Tính:
a) (-173) [214+(-25)]+214(173+25)+(-25)(-13)
b)1-3-5+7+9-11-13+15+...+2009-2011-2013+2015
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1) Ta có: \(\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{7}{25}\cdot\dfrac{5}{7}\right)\)
\(=\left(\dfrac{3}{4}\cdot\dfrac{5}{97}+\dfrac{1}{9}\cdot\dfrac{13}{47}\right)\cdot\left(\dfrac{1}{5}-\dfrac{1}{5}\right)\)
=0
2) Ta có: \(\dfrac{8}{17}\cdot\dfrac{4}{15}+\dfrac{8}{17}\cdot\dfrac{22}{15}-\dfrac{8}{15}\cdot\dfrac{9}{17}\)
\(=\dfrac{8}{17}\left(\dfrac{4}{15}+\dfrac{22}{15}-\dfrac{9}{15}\right)\)
\(=\dfrac{8}{17}\cdot\dfrac{15}{15}=\dfrac{8}{17}\)
3) Ta có: \(\dfrac{2021}{2}\cdot\dfrac{1}{3}+\dfrac{4042}{4}\cdot\dfrac{1}{5}+\dfrac{6063}{3}\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)+2021\cdot\dfrac{22}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{8}{15}+\dfrac{2021}{2}\cdot\dfrac{44}{15}\)
\(=\dfrac{2021}{2}\cdot\dfrac{52}{15}\)
\(=\dfrac{52546}{15}\)
4) Ta có: \(\dfrac{4}{7}\cdot\dfrac{2}{13}+\dfrac{8}{13}:\dfrac{7}{4}+\dfrac{4}{7}:\dfrac{13}{2}+\dfrac{4}{7}\cdot\dfrac{1}{13}\)
\(=\dfrac{4}{7}\left(\dfrac{2}{13}+\dfrac{8}{13}+\dfrac{2}{13}+\dfrac{1}{13}\right)\)
\(=\dfrac{4}{7}\)
a)
\(\begin{array}{l}A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\left( { - 4,6} \right)\\A = \frac{5}{{11}}.\left( {\frac{{ - 3}}{{23}}} \right).\frac{{11}}{5}.\frac{{ - 23}}{5}\\A = \frac{{5.\left( { - 3} \right).11.\left( { - 23} \right)}}{{11.23.5.5}}\\A = \frac{3}{5}\end{array}\)
b)
\(\begin{array}{l}B = \left( {\frac{{ - 7}}{9}} \right).\frac{{13}}{{25}} - \frac{{13}}{{25}}.\frac{2}{9}\\B = \frac{{13}}{{25}}.\left( {\frac{{ - 7}}{9} - \frac{2}{9}} \right)\\B = \frac{{13}}{{25}}.(-1)\\B = \frac{{-13}}{{25}}.\end{array}\)
a: =214*(13+17)=214*30=6420
b: =635(58-48)=635*10=6350
c: =100*19=1900
d: =4*32=128
`Answer:`
a. \(-\frac{17}{30}-\frac{11}{-15}+-\frac{7}{12}\)
\(=-\frac{17}{30}+\frac{11}{15}-\frac{7}{12}\)
\(=-\frac{17}{30}+\frac{22}{30}-\frac{7}{12}\)
\(=\frac{1}{6}-\frac{7}{12}\)
\(=-\frac{5}{12}\)
b. \(-\frac{5}{9}+\frac{5}{9}:\left(\frac{5}{3}-\frac{25}{12}\right)\)
\(=-\frac{5}{9}+\frac{5}{9}:\left(-\frac{5}{12}\right)\)
\(=\frac{5}{9}.\left(-1\right)+\frac{5}{9}.\left(-\frac{12}{5}\right)\)
\(=\frac{5}{9}.\left(-1-\frac{12}{5}\right)\)
\(=\frac{5}{9}.\frac{-17}{5}\)
\(=-\frac{17}{9}\)
c. \(-\frac{7}{25}.\frac{11}{13}+-\frac{7}{25}-\frac{2}{13}-\frac{18}{25}\)
\(=\frac{77}{325}-\left(\frac{7}{25}+\frac{18}{25}\right)-\frac{2}{13}\)
\(=\frac{77}{325}-1-\frac{2}{13}\)
\(=-\frac{452}{325}\)
-4/5 + 11/3 - 9/10 =(-4/5 + 11/3) -9/10 =43/15 -9/10 =59/30 5/6 + -7/15 - 13/30 =(5/6 + -7/15) -13/30 =11/30 - 13/30 =-1/15 CHÚC EM HỌC TỐT ^_^
b)(1-3-5+7)+(9-11-13+15)+...+(2009-2011-2013+2015)
=0+0+...+0
=0