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a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
mk làm câu c cho nó dễ
c)1/1.2+1/2.3+...+1/x.(x+1)=2009/2010
=1-1/2+1/2-1/3+...+1/x-1/x+1=2009/2010
=1-1/x+1=2009/2010
=1/x+1=1-2009/2010
=1/x+1=1/2010
=) x+1=2010
x =2010-1
x =2009
a/\(\frac{\left(2^3.5.7\right).\left(5^2.7^3\right)}{\left(2.5.7^2\right)^2}\)
=\(\frac{2^3.5^3.7^4}{2^2.5^2.7^4}\)
=2.5
=10
tung từng vế một thôi
bạn nhác quá éo chịu suy nghĩ
bài này dễ vl
Bài 1:
a, \(\frac{5}{1.6}+\frac{5}{6.11}+...+\frac{5}{\left(5x+1\right)\left(5x+6\right)}=\frac{2010}{2011}\)
\(1-\frac{1}{6}+\frac{1}{6}-\frac{1}{11}+...+\frac{1}{5x+1}-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(1-\frac{1}{5x+6}=\frac{2010}{2011}\)
\(\frac{1}{5x+6}=1-\frac{2010}{2011}\)
\(\frac{1}{5x+6}=\frac{1}{2011}\)
=> 5x + 6 = 2011
5x = 2011 - 6
5x = 2005
x = 2005 : 5
x = 401
b, \(\frac{7}{x}+\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{4}{5.9}+\frac{4}{9.13}+...+\frac{4}{41.45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{9}+\frac{1}{9}-\frac{1}{13}+...+\frac{1}{41}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\left(\frac{1}{5}-\frac{1}{45}\right)=\frac{29}{45}\)
\(\frac{7}{x}+\frac{8}{45}=\frac{29}{45}\)
\(\frac{7}{x}=\frac{29}{45}-\frac{8}{45}\)
\(\frac{7}{x}=\frac{7}{15}\)
=> x = 15
c, ghi lại đề
d, ghi lại đề
Bài 2:
\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
\(1+\frac{1}{3}+\frac{1}{6}+...+\frac{2}{x.\left(x+1\right)}=1\frac{2008}{2010}\)
\(\frac{2}{1.2}+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=1\frac{2008}{2010}\)
\(\Rightarrow\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{x\left(x+1\right)}=1\frac{2008}{2010}\):2
\(\Rightarrow1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2009}{2010}\)
\(\Rightarrow1-\frac{1}{x+1}=\frac{2009}{2010}\)
\(\Rightarrow1-\frac{2009}{2010}=\frac{1}{x+1}\)
\(\Rightarrow\frac{1}{2010}=\frac{1}{x+1}\)
\(\Rightarrow x=2009\)
nha !
Ta có :A=1+\(\frac{2}{6}\)+\(\frac{2}{12}\)+......+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{4018}{2010}\)
\(\Rightarrow\)A=\(\frac{2}{2.3}\)+\(\frac{2}{3.4}\)+...+\(\frac{2}{x\left(x+1\right)}\)=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2.3}\)+\(\frac{1}{3.4}\)+...+\(\frac{1}{x\left(x+1\right)}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{x}\)-\(\frac{1}{x+1}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=2(\(\frac{1}{2}\)-\(\frac{1}{x+1}\))=\(\frac{2008}{2010}\)
\(\Rightarrow\)A=\(\frac{1}{2}\)-\(\frac{1}{x+1}\)=\(\frac{502}{1005}\)
\(\Rightarrow\)\(\frac{1}{x+1}\)=\(\frac{1}{2010}\)\(\Rightarrow\)x+1=2010\(\Rightarrow\)x=2009