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\(\dfrac{9}{6}-\dfrac{5}{7}=\dfrac{63}{42}-\dfrac{30}{42}=\dfrac{33}{42}=\dfrac{11}{14}\)
\(\dfrac{33}{12}-\dfrac{3}{8}=\dfrac{66}{24}-\dfrac{9}{24}=\dfrac{57}{24}=\dfrac{19}{8}\)
\(1-\dfrac{3}{7}=\dfrac{7}{7}-\dfrac{3}{7}=\dfrac{4}{7}\)
\(2-\dfrac{3}{4}=\dfrac{8}{4}-\dfrac{3}{4}=\dfrac{5}{4}\)
9/5 + 4/7 + 6/5 + 3/7
= ( 9/5 + 6/5 ) + ( 4/7 + 3/7)
= 3 + 1
= 4
9/5 + 4/7 + 6/5 + 3/7 = (9/5 + 6/5) + (4/7 + 3/7)
= 15/5 + 7/7
= 3 + 1
= 4
1/2 x 45/33 x 1/9 x 11/6 = (1/2 x 1/9) x (45/33 x 11/6)
= 1/18 x 45/18
= 5/2
a: 1/3h=20p
b: 11/7*3/5=33/35
3/4:5/6+2/5=3/4*6/5+2/5=9/10+4/10=13/10
4/7+1/2*2/3=4/7+1/3=12/21+7/21=19/21
3-7/5=15/5-7/5=8/5
5/9:37=5/9*1/37=5/333
\(\dfrac{4}{7}-\dfrac{1}{3}< \dfrac{1}{2}-\dfrac{2}{9}\)
\(\dfrac{125}{100}-\dfrac{48}{64}< \dfrac{6}{7}-\dfrac{14}{28}\)
\(\dfrac{7}{5}-\dfrac{22}{25}=\dfrac{1}{5}+\dfrac{4}{10}\)\(\dfrac{45}{75}+\dfrac{22}{33}>\dfrac{4}{5}+\dfrac{4}{6}\)
\(\frac{17}{33}.\frac{2}{5}+\frac{3}{5}.\frac{17}{33}-\frac{17}{33}\)
\(=\frac{17}{33}\times\frac{2}{5}+\frac{3}{5}\times\frac{17}{33}-\frac{17}{33}\times1\)
\(=\frac{17}{33}\times\left(\frac{2}{5}+\frac{3}{5}-1\right)\)
\(=\frac{17}{33}\times\left(1-1\right)\)
\(=\frac{17}{33}\times0=\frac{17}{33}\)
a) Ta có: 4x-33=-5
\(\Leftrightarrow4x=28\)
hay x=7
Vậy: x=7
b) Ta có: \(2x+\dfrac{1}{4}=\dfrac{3}{2}\)
\(\Leftrightarrow2x=\dfrac{5}{4}\)
hay \(x=\dfrac{5}{8}\)
Vậy: \(x=\dfrac{5}{8}\)
1, \(x=-\dfrac{7}{3}-\dfrac{1}{3}=-\dfrac{8}{3}\)
2, \(x=\dfrac{1}{8}-\dfrac{3}{8}=-\dfrac{2}{8}=-\dfrac{1}{4}\)
3, \(x=\dfrac{6}{5}+\dfrac{4}{5}=\dfrac{10}{5}=2\)
4, \(x=\dfrac{7}{3}-\dfrac{2}{3}=\dfrac{5}{3}\)
5, \(x+\dfrac{7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{5}{3}-\dfrac{7}{3}=-\dfrac{2}{3}\)
\(=x+\dfrac{1}{3}=\dfrac{-7}{3}\Leftrightarrow x=\dfrac{-8}{3}\)
\(=\dfrac{1}{8}-x=\dfrac{3}{8}\Leftrightarrow x=\dfrac{-1}{4}\)
\(=x-\dfrac{4}{5}=\dfrac{6}{5}\Leftrightarrow x=2\)
\(=\dfrac{2}{3}+x=\dfrac{7}{3}\Leftrightarrow x=\dfrac{5}{3}\)
\(=x-\dfrac{-7}{3}=\dfrac{5}{3}\Leftrightarrow x=\dfrac{-2}{3}\)
