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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:
\(\frac{1}{n}-\frac{1}{n+a}=\frac{n+a}{n\left(n+a\right)}-\frac{n}{n\left(n+a\right)}=\frac{a}{n\left(n+a\right)}\)
\(\frac{1}{2.5}+\frac{1}{5.8}+...+\frac{1}{\left(3n-1\right).\left(3n+2\right)}=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right).\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\left(\frac{3n+2}{2.\left(3n+2\right)}-\frac{2}{2.\left(3n+4\right)}\right)\)
\(=\frac{1}{3}.\frac{3n}{2.\left(3n+2\right)}=\frac{n}{2.\left(3n+2\right)}\)
\(a)\) Đặt \(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{2010^2}\) ta có :
\(A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2009.2010}\)
\(A< \frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2009}-\frac{1}{2010}\)
\(A< 1-\frac{1}{2010}=\frac{2009}{2010}< 1\)
\(\Rightarrow\)\(A< 1\) ( đpcm )
Vậy \(A< 1\)
Chúc bạn học tốt ~