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\(\begin{equation} x = a_0 + \cfrac{1}{740_1 + \cfrac{1}{897654_2 + \cfrac{1}{672_3 + \cfrac{1}{100_4} } } } \end{equation}\)
Bài 1:
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{\left(x-1\right).x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4}.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{x-1}-\frac{1}{x}\right)\)\(=\frac{11}{48}\)
\(\frac{1}{4.}.\left(1-\frac{1}{x}\right)=\frac{11}{48}\)
\(1-\frac{1}{x}=\frac{11}{48}:\frac{1}{4}\)
\(1-\frac{1}{x}=\frac{11}{12}\)
\(\frac{1}{x}=1-\frac{11}{12}\)
\(\frac{1}{x}=\frac{1}{12}\)
Vậy x= 12
Bài 2 :
Xét vế trái ta có :
\(\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{\left(3n-1\right).\left(3n+2\right)}\)
\(=\frac{1}{3}.\left(\frac{3}{2.5}+\frac{3}{5.8}+...+\frac{3}{\left(3n-1\right)\left(3n+2\right)}\right)\)
\(=\frac{1}{3}.\left(\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+...+\frac{1}{3n-1}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}\left(\frac{1}{2}-\frac{1}{3n+2}\right)\)
\(=\frac{1}{3}.\frac{1}{2\left(3n+2\right)}=\frac{n}{2\left(3n+2\right)}\)
VẾ TRÁI ĐÚNG BẰNG VẾ PHẢI .ĐẲNG THỨC ĐÃ CHỨNG TỎ LÀ ĐÚNG
cHÚC BẠN HỌC TỐT ( -_- )
\(a)\) 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 ~
\(\frac{1}{2.4}+\frac{1}{4.6}+...+\frac{1}{\left(2x-2\right).2x}=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+...+\frac{2}{\left(2x-2\right).2x}\right)=\frac{1}{8}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{2x-2}-\frac{1}{2x}=\frac{1}{8}:\frac{1}{2}\)
\(\Rightarrow\frac{1}{2}-\frac{1}{2x}=\frac{1}{4}\)
\(\Rightarrow\frac{1}{2x}=\frac{1}{2}-\frac{1}{4}=\frac{1}{4}\)
\(\Leftrightarrow2x=4\)
\(\Leftrightarrow x=2\)
TL:
\(\frac{1}{2}\left(\frac{2}{2.4}+\frac{2}{4.6}+....+\frac{2}{\left(2x-2\right)2x}\right)=\frac{1}{8}\)
\(\frac{1}{2}-\frac{1}{4x}=\frac{1}{8}\)
\(\frac{1}{4x}=\frac{3}{8}\)
=>x=2/3
hc tốt
\(M=\frac{1}{2_{^{^2}}}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{n^2}\)
< \(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{\left(n-1\right).n}\)
= \(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{n-1}-\frac{1}{n}\)
=\(1-\frac{1}{n}<1\)
\(\Rightarrow M<1\)