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a ) \(\frac{2}{5}+\frac{-1}{6}-\frac{3}{4}-\frac{-2}{3}\)
= \(\frac{2}{5}+\frac{-1}{6}+\frac{-3}{4}+\frac{2}{3}\)
= \(\left(\frac{2}{5}+\frac{-3}{4}\right)+\left(\frac{-1}{6}+\frac{2}{3}\right)\)
= \(\left(\frac{8}{20}+\frac{-15}{20}\right)+\left(\frac{-1}{6}+\frac{4}{6}\right)\)
= \(\left(\frac{8+\left(-15\right)}{20}\right)+\left(\frac{\left(-1\right)+4}{6}\right)\)
= \(\frac{-7}{20}+\frac{1}{2}\)
= \(\frac{-7}{20}+\frac{10}{20}=\frac{\left(7\right)+10}{20}=\frac{3}{20}\)
tk mk nha
đang âm rất nhiều rồi , giúp nha !!!!!
a)\(\left(4\frac{5}{37}-3\frac45+8\frac{15}{29}\right)-\left(3\frac{5}{57}-6\frac{14}{29}\right)\)
=\(4\frac{5}{37}-3\frac45+8\frac{15}{29}-3\frac{5}{37}+6\frac{14}{29}\)
=\(\left(4\frac{5}{37}-3\frac{5}{37}\right)+\left(8\frac{15}{29}+6\frac{14}{29}\right)-3\frac45\)
=\(\left\lbrack\left(4-3\right)+\left(\frac{5}{37}-\frac{5}{37}\right)\right\rbrack+\left\lbrack\left(8+6\right)+\left(\frac{15}{29}\right.\right.\)+\(\frac{14}{29})\) -\(\frac{19}{5}\)
=\(1+0+14+1-\frac{19}{5}\)
=\(15+1-\frac{19}{5}\)
=\(16-\frac{19}{5}\)
=\(\frac{80}{5}-\frac{19}{5}\)
=\(\frac{61}{5}\)
a) \(\frac{2}{3}=\frac{8}{12}\) ; \(\frac{1}{4}=\frac{3}{12}\)
mà 8 > 3 ⇒ \(\frac{8}{12}>\frac{3}{12}\)⇒\(\frac{2}{3}>\frac{1}{4}\)
b) \(\frac{7}{10}\) và \(\frac{7}{8}\); mà 10 > 8 ⇒ \(\frac{7}{10}< \frac{7}{8}\)
c) \(\frac{6}{7}=\frac{30}{35}\); \(\frac{3}{5}=\frac{21}{35}\)
mà 30 > 21 ⇒ \(\frac{30}{35}>\frac{21}{35}\)⇒\(\frac{6}{7}>\frac{3}{5}\)
d) \(\frac{14}{21}=\frac{2}{3}\); \(\frac{60}{72}=\frac{5}{6}\)
\(\frac{2}{3}=\frac{4}{6}\) ⇒ \(\frac{2}{3}< \frac{5}{6}\)⇒ \(\frac{14}{21}< \frac{60}{72}\)
e) \(\frac{38}{133}=\frac{2}{7}\); \(\frac{129}{344}=\frac{3}{8}\)
\(\frac{2}{7}=\frac{16}{56}\) ; \(\frac{3}{8}=\frac{21}{56}\) mà 16<21 ⇒ \(\frac{16}{56}< \frac{21}{56}\)⇒ \(\frac{38}{133}< \frac{129}{344}\)
f) \(\frac{11}{54}=\frac{22}{108}\)và \(\frac{22}{37}\) mà 108 > 37 ⇒ \(\frac{22}{108}< \frac{22}{37}\)⇒ \(\frac{11}{54}< \frac{22}{37}\)
\(\frac{-2}{5}+\frac{3}{10}-\frac{3}{5}+\frac{7}{10}-1\frac{1}{2}\)
=\(\frac{-4}{10}+\frac{3}{10}-\frac{6}{10}+\frac{7}{10}-\frac{15}{10}\)
=\(\frac{-15}{10}\)
=\(\frac{-3}{2}\)
\(\frac{-2}{5}+\frac{3}{10}-\frac{3}{5}+\frac{7}{10}-1\frac{1}{2}\)
\(=\frac{-4}{10}+\frac{3}{10}-\frac{6}{10}+\frac{7}{10}-\frac{15}{10}\)
\(=\frac{-3}{2}\)
a) \(A=\frac{-7}{813}+496.\left(\frac{-7}{813}\right)+\left(\frac{-7}{813}\right).316\)
\(=\frac{-7}{813}.\left(1+496+316\right)\)
\(=\frac{-7}{813}.813\)
\(=-7\)
b) \(B=\frac{-9}{10}.\frac{5}{14}+\frac{1}{10}.\left(\frac{-9}{2}\right)+\frac{1}{7}.\left(\frac{-9}{10}\right)\)
\(=\frac{-9}{10}.\left(\frac{5}{14}+\frac{1}{2}+\frac{1}{7}\right)\)
\(=\frac{-9}{10}.1\)
\(=\frac{-9}{10}\)
a
\(5\frac{4}{7}:x+=13\)
\(\frac{39}{7}:x=13\)
\(x=\frac{39}{7}:13\)
\(x=\frac{3}{7}\)
\(\frac{4}{7}x=\frac{9}{8}-0,125\)
\(\frac{4}{7}x=1\)
\(x=1:\frac{4}{7}\)
\(x=\frac{7}{4}=1\frac{3}{4}\)
\(\frac{\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{7}{5}+1+\frac{7}{17}}\div\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{\frac{3}{5}+\frac{3}{7}+\frac{3}{11}}{\frac{7}{5}+\frac{7}{7}+\frac{7}{17}}\div\frac{\frac{2}{6}-\frac{2}{8}+\frac{2}{10}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(=\frac{3\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}{7\left(\frac{1}{5}+\frac{1}{7}+\frac{1}{11}\right)}\div\frac{2\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(=\frac{3}{7}\div\frac{2}{7}\)
\(=\frac{3}{7}\times\frac{7}{2}=\frac{3}{2}\)
\(\frac{130}{120}\)
\(\frac{130}{120}\)=\(\frac{13}{12}\)