=>1/2^2+...+1/1990^2<1/1990<3/4 

xét vế trái

=(1+1/3+1/5+...+1/1989)-(1/2+1/4+...+1/1990)

=(1+1/2+1/3+1/4+...+1/1990)-2.(1/2+1/4+...+1/1990)

=(1+1/2+1/3+1/4+...+1/1990)-!1+1/2+1/3+1/4+...+1/995)

=1/996+1/997+.../1+1990

vậy 1-1/2+1/3-1/4+...-1/1990=1/996+1/997+...+1/1990

cmr 1-$\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+.......-\frac{1}{1990}=\frac{1}{996}+\frac{1}{997}+\frac{1}{998}+.......+\frac{1}{1990}$

dòng dấu = thứ 3 sửa ! thành ( nha

Đặt \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}=A\)

ta có :\(\frac{1}{2^2}=\frac{1}{2\cdot2}=\frac{1}{4}\)

\(\frac{1}{3^2}=\frac{1}{3\cdot3}< \frac{1}{2\cdot3}\)

\(...\)

\(\frac{1}{1990^2}=\frac{1}{1990\cdot1990}< \frac{1}{1989\cdot1990}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2\cdot3}+...+\frac{1}{1989\cdot1990}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1989}-\frac{1}{1990}\)

\(\Rightarrow A< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)

\(\Rightarrow A< \frac{3}{4}\left(ĐPCM\right)\)

Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)

hk tốt #

Ta có \(\frac{1}{3^2}< \frac{1}{2.3};\frac{1}{4^2}< \frac{1}{3.4};...;\frac{1}{1990^2}< \frac{1}{1989.1990}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{1}{2^2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{1989.1990}\)

                                                                     \(< \frac{1}{4}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{1989}-\frac{1}{1990}\)

                                                                    \(< \frac{1}{4}+\frac{1}{2}-\frac{1}{1990}=\frac{3}{4}-\frac{1}{1990}< \frac{3}{4}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{1990^2}< \frac{3}{4}\)

\(\Rightarrow\)Bài toán được chứng minh

dễ thì đăng đáp an đi

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