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Câu 1:
a: Phân số lớn nhất là 11/12
b: Phân số nhỏ nhất là 2/3
Câu 2:
a: \(\Leftrightarrow y=\dfrac{1}{2}-\dfrac{1}{4}=\dfrac{1}{4}\)
b: \(\Leftrightarrow y=\dfrac{8}{12}-\dfrac{3}{12}+\dfrac{5}{12}=\dfrac{10}{12}=\dfrac{5}{6}\)
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a, 21/11-9/19-3/13-12/38-24/22 =21/11-(9/19-3/13)-12/38-24/22 =207/143-12/38-24/22 =3075/2717-24/22 =111/2717 b,2/3+3/4+4/5+12/9+28/16+48/15 =17/12+4/5+12/9+28/16+48/15 =133/60+12/9+28/16+48/15 =71/20+28/16+48/15 =53/10+48/15 =17/2 c, 4/5-1/2-1/4-1/8-1/16-1/36 =3/10-1/4-1/8-1/36 =1/20-1/8-1/16-1/36 =-3/40-1/16-1/36 =-11/60-1/36=-119/720
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(x-1/2)=19/4:5/3
(x-1/2)=19/4x3/5
x-1/2=57/20
x=57/20+1/2
x=67/20
tk cho mk nha
\(\frac{19}{4}:\left(x-\frac{1}{2}\right)=\frac{5}{3}\)
\(x-\frac{1}{2}=\frac{19}{4}:\frac{5}{3}=\frac{57}{20}\)
\(x=\frac{57}{20}+\frac{1}{2}\)
\(x=\frac{67}{20}\)
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a) \(\dfrac{5}{6}-\dfrac{2}{14}\)
\(=\dfrac{5}{6}-\dfrac{1}{7}\)
\(=\dfrac{35}{42}-\dfrac{6}{42}\)
\(=\dfrac{29}{42}\)
b) \(\dfrac{5}{20}-\dfrac{1}{6}\)
\(=\dfrac{1}{4}-\dfrac{1}{6}\)
\(=\dfrac{3}{12}-\dfrac{2}{12}\)
\(=\dfrac{1}{12}\)
c) \(\dfrac{5}{9}-\dfrac{3}{12}\)
\(=\dfrac{5}{9}-\dfrac{1}{4}\)
\(=\dfrac{20}{36}-\dfrac{9}{36}\)
\(=\dfrac{11}{36}\)
d) \(8-\dfrac{4}{6}\)
\(=\dfrac{48}{6}-\dfrac{4}{6}\)
\(=\dfrac{44}{6}=\dfrac{22}{3}\).
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Kết quả là \(\frac{1}{10}\)
Vì \(1-\frac{1}{2}=\frac{1}{2}\)nên \(1-\frac{1}{10}=\frac{1}{10}\)
k mình nha
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a x 4/7 + 1/3 x a = 2/3
<=>ax(4/7+1/3)=2/3
<=>ax19/21=2/3
<=>a=14/19
Vậy x=14/19
\(a\times\frac{4}{7}+\frac{1}{3}\times a=\frac{2}{3}\)
\(a\times\left(\frac{4}{7}+\frac{1}{3}\right)=\frac{2}{3}\)
\(a\times\frac{19}{21}=\frac{2}{3}\)
\(a=\frac{2}{3}:\frac{19}{21}\)
\(a=\frac{14}{19}\)
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Sao cx gặp bài phân số không -_-
\(a)\frac{19}{21}-\frac{1}{3}+\frac{3}{7}=\frac{19}{21}-(\frac{1}{3}+\frac{3}{7})=\frac{19}{21}-\frac{16}{21}=\frac{3}{21}\)
\(b)1+\frac{2}{3}-\frac{5}{3}=\frac{3}{3}+\frac{2}{3}-\frac{5}{3}=\frac{5}{3}-\frac{5}{3}=0\)
\(c)\frac{7}{12}+\frac{1}{3}-\frac{3}{4}=\frac{7}{12}+\frac{4}{12}-\frac{9}{12}=\frac{7+4-9}{12}=\frac{2}{12}=\frac{1}{6}\)
\(d)\frac{5}{7}-\frac{4}{7}\cdot\frac{1}{4}=\frac{5}{7}-\frac{1}{7}=\frac{4}{7}\)
\(e)\frac{7}{12}\cdot\frac{4}{3}+\frac{2}{9}=\frac{7}{9}+\frac{2}{9}=\frac{9}{9}=1\)
A. \(\frac{19}{21}-\frac{1}{3}+\frac{3}{7}=\frac{4}{7}+\frac{3}{7}=\frac{7}{7}=1\)
B. \(1+\frac{2}{3}-\frac{5}{3}=\frac{5}{3}-\frac{5}{3}=0\)
C. \(\frac{7}{12}+\frac{1}{3}-\frac{3}{4}=\frac{11}{12}-\frac{3}{4}=\frac{1}{6}\)
D. \(\frac{5}{7}-\frac{4}{7}\times\frac{1}{4}=\frac{5}{7}-\frac{1}{7}=\frac{4}{7}\)
E. \(\frac{7}{12}\times\frac{4}{3}+\frac{2}{9}=\frac{7}{9}+\frac{2}{9}=\frac{9}{9}=1\)
\(Linh\)\(Yukiko\)