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Chứng minh rằng
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}\) < \(\frac{3}{4}\)
Vì \(\frac{1}{2^2}< \frac{3}{4}\)
\(\frac{1}{3^2}< \frac{3}{4}\)
...
\(\frac{1}{1990^2}< \frac{3}{4}\)
=> Tổng đó bé hơn \(\frac{3}{4}\)
\(\frac{1}{2^2}< \frac{1}{2}\left(1-\frac{1}{3}\right)\)
\(\frac{1}{1990^2}< \frac{1}{2}\left(\frac{1}{1989}-\frac{1}{1991}\right)\)
\(VP< \frac{1}{2}\left(1-\frac{1}{1991}\right)=\frac{1990}{2.1991}=\frac{995}{1991}< \frac{3}{4}\)
a.\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\Rightarrow\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}-\frac{x+1}{13}-\frac{x+1}{14}=0\)
\(\Rightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
Mà: \(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\Rightarrow x+1=0\Rightarrow x=-1\)
b.
\(\frac{x+4}{1990}+\frac{x+3}{1991}=\frac{x+2}{1992}+\frac{x+1}{1993}\Rightarrow2+\frac{x+4}{1990}+\frac{x+3}{1991}=2+\frac{x+2}{1992}+\frac{x+1}{1993}\)
\(\Rightarrow\left(1+\frac{x+4}{1990}\right)+\left(1+\frac{x+3}{1991}\right)=\left(1+\frac{x+2}{1992}\right)+\left(1+\frac{x+1}{1993}\right)\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}=\frac{x+1994}{1992}+\frac{x+1994}{1993}\)
\(\Rightarrow\frac{x+1994}{1990}+\frac{x+1994}{1991}-\frac{x+1994}{1992}-\frac{x+1994}{1993}=0\)
\(\Rightarrow\left(x+1994\right)\left(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\right)=0\)
\(\frac{1}{1990}+\frac{1}{1991}-\frac{1}{1992}-\frac{1}{1993}\ne0\Rightarrow x+1994=0\Rightarrow x=-1994\)
\(A=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}\)
\(A< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)
\(A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)
\(A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}< \frac{1}{4}+\frac{1}{2}\)
\(A< \frac{1}{4}+\frac{2}{4}=\frac{3}{4}\left(đpcm\right)\)
\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1990^2}<\frac{1}{2^2-1}+\frac{1}{3^2-1}+...+\frac{1}{1990^2-1}=\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{1989.1991}=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{1989}-\frac{1}{1991}\right)\)
\(=\frac{1}{2}.\left(1-\frac{1}{1991}\right)=\frac{1}{2}.\frac{1990}{1991}=\frac{995}{1991}<\frac{1}{2}<\frac{3}{4}\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1990^2}<\frac{3}{4}\)
Có :
\(\frac{1}{3^2}< \frac{1}{2.3}\)
...
\(\frac{1}{1990^2}< \frac{1}{1989.1990}\)
=> \(M< \frac{1}{2^2}+\frac{1}{2.3}+...+\frac{1}{1989.1990}\)
=> \(M< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}\)
=> \(M< \frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)
Vậy M < 3/4