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Q
7 tháng 8 2017
\(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{90}+\dfrac{1}{72}+\dfrac{1}{56}+\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9\cdot10}+\dfrac{1}{9\cdot8}+\dfrac{1}{7\cdot8}+\dfrac{1}{7\cdot6}+\dfrac{1}{5\cdot6}-\dfrac{1}{5\cdot4}-\dfrac{1}{3\cdot4}-\dfrac{1}{3\cdot2}-\dfrac{1}{1\cdot2}\right)\)
\(=\dfrac{9}{10}-\left(\dfrac{1}{9}-\dfrac{1}{10}+\dfrac{1}{8}-\dfrac{1}{9}+...+1-\dfrac{1}{2}\right)\)
\(=\dfrac{9}{10}-\left(1-\dfrac{1}{10}\right)\)
\(=\dfrac{9}{10}-\dfrac{9}{10}\)
\(=0\)
JN
6 tháng 8 2018
a , \(\frac{7}{8}:\frac{1}{6}+\frac{7}{8}.\frac{-7}{18}\)
= \(\frac{21}{4}+\frac{-49}{144}=\frac{707}{144}\)
b, -1 : (-5) + \(\frac{1}{15}-\frac{-1}{15}\)
= \(\frac{1}{5}+0=\frac{1}{5}\)
c, \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{10-9}{10.9}-\frac{9-8}{9.8}-\frac{8-7}{8.7}-\frac{7-6}{7.6}-\frac{6-5}{6.5}-\frac{5-4}{5.4}-\frac{4-3}{4.3}-\frac{3-2}{3.2}.\frac{2-1}{2.1}\)
= \(\frac{9}{10}-1-\frac{1}{10}-1-\frac{1}{9}-1-\frac{1}{8}-1-\frac{1}{7}-1-\frac{1}{6}-1-\frac{1}{5}-1-\frac{1}{4}-1-\frac{1}{3}-1-\frac{1}{2}\)
= \(\frac{9}{10}-\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}\right)\)
= \(\frac{9}{10}-9-1,928=\frac{9}{10}-7,071=-6.171\)
HM
8 tháng 9 2016
em lớp 6 nha
B= 1/2 + 1/6 + 1/12 +1/20 + 1/30 + 1/42 + 1/56 + 1/72
B= 1/1*2 + 1/2*3 + 1/3*4 + 1/4*5 + 1/5*6 + 1/6*7 + 1/7*8 + 1/8*9
B=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
B=1+0-0-0-0-0-0-0-1/9
B=1-1/9
B=8/9
k em nha
24 tháng 6 2017
a)Ta có : \(a=2005\)
\(a+b+c\ne0\)
Áp dụng tính chất dãy tỉ số bằng nhau ta có :
\(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}=\dfrac{2005}{b}+\dfrac{b}{c}+\dfrac{c}{2005}=\dfrac{2005+b+c}{b+c+2005}=1\)
\(\Rightarrow\dfrac{b}{2005}=1\Rightarrow b=2005\)
\(\Rightarrow\dfrac{c}{2005}=1\Rightarrow c=\dfrac{1}{2005}\)
Vậy .................................
Q
24 tháng 6 2017
a) \(\dfrac{a}{b} = \dfrac{b}{c} = \dfrac{c}{a} = \dfrac{{a + b + c}}{{b + c + a}} = 1\)
\(\Rightarrow a = b = c\)
Mà \(a=2005\)
\(\Rightarrow b=c=2005\)
b) \(\dfrac{9}{10}-\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{30}-\dfrac{1}{20}-\dfrac{1}{12}-\dfrac{1}{6}-\dfrac{1}{2}\)
\(=\dfrac{2268-28-36-45-60-84-126-210-420-1260}{2520}\)
\(=\dfrac{0}{2520}\)
\(=0\)
Ta có : \(C=\frac{1}{12}+\frac{1}{30}+\frac{1}{56}+...+\frac{1}{2652}=\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+..+\frac{1}{51.52}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{51}-\frac{1}{52}\)
\(=\left(\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{51}+\frac{1}{52}\right)-2\left(\frac{1}{8}+\frac{1}{10}+...+\frac{1}{52}\right)\)
\(=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{51}+\frac{1}{52}-\left(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{26}\right)\)\(=\frac{1}{27}+\frac{1}{28}+...+\frac{1}{52}\)
Khi đó ta không thể chứng minh C < 1/4 vì sở dĩ \(\frac{1}{27}+\frac{1}{28}+...+\frac{1}{34}>\frac{1}{4}\)(bạn thử lấy máy tính tính xem)