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8)\(\frac{4}{9}:\left(-\frac{1}{7}\right)+6\frac{5}{9}:\left(-\frac{1}{7}\right)\)
=\(\frac{4}{9}:\left(-\frac{1}{7}\right)+\frac{59}{9}:\left(-\frac{1}{7}\right)\)
=\(\left(\frac{4}{9}+\frac{59}{9}\right).\left(-7\right)\)
=7.(-7)
=-49
Bài 1:
a: \(=17+\dfrac{2}{31}-\dfrac{15}{17}-6-\dfrac{2}{31}=11-\dfrac{15}{17}=\dfrac{172}{17}\)
b: \(=31+\dfrac{6}{13}+5+\dfrac{9}{41}-36-\dfrac{9}{41}-36-\dfrac{6}{13}\)
=36
c: \(=27+\dfrac{51}{59}-7-\dfrac{51}{59}+\dfrac{1}{3}=20+\dfrac{1}{3}=\dfrac{61}{3}\)
Bài 1:
a: \(=\dfrac{1}{2}-\dfrac{7}{13}-\dfrac{1}{3}-\dfrac{6}{13}+\dfrac{1}{3}+\dfrac{4}{3}=\dfrac{4}{3}-1+\dfrac{1}{2}=\dfrac{1}{3}+\dfrac{1}{2}=\dfrac{5}{6}\)
b: \(=\dfrac{3}{4}+\dfrac{2}{5}+\dfrac{1}{9}-1-\dfrac{2}{5}+\dfrac{5}{4}=2-1+\dfrac{1}{9}=\dfrac{10}{9}\)
c: \(=\left(\dfrac{-3}{2}\cdot\dfrac{4}{3}\right)\cdot\dfrac{-9}{2}-\dfrac{1}{2}=9-\dfrac{1}{2}=8.5\)
\(A=\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(A=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(A=\dfrac{8}{9}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)\)
\(A=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(A=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)=\dfrac{8}{9}-\dfrac{8}{9}=0\)
\(\dfrac{8}{9}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}\right)\)
\(=\dfrac{8}{9}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\left(1-\dfrac{1}{9}\right)\)
\(=\dfrac{8}{9}-\dfrac{8}{9}\)
= 0
b)\(\frac{1}{9}.\frac{2}{145}-4\frac{1}{3}.\frac{2}{145}+\frac{2}{145}\)
\(=\frac{2}{145}.\left(\frac{1}{9}-\frac{13}{3}+1\right)\)
\(=\frac{2}{145}.\left(-\frac{29}{9}\right)\)
\(=\frac{-2}{45}\)
Giải:
a) \(\dfrac{1}{3}x+\dfrac{1}{5}-\dfrac{1}{2}x=1\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{1}{5}-\dfrac{1}{6}x=\dfrac{5}{4}\)
\(\Leftrightarrow\dfrac{1}{6}x=\dfrac{-21}{20}\)
\(\Leftrightarrow x=\dfrac{-63}{10}\)
Vậy ...
b) \(\dfrac{3}{2}\left(x+\dfrac{1}{2}\right)-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{3}{2}x+\dfrac{3}{4}-\dfrac{1}{8}x=\dfrac{1}{4}\)
\(\Leftrightarrow\dfrac{11}{8}x=\dfrac{-1}{2}\)
\(\Leftrightarrow x=\dfrac{-4}{11}\)
Vậy ...
Các câu sau làm tương tự câu b)