Tìm x biết :
\(2x+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)
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28 tháng 5 2022
a: \(\dfrac{2032-x}{25}+\dfrac{2053-x}{23}+\dfrac{2070-x}{21}+\dfrac{2083-x}{19}-10=0\)
\(\Leftrightarrow\left(\dfrac{2032-x}{25}-1\right)+\left(\dfrac{2053-x}{23}-2\right)+\left(\dfrac{2070-x}{21}-3\right)+\left(\dfrac{2083-x}{19}-4\right)=0\)
=>2007-x=0
hay x=2007
b: \(\Leftrightarrow x+\left(1+1+1+1+1+1+1\right)+\left(\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=0\)
\(\Leftrightarrow x+7+\left(\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=0\)
=>x+7+1/3-1/10=0
hay x=-217/30
16 tháng 11 2018
1/
a) ta có \(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{97.100}=\dfrac{1}{3}.\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{97.100}\right)\)
\(=\dfrac{1}{3}.\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{97}-\dfrac{1}{100}\right)\)
\(=\dfrac{1}{3}.\dfrac{99}{100}=\dfrac{33}{100}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33x}{2009}\)
⇒ \(\dfrac{33}{100}=\dfrac{0,33}{2009}.x\Rightarrow x=\dfrac{33}{100}:\dfrac{0,33}{2009}=2009\)
16 tháng 11 2018
b,1 + 1/3 + 1/6 + 1/10 + ... + 2/x(x+1)=1 1991/1993
2 + 2/6 + 2/12 + 2/20 + ... + 2/x(x+1) = 3984/1993
2.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/x(x+1) = 3984/1993
2.(1 − 1/2 + 1/2 − 1/3 + ... + 1/x − 1/x+1)=3984/1993
2.(1 − 1/x+1) = 3984/1993
1 − 1/x + 1= 3984/1993 :2
1 − 1/x+1 = 1992/1993
1/x+1 = 1 − 1992/1993
1/x+1=1/1993
<=>x+1 = 1993
<=>x+1=1993
<=> x+1=1993
<=> x = 1993-1
<=> x = 1992
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
10 tháng 7 2023
\(\dfrac{1}{3}-\dfrac{1}{12}-\dfrac{1}{20}-\dfrac{1}{30}-\dfrac{1}{42}-\dfrac{1}{56}-\dfrac{1}{72}-\dfrac{1}{90}-\dfrac{1}{110}=x-\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3.4}\) - \(\dfrac{1}{4.5}\) - \(\dfrac{1}{5.6}\) - \(\dfrac{1}{6.7}\) - \(\dfrac{1}{7.8}\)- \(\dfrac{1}{8.9}\) - \(\dfrac{1}{9.10}\) - \(\dfrac{1}{10.11}\) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3.4}\) + \(\dfrac{1}{4.5}\) + \(\dfrac{1}{5.6}\) + \(\dfrac{1}{6.7}\)+ \(\dfrac{1}{7.8}\) + \(\dfrac{1}{8.9}\) + \(\dfrac{1}{9.10}\) + \(\dfrac{1}{10.11}\) =\(x\)-\(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{4}\) + \(\dfrac{1}{4}\) - \(\dfrac{1}{5}\) +...+ \(\dfrac{1}{9}\) - \(\dfrac{1}{10}\) + \(\dfrac{1}{10}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - (\(\dfrac{1}{3}\) - \(\dfrac{1}{11}\)) = \(x\) - \(\dfrac{5}{13}\)
\(\dfrac{1}{3}\) - \(\dfrac{1}{3}\) + \(\dfrac{1}{11}\) = \(x\) - \(\dfrac{5}{13}\)
\(x-\dfrac{5}{13}=\dfrac{1}{11}\)
\(x\) = \(\dfrac{1}{11}\) + \(\dfrac{5}{13}\)
\(x\) = \(\dfrac{68}{143}\)
PT
3
13 tháng 8 2015
\(2x+1+\frac{1}{6}+1+\frac{1}{12}+..+1+\frac{1}{90}=10\)
=> 2x + 8 + \(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}=10\)
=> 2x + \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}=10-8\)
\(2x+1-\frac{1}{10}=2\)
=> 2x + \(\frac{9}{10}=2\)
=> 2x = 2 - 9/10
=>2x = 11/10
=> x = 11/10 : 2
x = 11/20
NQ
0
2 tháng 8 2023
=9/10-(1/2+1/6+...+1/90)
=9/10-(1-1/2+1/2-1/3+...+1/9-1/10)
=9/10-9/10=0
TT
11 tháng 11 2017
\(\dfrac{3}{2}+\dfrac{13}{6}+\dfrac{37}{12}+\dfrac{81}{20}+\dfrac{151}{30}+\dfrac{253}{42}+\dfrac{393}{56}+\dfrac{577}{72}+\dfrac{811}{90}\)
\(=\dfrac{459}{10}\)
NN
1
16 tháng 7 2016
2x+7/6+13/12+21/20+31/30+43/42+57/56+73/72+91/90=10
2x+1+1/6+1+1/12+1+1/20+1+1/30+1+1/42+1+1/56+1+1/72+1+1/90=10
2x+(1+1+1+1+1+1+1+1)+(1/2.3+1/3.4+1/4.5+1/5.6+1/6.7+1/7.8+1/8.9+1/9.10)=10
2x+8+(1-1/2+1/2-1/3+...+1/9-1/10)=10
2x+1-1/10=10-8
2x+9/10=2
2x=2-9/10
2x=11/10
x=11/10/2
x=11/20
\(2x+\dfrac{7}{6}+\dfrac{13}{12}+\dfrac{21}{20}+\dfrac{31}{30}+\dfrac{43}{42}+\dfrac{57}{56}+\dfrac{73}{72}+\dfrac{91}{90}=10\)
\(2x+\left(1+\dfrac{1}{6}\right)+\left(1+\dfrac{1}{12}\right)+\left(1+\dfrac{1}{20}\right)+\left(1+\dfrac{1}{30}\right)+\left(1+\dfrac{1}{42}\right)+\left(1+\dfrac{1}{56}\right)+\left(1+\dfrac{1}{72}\right)+\left(1+\dfrac{1}{90}\right)=10\)
\(2x+8+\left(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{56}+\dfrac{1}{72}+\dfrac{1}{90}\right)=10\)
\(2x+8+\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+\dfrac{1}{9.10}\right)=10\)
\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)
\(2x+8+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)
~ Chúc bạn học tốt ~
2x+\(\dfrac{7}{2.3}+\dfrac{13}{3.4}+\dfrac{21}{4.5}+...+\dfrac{91}{9.10}=10\)
\(2x+7\left(\dfrac{1}{2.3}\right)+13\left(\dfrac{1}{3.4}\right)+...+91\left(\dfrac{1}{9.10}\right)=10\)
\(2x+\left(7+13+...+91\right)\left(\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\right)=10\)
\(2x+336+\left(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\right)=10\)
\(2x+336+\left(\dfrac{1}{2}-\dfrac{1}{10}\right)=10\)
\(2x+336.\dfrac{2}{5}=10\)
\(2x+\dfrac{672}{5}=10\)
\(2x=10-\dfrac{672}{5}\)
\(2x=-\dfrac{622}{5}\)
\(x=-\dfrac{311}{5}\)
Tick nha 0o0^^^Nhi^^^0o0