Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(C=0,5+\frac{1}{3}+0,4+\frac{5}{7}+\frac{1}{6}-\frac{4}{35}+\frac{1}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{3}+\frac{2}{5}+\frac{1}{6}\right)+\left(\frac{5}{7}-\frac{4}{35}\right)+\frac{1}{41}\)
\(=\frac{15+10+12+5}{30}+\frac{25-4}{35}+\frac{1}{41}\)
\(=\frac{7}{5}+\frac{3}{5}+\frac{1}{41}\)
\(=2+\frac{1}{41}=\frac{83}{41}\)
\(D=\frac{1}{90}-\frac{1}{72}-\frac{1}{56}-\frac{1}{42}-\frac{1}{30}-\frac{1}{12}-\frac{1}{6}-\frac{1}{2}\)
\(=\left(\frac{1}{90}-\frac{1}{30}-\frac{1}{6}-\frac{1}{2}\right)+\left(-\frac{1}{72}-\frac{1}{12}\right)-\frac{1}{56}-\frac{1}{42}\)
\(=\frac{1-2-15-45}{90}+\frac{-1-6}{72}-\frac{1}{56}-\frac{1}{42}\)
\(=-\frac{61}{90}-\frac{7}{72}-\frac{1}{56}-\frac{1}{42}\)
\(=\frac{-1708-245-45-60}{2520}\)
\(=-\frac{49}{60}\)
Nghịch đảo của C là \(\frac{41}{83}\), nghịch đảo của D là \(-\frac{60}{49}\)
\(\frac{41}{83}\cdot\left(-\frac{60}{49}\right)=-\frac{2460}{4067}\)
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, \(\dfrac{1}{2}\) - ( - \(\dfrac{1}{3}\) ) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= \(\dfrac{5}{6}\) + \(\dfrac{1}{23}\) + \(\dfrac{1}{6}\)
= 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
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}+\frac{109}{110}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}+1-\frac{1}{110}\)
\(=10-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}+\frac{1}{110}\right)\)
\(=10-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}+\frac{1}{10.11}\right)\)
\(=10-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\right)\)
\(=10-\left(1-\frac{1}{10}\right)\)
\(=\frac{91}{10}\)
\(\frac{1}{2}+\frac{5}{6}+\frac{11}{12}+\frac{19}{20}+...+\frac{89}{90}\)
\(=1-\frac{1}{2}+1-\frac{1}{6}+1-\frac{1}{12}+1-\frac{1}{20}+...+1-\frac{1}{90}\)
\(=9-\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\right)\)
\(=9-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right)\)
\(=9-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(=9-\left(1-\frac{1}{10}\right)\)
\(=9-\frac{9}{10}=\frac{81}{10}\)
