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c) Ta có: \(\dfrac{3}{5}+\dfrac{-5}{20}+\dfrac{30}{75}+\dfrac{-7}{4}\)
\(=\dfrac{3}{5}+\dfrac{2}{5}+\dfrac{-1}{4}+\dfrac{-7}{4}\)
\(=1-2=-1\)
Giải:
a)-1/12+4/3=-1/12+16/12=15/12=5/4
b)(-4/14-3/15)-(1/5-20/35-(-1)).7
=-17/35-22/35.7
=-17/35-22/5
=-171/35
c)3/5+-5/20+30/75+-7/4
=3/5+-1/4+2/5+-7/4
=(3/5+2/5)+(-1/4+-7/4)
=1+-2
=-1
d)5/6.-12/14+7/13
=-5/7+7/13
=-16/91
e)2/-9-5/-36-1/4
=-1/12-1/4
=-1/3
f)2/23+-5/12+7/18+21/23+-7/12
=(2/23+21/23)+(-5/12+-7/12)+7/18
=1+-1+7/18
=7/18
Lời giải:
a.
$-18: \frac{3}{5}=-18.\frac{5}{3}=-30$
b.
$\frac{3}{4}:(-9)=\frac{3}{4}.\frac{-1}{9}=\frac{-1}{12}$
c.
$\frac{13}{20}-\frac{6}{7}: \frac{10}{21}=\frac{13}{20}-\frac{6}{7}.\frac{21}{10}$
$=\frac{13}{20}-\frac{9}{5}=\frac{13}{20}-\frac{36}{20}=\frac{-23}{20}$
d.
$\frac{-21}{5}: (\frac{7}{3}.\frac{7}{5})=\frac{-21}{5}: \frac{49}{15}$
$=\frac{-21}{5}.\frac{15}{49}=\frac{-9}{7}$
e.
$(\frac{-2}{5}+\frac{1}{4}): (1-\frac{2}{5})$
$=\frac{-3}{20}: \frac{3}{5}=\frac{-1}{4}$
a; - \(\dfrac{10}{13}\) + \(\dfrac{5}{17}\) - \(\dfrac{3}{13}\) + \(\dfrac{12}{17}\) - \(\dfrac{11}{20}\)
= - (\(\dfrac{10}{13}\) + \(\dfrac{3}{13}\)) + (\(\dfrac{5}{17}\) + \(\dfrac{12}{17}\)) - \(\dfrac{11}{20}\)
= - 1 + 1 - \(\dfrac{11}{20}\)
= 0 - \(\dfrac{11}{20}\)
= - \(\dfrac{11}{20}\)
b; \(\dfrac{3}{4}\) + \(\dfrac{-5}{6}\) - \(\dfrac{11}{-12}\)
= \(\dfrac{9}{12}\) - \(\dfrac{10}{12}\) + \(\dfrac{11}{12}\)
= \(\dfrac{10}{12}\)
= \(\dfrac{5}{6}\)
c; [13.\(\dfrac{4}{9}\) + 2.\(\dfrac{1}{9}\)] - 3.\(\dfrac{4}{9}\)
= [\(\dfrac{52}{9}\) + \(\dfrac{2}{9}\)] - \(\dfrac{4}{3}\)
= \(\dfrac{54}{9}\) - \(\dfrac{4}{3}\)
= \(\dfrac{14}{3}\)
\(\frac{3}{5}-\frac{-7}{10}-\frac{13}{-20}=\frac{3}{5}+\frac{7}{10}+\frac{13}{20}=\frac{12}{20}+\frac{14}{20}+\frac{13}{20}=\frac{39}{20}\)
\(\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}-\frac{-1}{6}=\frac{1}{2}+\frac{-1}{3}+\frac{1}{4}+\frac{1}{6}=\frac{6+(-4)+3+2}{12}=\frac{7}{12}\)
\(\frac{9}{4}.\frac{8}{27}.\frac{5}{7}=\frac{9.8.5}{4.27.7}=\frac{1.2.5}{1.3.7}=\frac{10}{21}\)
\(\frac{2}{5}.(\frac{2}{3}-\frac{1}{4})+\frac{1}{2}=\frac{2}{5}.(\frac{8}{12}-\frac{3}{12})+\frac{1}{2}=\frac{2}{5}.\frac{5}{12}+\frac{1}{2}=\frac{1}{6}+\frac{1}{2}=\frac{1}{6}+\frac{3}{6}=\frac{4}{6}=\frac{2}{3}\)
\((\frac{1}{3}-\frac{1}{6}):(\frac{1}{3}+\frac{1}{6})=(\frac{2}{6}-\frac{1}{6}):(\frac{2}{6}+\frac{1}{6})=\frac{1}{6}:\frac{3}{6}=\frac{1}{6}.\frac{6}{3}=\frac{1.6}{6.3}=\frac{1.1}{1.3}=\frac{1}{3}\)
Hok tốt
a; - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) - (- \(\dfrac{1}{6}\)) + (- \(\dfrac{2}{5}\))
= - \(\dfrac{2}{3}\) + \(\dfrac{3}{4}\) + \(\dfrac{1}{6}\) - \(\dfrac{2}{5}\)
= \(-\dfrac{40}{60}\) + \(\dfrac{45}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{5}{60}\) + \(\dfrac{10}{60}\) - \(\dfrac{24}{60}\)
= \(\dfrac{15}{60}\) - \(\dfrac{24}{60}\)
= - \(\dfrac{3}{20}\)
b; (- \(\dfrac{2}{3}\)) + (- \(\dfrac{1}{5}\)) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) - \(\dfrac{-7}{10}\)
= - \(\dfrac{2}{3}\) - \(\dfrac{1}{5}\) + \(\dfrac{3}{4}\) - \(\dfrac{5}{6}\) + \(\dfrac{7}{10}\)
= - \(\dfrac{40}{60}\) - \(\dfrac{12}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{52}{60}\) + \(\dfrac{45}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{7}{60}\) - \(\dfrac{50}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{57}{60}\) + \(\dfrac{42}{60}\)
= - \(\dfrac{1}{4}\)
\(\dfrac{1}{3}+\left(\dfrac{-1}{3.7}+\dfrac{-3.4}{3.7}\right)+\dfrac{5}{7}+\dfrac{1}{3.2}+\dfrac{1}{2.7}+\dfrac{-1}{2}=\dfrac{1}{3}-\dfrac{1}{3}.\dfrac{1}{7}-\dfrac{4}{7}+\dfrac{5}{7}+\dfrac{1}{3}.\dfrac{1}{2}+\dfrac{1}{2}.\dfrac{1}{7}-\dfrac{1}{2}=\left[\dfrac{1}{3}.\dfrac{6}{7}+\dfrac{1}{7}.\dfrac{3}{2}-\dfrac{1}{3}\right]=\dfrac{1}{3}.\dfrac{-1}{7}+\dfrac{1}{7}.\dfrac{3}{2}=\dfrac{1}{7}\left\{\dfrac{-2}{6}+\dfrac{9}{6}\right\}.=\dfrac{1}{6}\)
cau b:triet. het bang 1/7