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Bài1
a) 25/42 - 20/63 =5/18
b) 9/50 - 13/75 - 1/6 = -4/25
c) 2/15 - 2/65 - 4/39 = 0
Bài2
a) x + 7/12 =17/18-1/9 b) 29/30 - (18/23 + x)=7/69
x + 7/12 = 5/6 18/23 + x =29/30 - 7/69
x =5/6 - 7/12 18/23 +x = 199/230
x = 1/4 x = 199/230 - 18/23
x= 19/230
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{2012}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{2012}\)
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-...-\frac{1}{x+1}=\frac{2011}{2012}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2011}{2012}\)
\(\Rightarrow\frac{1}{x+1}=1-\frac{2011}{2012}\)
\(\Rightarrow\frac{1}{x+1}=\frac{1}{2012}\)
\(\Rightarrow x+1=2012\)
\(\Rightarrow x=2011\)
\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{2012}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=\frac{2011}{2012}\)
\(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2011}{2012}\)
\(1-\frac{1}{x+1}=\frac{2011}{2012}\)
\(\frac{1}{x+1}=1-\frac{2011}{2012}=\frac{1}{2012}\)
\(\Leftrightarrow x+1=2012\)
\(\Leftrightarrow x=2011\)
Vậy ...
P/s: Hoq chắc :<
Tìm x biết:
\(\frac{x}{3}-\frac{3}{4}=\frac{1}{12}\)
\(\frac{x}{3}=\frac{1}{12}+\frac{3}{4}\)
\(\frac{x}{3}=\frac{5}{6}\)
\(x=\frac{5}{6}.3\)
\(x=\frac{5}{2}\)
Vậy \(x=\frac{5}{2}\)
\(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\frac{13}{23}+x=\frac{29}{30}-\frac{7}{69}\)
\(\frac{13}{23}+x=\frac{199}{230}\)
\(x=\frac{199}{230}-\frac{13}{23}\)
\(x=\frac{3}{10}\)
Vậy \(x=\frac{3}{10}\)
Bài 2: tính
\(\frac{1}{30}+\frac{1}{42}+\frac{1}{56}+\frac{1}{72}+\frac{1}{90}+\frac{1}{110}\)
\(=\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}+\frac{1}{9.10}+\frac{1}{10.11}\)
\(=\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}+\frac{1}{9}-\frac{1}{10}+\frac{1}{10}-\frac{1}{11}\)
\(=\frac{1}{5}-\frac{1}{11}\)
\(=\frac{6}{55}\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{49.50}\)
\(=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{49}-\frac{1}{50}\)
\(=\frac{1}{1}-\frac{1}{50}\)
\(=\frac{49}{50}\)
Bài 2:
1/30+1/42+1/56+1/72+1/90+1/110
=1/5.6+1/6.7+1/7.8+1/8.9+1/9.10+1/10.11
=1/5-1/6+1/6-1/7+1/7-1/8+1/8-1/9+1/9-1/10+1/10-1/11
=1/5-1/11=6/55
b)1/1.2+1/2.3+...+1/49.50
=1-1/2+1/2-1/3+...+1/49-1/50
=1-1/50
=49/50
a)\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)
đề sai
b)\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\).Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(x=-2004\)
c)\(\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0\)
\(\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0\)
\(x-200=0\).Do\(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0\)
\(x=200\)
d)chịu
a) \(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
\(\Rightarrow x+\frac{7}{12}=\frac{5}{6}\)
\(\Rightarrow x=\frac{5}{6}-\frac{7}{12}\)
\(\Rightarrow x=\frac{1}{4}\)
b) \(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\Rightarrow\frac{13}{23}+x=\frac{29}{30}-\frac{7}{69}\)
\(\Rightarrow\frac{13}{23}+x=\frac{199}{230}\)
\(\Rightarrow x=\frac{199}{230}-\frac{13}{23}\)
\(\Rightarrow x=\frac{3}{10}\)
a)\(x+\frac{7}{12}=\frac{17}{18}-\frac{1}{9}\)
\(x+\frac{7}{12}=\frac{5}{6}\)
\(x=\frac{5}{6}-\frac{7}{12}\)
\(x=\frac{1}{4}\)
b)\(\frac{29}{30}-\left(\frac{13}{23}+x\right)=\frac{7}{69}\)
\(\left(\frac{13}{23}+x\right)=\frac{29}{30}-\frac{7}{69}\)
\(\left(\frac{13}{23}+x\right)=\frac{199}{230}\)\(x=\frac{199}{230}-\frac{13}{23}\)
\(x=\frac{3}{10}\)
a) \(\frac{29}{30}\)- (\(\frac{13}{23}\)+X)=\(\frac{7}{69}\)
\(\frac{13}{23}\)+X=\(\frac{29}{30}\)-\(\frac{7}{69}\)
\(\frac{13}{23}\)+X=\(\frac{199}{230}\)
X=\(\frac{199}{230}\)-\(\frac{13}{23}\)
X=\(\frac{3}{10}\)
b)1/2+1/6+1/12+...+1/x(x+1)=2011/2012
=>1/1.2+1/2.3+1/3.4+...+1/x(x+1)=2011/2012
=>1-1/2+1/2-1/3+1/3+1/4+...+1/x+1/x+1=2011/2012
=>1-1/x+1=2011/2012
=>1/x+1=1-2011-2012
=>1/x+1=2012/2012-2011/2012
1/x+1=1/2012
=>x+1=2012
=>x=2011
a) 3/10