A=1/2+1/3+0,4+5/7+1/6-4/35=?
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a. \(\dfrac{-2}{3}+\dfrac{-1}{5}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-4}{6}+\dfrac{-2}{10}+\dfrac{3}{4}-\dfrac{5}{6}-\dfrac{7}{10}\)
= \(\dfrac{-3}{2}+\dfrac{1}{2}+\dfrac{3}{4}\)
= (-1) + \(\dfrac{3}{4}\)
= \(\dfrac{-4}{4}+\dfrac{3}{4}\)
= \(\dfrac{-1}{4}\)
a)\(3-\left(\frac{1}{4}+\frac{2}{3}\right)-\left(5+\frac{1}{3}-\frac{6}{5}\right)-\left(6-\frac{7}{4}+\frac{3}{2}\right)\)
=\(3-\frac{1}{4}-\frac{2}{3}-5-\frac{1}{3}+\frac{6}{5}-6+\frac{7}{4}-\frac{3}{2}\)
=\(\left(3-5-6\right)+\left(\frac{-1}{4}+\frac{7}{4}\right)+\left(\frac{-2}{3}-\frac{1}{3}\right)+\left(\frac{6}{5}-\frac{3}{2}\right)\)
=\(-8+\frac{3}{2}-1-\frac{3}{10}\)
=\(\left(-8-1\right)+\left(\frac{3}{2}-\frac{3}{10}\right)\)
=-9+\(\frac{6}{5}\)
=\(\frac{-39}{5}\)
a)
1/12 + 1/6 + 1/2 = (1+2+6)/12 = 9/12 = 3/4
1/30 + 1/20 = (2+3)/60 = 5/60 = 1/12
1/56 + 1/42 = 1/7(1/8+1/6) = 1/7 .(3+4)/24 = 1/24
8/9- 1/72 = (8.8 - 1)/72 = 63/72 = 7/8
1/12 + 1/24 = (2+1)/24 = 1/8
7/8 - 1/8 = 6/8 = 3/4
3/4 - 3/4 = 0
b)
\(0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{3}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)\)
\(=\frac{15+10+12+5}{30}+\frac{25-4}{35}\)
\(=\frac{7}{5}+\frac{3}{5}\)
\(=2\)
a) Ta có: \(D=\dfrac{1}{3}+\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{24}+\dfrac{1}{48}+\dfrac{1}{96}\)
\(=\dfrac{2}{3}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{12}+\dfrac{1}{12}-\dfrac{1}{24}+\dfrac{1}{24}-\dfrac{1}{48}+\dfrac{1}{48}-\dfrac{1}{96}\)
\(=\dfrac{2}{3}-\dfrac{1}{96}\)
\(=\dfrac{63}{96}=\dfrac{21}{32}\)
b)
Sửa đề: \(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\)
Ta có: \(E=\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+...+\dfrac{1}{2048}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot E=\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}+\dfrac{1}{32}+...+\dfrac{1}{4096}\)
\(\Leftrightarrow\dfrac{1}{2}\cdot E=\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{16}+...+\dfrac{1}{2048}-\dfrac{1}{4096}\)
\(\Leftrightarrow\dfrac{E}{2}=\dfrac{1}{2}-\dfrac{1}{4096}=\dfrac{2047}{4096}\)
hay \(E=\dfrac{2047}{2048}\)
Giải:
A=1/10+1/40+1/88+1/154+1/238+1/340
A=1/2.5+1/5.8+1/8.11+1/11.14+1/14.17+1/17.20
A=1/2-1/5+1/5-1/8+1/8-1/11+1/11-1/14+1/14-1/17+1/17-1/20
A=1/2-1/20
A=9/20
D=1/3+1/6+1/12+1/24+1/48
D=1/3+1/2.3+1/3.4+1/4.6+1/6.8
D=1/3+1/2-1/3+1/3-1/4+1/2.(2/4.6+2/6.8)
D=1/3+1/2-1/4+1/2.(1/4-1/6+1/6-1/8)
D=1/3+1/4+1/2.(1/4-1/8)
D=1/3+1/4+1/2.1/8
D=1/3+1/4+1/16
D=31/48
F=0,5-1/3-0,4-4/7-1/6+4/35-1/41
F=1/2-1/3-2/5-4/7-1/6+4/35-1/41
F=1/6-(-6/35)-1/6+4/35-1/41
F=(1/6-1/6)+(6/35+4/35)-1/41
F=0+2/7-1/41
F=2/7+1/41
F=75/287
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\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{4}\)
\(=\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{4}\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{4}\right)+\left(\dfrac{1}{3}+\dfrac{1}{6}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)\)
\(=\dfrac{3}{4}+\dfrac{1}{2}+1\)
\(=\dfrac{9}{4}\)
\(0,5+\dfrac{1}{3}+0,4+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{4}\)
=\(\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{2}{5}+\dfrac{5}{7}+\dfrac{1}{6}-\dfrac{4}{35}+\dfrac{1}{4}\)
=\(\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{4}+\dfrac{1}{3}\right)+\left(\dfrac{2}{5}+\dfrac{5}{7}-\dfrac{4}{35}\right)\)
=\(\dfrac{5}{4}+1\)
=\(\dfrac{9}{4}\)
A=\(\frac{1}{2}+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
\(A=\frac{3}{6}+\frac{2}{6}+\frac{2}{5}+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}\)
\(A=\left(\frac{3}{6}+\frac{2}{6}+\frac{1}{6}\right)+\left(\frac{2}{5}+\frac{5}{7}-\frac{4}{35}\right)\)
\(A=1+\left(\frac{14}{35}+\frac{25}{35}-\frac{4}{35}\right)\)
\(A=1+1\)
\(A=2\)