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p) \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(=-1+1+\frac{-2}{11}\)
\(=-\frac{2}{11}\)
q) \(\frac{5}{13}+\frac{-5}{17}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{17}\)
\(=1+\left(-1\right)+\frac{-5}{17}=\frac{-5}{17}\)
r) \(\frac{1}{5}+\frac{-2}{9}+\frac{-7}{9}+\frac{4}{5}+\frac{16}{17}\)
\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+\frac{-7}{9}\right)+\frac{16}{17}\)
\(=1+\left(-1\right)+\frac{16}{17}=\frac{16}{17}\)
M=3/1.2-5/2.3+7/3.4-9/4.5+11/5.6-13/6.7+15/7.8+17/8.9
=(1/1.1+2/1.2)-(2/2.3+3/2.3)+(3/3.4+4/3.4)-(4/4.5+5/4.5)+...+(8/8.9+9/8.9)(phần ... là làm tương tự nhé)
=1/2+1-(1/3+1/2)+(1/4+1/3)-(1/5+1/4)+...+(1/9+1/8)(phần ... là làm tương tự nhé)
=1+(1/2-1/2)+(1/3-1/3)+(1/4-1/4)+...+(1/8-1/8)-1/9
=1+0+0+0+...+0-1/9
=1-1/9
=8/9
b,\(\frac{2}{3}\)+\(\frac{1}{3}\).(\(\frac{-2}{3}\)+\(\frac{5}{6}\)):\(\frac{2}{3}\)
=\(\frac{2}{3}\)+\(\frac{1}{3}\).(\(\frac{-4}{6}\)+\(\frac{5}{6}\)):\(\frac{2}{3}\)
=\(\frac{2}{3}\)+\(\frac{1}{3}\).\(\frac{1}{6}\).\(\frac{3}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{18}\).\(\frac{3}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{6}\).\(\frac{1}{2}\)
=\(\frac{2}{3}\)+\(\frac{1}{12}\)
=\(\frac{8}{12}\)+\(\frac{1}{12}\)
=\(\frac{9}{12}\)=\(\frac{3}{4}\)
a,\(=\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{-4}{9}+\frac{7}{15}\)
\(\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
=-1+1+-2/11
=0+-2/11
=-2/11
b,\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{40}\right)+\frac{-5}{17}\)
=1+-1+-5/17
=0+-5/17
=-5/17
c,\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+-\frac{7}{9}\right)+\frac{16}{17}\)
=1+-1+16/17
=0+16/17
=16/17
d,\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{99}-\frac{1}{101}\)
\(=\frac{1}{3}-\frac{1}{101}=\frac{98}{303}\)
a.\(\frac{-5}{9}\)+\(\frac{8}{15}\)+\(\frac{-2}{11}\)+\(\frac{4}{-9}\)+\(\frac{7}{15}\)
=\(\frac{-5}{9}\)+\(\frac{4}{-9}\)+\(\frac{8}{15}\)+\(\frac{7}{15}\)+\(\frac{-2}{11}\)
=(\(\frac{-5}{9}\)+\(\frac{-4}{9}\))+(\(\frac{8}{15}\)+\(\frac{7}{15}\))+\(\frac{-2}{11}\)
=(-1)+1+\(\frac{-2}{11}\)
=0+\(\frac{-2}{11}\)
=\(\frac{-2}{11}\).
\(M=\frac{17}{5}\cdot\frac{-31}{125}\cdot\frac{1}{2}\cdot\frac{10}{17}\cdot\frac{-1}{2^3}\)
\(M=\frac{17}{5}\cdot\frac{-31}{125}\cdot\frac{1}{2}\cdot\frac{10}{7}\cdot\frac{-1}{8}\)
\(M=\left(\frac{17}{5}\cdot\frac{10}{17}\cdot\frac{1}{2}\right)\cdot\frac{-31}{125}\cdot\frac{-1}{8}\)
\(M=1\cdot\frac{31}{1000}=\frac{31}{1000}\)
\(P=\frac{6}{7}\cdot\frac{8}{13}+\frac{6}{9}\cdot\frac{9}{7}-\frac{3}{13}\cdot\frac{6}{7}=\frac{6}{7}\cdot\frac{8}{13}+\frac{6}{7}\cdot1-\frac{3}{13}\cdot\frac{6}{7}\)
\(=\frac{6}{7}\left(\frac{8}{13}+1-\frac{3}{13}\right)=\frac{6}{7}\left(\frac{8}{3}+\frac{13}{13}-\frac{3}{13}\right)=\frac{6}{7}\cdot\frac{18}{13}=\frac{108}{91}\)
c. N= \(\frac{-5}{7}\cdot\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{12}{7}\)
N=1
a)\(\frac{-10}{13}+\frac{8}{17}-\frac{3}{13}+\frac{12}{17}-\frac{11}{20}\)
= \(\frac{-10}{13}+\frac{8}{17}+\frac{-3}{13}+\frac{12}{17}+\frac{-11}{20}\)
=\(\left(\frac{-10}{13}+\frac{-3}{13}\right)+\left(\frac{8}{17}+\frac{12}{17}\right)+\frac{-11}{20}\)
=\(\frac{-13}{13}+\frac{20}{17}+\frac{-11}{20}\)
= \(\frac{-127}{340}\)
b) \(\frac{3}{4}+\frac{-5}{6}-\frac{11}{-12}\)
= \(\frac{3}{4}+\frac{-5}{6}+\frac{11}{12}\)
= \(\frac{9}{12}+\frac{-10}{12}+\frac{11}{12}\)
=\(\frac{10}{12}=\frac{5}{6}\)
c) \(\left[13.\frac{4}{9}+2.\frac{1}{9}\right]-3.\frac{4}{9}\)
= \(13+2.\left(\frac{4}{9}+\frac{1}{9}\right)-3.\frac{4}{9}\)
=\(15.\frac{5}{9}-3.\frac{4}{9}\)
=\(\left[15-3.\left(\frac{5}{9}-\frac{4}{9}\right)\right]\)
=\(12.\frac{1}{9}\)
=\(\frac{4}{3}\)
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\(B=\frac{3+\frac{8}{12}+\frac{9}{13}-\frac{12}{17}}{4+\frac{8}{12}+\frac{12}{13}-\frac{16}{17}}+\frac{4+\frac{16}{60}-\frac{24}{213}-\frac{32}{11}}{5+\frac{20}{61}-\frac{36}{213}-\frac{40}{11}}\)
\(\Leftrightarrow B=\frac{3\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}{4\left(1+\frac{8}{12}+\frac{3}{13}-\frac{4}{17}\right)}+\frac{4\left(1+\frac{4}{60}-\frac{6}{213}-\frac{8}{11}\right)}{5\left(1+\frac{4}{60}+\frac{6}{213}-\frac{8}{11}\right)}\)
\(\Leftrightarrow B=\frac{3}{4}+\frac{4}{5}\)
\(\Leftrightarrow B=\frac{15}{20}+\frac{16}{20}\)
\(\Leftrightarrow B=\frac{31}{20}\)
a) \(\frac{31}{17}+\frac{-5}{13}+\frac{-8}{13}-\frac{14}{17}\)
\(=\left(\frac{31}{17}-\frac{14}{17}\right)+\left(\frac{-5}{13}+\frac{-8}{13}\right)\)
\(=1+\left(-1\right)\)
\(=0\)
b) \(\frac{-5}{7}.\frac{2}{11}+\frac{-5}{7}.\frac{9}{11}+\frac{5}{7}\)
\(=\frac{-5}{7}.\left(\frac{2}{11}+\frac{9}{11}\right)+\frac{5}{7}\)
\(=\frac{-5}{7}.1+\frac{5}{7}\)
\(=\frac{-5}{7}+\frac{5}{7}\)
\(=0\)
chắc thế này :
\(\frac{9}{17}.\frac{21}{13}+\frac{5}{17}.\frac{9}{13}-\frac{8}{17}.2\)
\(=\frac{9}{17}.\frac{21}{13}+\frac{5}{13}.\frac{9}{17}-\frac{8}{17}.2\)
\(=\frac{9}{17}.\left(\frac{21}{13}+\frac{5}{13}\right)-\frac{8}{17}.2\)
\(=\frac{9}{17}.2-\frac{8}{17}.2\)
\(=2.\left(\frac{9}{17}-\frac{8}{17}\right)\)
\(=2.\frac{1}{17}\)
\(=\frac{2}{17}\)
1/17.(189/13+45/13)-16/17=1/17.18-16/17=18/17-16/17=2/17