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\(7\frac{5}{9}-\left(3\frac{5}{9}-2\frac{3}{4}\right)=7\frac{5}{9}-3\frac{5}{9}+2\frac{3}{4}=4+2\frac{3}{4}=6\frac{3}{4}\)
a,\(\frac{5}{6}:\frac{-7}{12}\) b, \(\frac{-21}{24}:\frac{-14}{8}\)
d,\(\frac{4}{5}:\frac{-8}{15}\)e,\(\frac{3}{5}+\frac{-7}{4}\)
g,\(\frac{5}{12}-\frac{-7}{12}\)h,\(\frac{-15}{16}.\frac{8}{-25}\)
\(A=\frac{-7}{21}+\left(1+\frac{1}{3}\right)\)
\(A=\frac{-7}{21}+1+\frac{1}{3}\)
\(A=\left(\frac{-7}{21}+\frac{1}{3}\right)+1\)
\(A=0+1\\ A=1\)
\(B=\frac{2}{15}+\left(\frac{5}{9}+-\frac{6}{9}\right)\)
\(B=\frac{2}{15}+\frac{-1}{9}\)
\(B=\frac{1}{45}\)
\(C=\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(C=\frac{-1}{5}+\left(\frac{3}{12}+\frac{-3}{4}\right)\)
\(C=\frac{-1}{5}+\frac{-1}{2}\)
\(C=\frac{-7}{10}\)
\(A=\frac{-7}{21}+\left(1+\frac{1}{3}\right)\)
\(A=\frac{-1}{3}+1+\frac{1}{3}\)
\(A=1\)
\(B=\frac{2}{15}+\left(\frac{5}{9}+\frac{-6}{9}\right)\)
\(B=\frac{2}{15}+\frac{5}{9}+\frac{-6}{9}\)
\(B=\frac{2}{15}+\frac{5}{9}+\frac{-10}{15}\)
\(B=\frac{-8}{15}+\frac{5}{9}\)
\(B=\frac{1}{45}\)
\(C=\left(\frac{-1}{5}+\frac{3}{12}\right)+\frac{-3}{4}\)
\(C=\frac{-1}{5}+\frac{3}{4}+\frac{-3}{4}\)
\(C=\frac{-1}{5}\)
A =\(\frac{2}{7}\times\frac{7}{15}-\frac{2}{3}\times\frac{5}{27}\) =\(\frac{14}{105}-\frac{10}{81}\)=\(\frac{2}{15}-\frac{10}{81}\)=\(\frac{162}{1215}-\frac{150}{1215}\)=\(\frac{12}{1215}\)
\(\frac{-4}{9}\times\frac{7}{15}+\frac{4}{9}\times\frac{5}{27}\)=\(\frac{-28}{135}+\frac{20}{243}\)=\(\frac{-6804}{32805}+\frac{2700}{32805}\)=\(\frac{-4104}{32805}\)=\(\frac{-152}{1215}\)
A=\(\frac{2}{7}\times\frac{7}{15}-\frac{2}{3}\times\frac{5}{27}\)chia cho\(\frac{-4}{9}\times\frac{7}{15}+\frac{4}{9}\times\frac{5}{27}\)
= \(\frac{12}{1215}:\frac{-152}{1215}\)
=\(\frac{12}{1215}\times\frac{1215}{-152}\)
=\(\frac{14580}{-184680}\)
\(\frac{14580}{-184680}\)rút gọn bằng\(\frac{-3}{38}\)
a/ A= 1-3+5-7+9-11+......+97-99
= -2+(-2)+(-2)+......+(-2)
= (-2).25=-50
b/B=-1-2-3-4-...-100
=-(1+2+3+4+...+100)
=-5050
c/C=1-2+3-4+5-6+......+99-100
= -1+(-1)+(-1)+.............+(-1)
=(-1).50=-50
d/D=1-2-3+4+5-6-7+8+9-....+94-95
= (1-2-3+4)+(5-6-7+8)+.......+(92-93-94+95)
= 0+0+0+...+0=0
a) \(\left(\dfrac{1}{6}+\dfrac{5}{9}\right)+\dfrac{4}{9}\)
\(=\dfrac{1}{6}+\dfrac{5}{9}+\dfrac{4}{9}\)
\(=\dfrac{1}{6}+1\)
\(=\dfrac{7}{6}\)
b) \(\dfrac{3}{17}+\left(\dfrac{14}{17}-\dfrac{2}{3}\right)\)
\(=\dfrac{3}{17}+\dfrac{14}{17}-\dfrac{2}{3}\)
\(=1-\dfrac{2}{3}\)
\(=\dfrac{1}{3}\)
c) \(\left(\dfrac{3}{2}-\dfrac{2}{3}\right)+\dfrac{7}{6}\)
\(=\left(\dfrac{9}{6}-\dfrac{4}{6}\right)+\dfrac{7}{6}\)
\(=\dfrac{13}{6}+\dfrac{7}{6}\)
\(=\dfrac{20}{6}\)
\(7\frac{5}{9}-\left(3\frac{5}{9}-2\frac{3}{4}\right)=7\frac{5}{9}-3\frac{5}{9}+2\frac{3}{4}=4+2\frac{3}{4}=6\frac{3}{4}\)