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a ) \(-\frac{6}{7}< \frac{3}{7}< \frac{18}{7}\)
b ) \(\frac{17}{35}>\frac{17}{-35}\)
c ) \(\frac{17}{35}>\frac{17}{53}\)
d ) \(\frac{12}{7}< \frac{17}{5}\)
\(A=\frac{2}{7}+\frac{-3}{8}+\frac{11}{7}+\frac{1}{3}+\frac{1}{7}+\frac{5}{-8}\)
\(A=\left(\frac{2}{7}+\frac{11}{7}+\frac{1}{7}\right)+\left(\frac{-3}{8}+\frac{5}{-8}\right)+\frac{1}{3}\)
\(A=2-1+\frac{1}{3}\)
\(A=\frac{4}{3}\)
\(B=\frac{3}{17}+\frac{-5}{13}+\frac{-18}{35}+\frac{14}{17}+17\)
\(B=\left(\frac{3}{17}+\frac{14}{17}\right)+\frac{-5}{13}+\frac{-18}{35}+17\)
\(B=1+\frac{-5}{13}+\frac{-18}{35}+17\)
\(B=18+\frac{-5}{13}+\frac{-18}{35}\)
\(B=\frac{7781}{455}\)
\(C=\left|-3\left(\dfrac{-13}{15}-\dfrac{17}{21}\right)\right|-\left|\dfrac{-13}{15}+\dfrac{17}{7}\right|+\left(-12+\dfrac{35}{3}\right):\left|-\dfrac{7}{6}\right|\\ =\left|-3.-\dfrac{176}{105}\right|-\left|-\dfrac{6}{35}\right|+\left(-\dfrac{1}{3}\right):\dfrac{7}{6}\\ =\dfrac{176}{35}-\dfrac{6}{35}-\dfrac{1}{3}:\dfrac{7}{6}\\ =\dfrac{176}{35}-\dfrac{6}{35}-\dfrac{2}{7}\\ =\dfrac{170}{35}-\dfrac{2}{7}=\dfrac{32}{7}.\)
a)\(-\frac{6}{7}< \frac{3}{7}< \frac{18}{7}\)
b) \(\frac{17}{35}>\frac{17}{53}\)
c) \(\frac{17}{35}< \frac{17}{-35}\)
d) Ta có : \(\frac{12}{7}=\frac{60}{35}\)
\(\frac{17}{5}=\frac{85}{35}\)
\(\Rightarrow\frac{12}{7}< \frac{17}{5}\)
a)\(\frac{3}{7}\)>\(\frac{18}{7}\)>\(\frac{-6}{7}\)
b)\(\frac{17}{35}\)>\(\frac{17}{53}\)
c)\(\frac{17}{35}\)<\(\frac{17}{-35}\)
d)\(\frac{12}{7}\)<\(\frac{17}{5}\)
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