \(\frac{21}{4}\cdot\frac{3}{8}+\frac{43}{4}\cdot\frac{3}{8}-4\cdot\frac{1}{2}\)
\(=\frac{3}{8}\left(\frac{21}{4}+\frac{43}{4}\right)-2\)
\(=\frac{3}{8}\cdot16-2\)
\(=6-2=4\)
c) Ta có: \(\frac{-5}{9}+\frac{7}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{8}{15}\)
\(=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{7}{15}+\frac{8}{15}\right)+\frac{-2}{11}\)
\(=-1+1+\frac{-2}{11}\)
\(=\frac{-2}{11}\)
d) Ta có: \(125\%\cdot\left(\frac{-1}{2}\right)^2:\left(1\frac{5}{6}-1.5\right)+2016^0\)
\(=\frac{5}{4}\cdot\frac{1}{4}:\left(\frac{11}{6}-\frac{3}{2}\right)+1\)
\(=\frac{5}{16}\cdot3+1\)
\(=\frac{15}{16}+\frac{16}{16}=\frac{31}{16}\)
a) \(x-\frac{2}{5}=\frac{5}{7}\)
\(x=\frac{2}{5}+\frac{5}{7}\)
\(x=\frac{14}{35}+\frac{25}{35}=\frac{39}{35}\)
b)
\(\frac{-2}{5}x=\frac{4}{15}\)
\(x=\frac{4}{15}:-\frac{2}{5}\)
\(x=\frac{4}{15}\cdot-\frac{5}{2}=-\frac{2}{3}\)
c) \(2x\left(x-\frac{1}{7}\right)=2x^2-\frac{2x}{7}\)
d) \(\frac{1}{2}+\frac{3}{4}x=\frac{1}{4}\)
\(\frac{3}{4}x=\frac{1}{4}-\frac{1}{2}\)
\(\frac{3}{4}x=-\frac{1}{4}\)
\(x=-\frac{1}{4}\cdot\frac{4}{3}=-\frac{1}{3}\)
f) \(\frac{11}{12}-\left(\frac{2}{5}+x\right)=\frac{2}{5}\)
\(\frac{2}{5}+x=\frac{11}{12}-\frac{2}{5}=\frac{31}{60}\)
\(x=\frac{31}{60}-\frac{2}{5}=\frac{7}{60}\)
a;-3/4
b;-0.9
c;7/6
d;-1/50