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a, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 1 + \(\dfrac{1}{23}\)
= \(\dfrac{24}{23}\)
b, \(\dfrac{11}{24}\) - \(\dfrac{5}{41}\) + \(\dfrac{13}{24}\) + 0,5 - \(\dfrac{36}{41}\)
= (\(\dfrac{11}{24}\) + \(\dfrac{13}{24}\)) - ( \(\dfrac{5}{41}\) + \(\dfrac{36}{41}\)) + 0,5
= 1 - 1 + 0,5
= 0,5
c,\(-\dfrac{1}{12}-\left(\dfrac{1}{6}-\dfrac{1}{4}\right)\)
=\(-\dfrac{1}{12}-\left(-\dfrac{1}{12}\right)\)
=0
d, \(\dfrac{1}{6}-\left[\dfrac{1}{6}-\left(\dfrac{1}{4}+\dfrac{9}{12}\right)\right]\)
= \(\dfrac{1}{6}-\left[\dfrac{1}{6}-1\right]\)
= \(\dfrac{1}{6}-\left(-\dfrac{5}{6}\right)\)
= 1
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
Chúc bạn học tốt!
C=0.5+1/3+0.4+5/7+1/6-4/35+1/41
C=1/2+1/3+2/5+5/7+1/6-4/35+1/41
C=(1/2+1/6+1/3)+(2/5+5/7-4/35)+1/41
C=1+1-1/41
C=2-1/41
=>C=81/41
D=1/90-1/72-1/56-1/42-1/30-1/20--1/12-1/6-1/2
=> D=1/9*10-1/8*9-1/7*8-1/6*7-1/5*6-1/4*5-1/3*4-1/2*3-1/1*2
=>D=-(1/1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10)
=>D=-(1-1/10)
=>D=-9/10
ai k mh mh k lại
a: \(=\dfrac{15}{4}:\dfrac{2-7}{16}-\dfrac{5}{9}\cdot\left(\dfrac{3}{5}+\dfrac{4}{5}\right)\)
\(=\dfrac{15}{4}\cdot\dfrac{16}{-5}-\dfrac{5}{9}\cdot\dfrac{7}{5}\)
\(=\dfrac{-240}{20}-\dfrac{7}{9}=-12-\dfrac{7}{9}=\dfrac{-115}{9}\)
c: \(=\dfrac{1}{5}-\dfrac{7}{2}-\dfrac{3}{2}+\dfrac{5}{4}\)
\(=\dfrac{4+25}{20}-5=\dfrac{29}{20}-\dfrac{100}{20}=\dfrac{-71}{20}\)
d: \(=\dfrac{12}{17}\left(1-\dfrac{1}{15}-\dfrac{4}{5}+1\right)=\dfrac{12}{17}\cdot\dfrac{17}{15}=\dfrac{12}{15}=\dfrac{4}{5}\)
e: \(=\dfrac{2}{15}\cdot\dfrac{9}{8}-\dfrac{7}{4}\cdot\dfrac{9}{8}+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\left(\dfrac{2}{15}-\dfrac{7}{4}\right)+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\cdot\dfrac{8-105}{60}+\dfrac{25}{36}\)
\(=\dfrac{9}{8}\cdot\dfrac{-97}{60}+\dfrac{25}{36}=\dfrac{-1619}{1440}\)
\(C=0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)+\frac{1}{41}\)
\(=\frac{15+10+12+5}{30}+\frac{25-4}{35}+\frac{1}{41}\)
\(=\frac{7}{5}+\frac{3}{5}+\frac{1}{41}\)
\(=2+\frac{1}{41}=\frac{83}{41}\)
\(D=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\left(\frac{1}{90}-\frac{1}{30}-\frac{1}{6}-\frac{1}{2}\right)+\left(-\frac{1}{72}-\frac{1}{12}\right)-\frac{1}{56}-\frac{1}{42}\)
\(=\frac{1-2-15-45}{90}+\frac{-1-6}{72}-\frac{1}{56}-\frac{1}{42}\)
\(=-\frac{61}{90}-\frac{7}{72}-\frac{1}{56}-\frac{1}{42}\)
\(=\frac{-1708-245-45-60}{2520}\)
\(=-\frac{49}{60}\)
Nghịch đảo của C là \(\frac{41}{83}\), nghịch đảo của D là \(-\frac{60}{49}\)
\(\frac{41}{83}\cdot\left(-\frac{60}{49}\right)=-\frac{2460}{4067}\)
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 \(E=\dfrac{2047}{2048}\)