\(\frac{1}{2^2}+\frac{1}{4^2}+\frac{1}{6^2}+...+\frac{1}{2016^2}\)

\(=\frac{1}{2^2}.\left(1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{1008^2}\right)< \frac{1}{2^2}.\left(1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{1007.1008}\right)\)

                                                                         \(< \frac{1}{4}.\left(1+1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{1007}-\frac{1}{1008}\right)\)

                                                                           \(< \frac{1}{4}.\left(2-\frac{1}{1008}\right)< \frac{1}{4}.2=\frac{1}{2}\)

=> đpcm

đặt A=1/3²+1/4²+1/5²+……1/100²

B=1/2.3+1/3.4+...+1/99.100

=1/2-1/3...+1/99-1/100

=1/2-1/100<1/2 (1)

mà A=1/3²+1/4²+1/5²+……1/100²<B=1/2.3+1/3.4+...+1/99.100 (2)

kết hợp từ (1),(2)ta được A<B<1/2

=>A<1/2

1) 1-2+3-4+...+99-100

=(1-2)+(3-4)+...+(99-100)

=  -1.(100:2)

= -50

2) 2-4+6-8+...+48-50

=(2-4)+(6-8)+...+(48-50)

=  -2.(50:2)

= -50

3) 

=(-1+3-5)+...+(-95+97-99)

= -3.(99:3)

=-99

4)

=(1+2-3-4+5)+...+(-96+97+98-99-100)

=  1.(100:5)

= 20

Chúc bạn học tốt

A = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\) + .....+ \(\dfrac{1}{1002^2}\)

A = \(\dfrac{1}{2^2.1^2}\) + \(\dfrac{1}{2^2.2^2}\) + \(\dfrac{1}{2^2.3^2}\)+......+\(\dfrac{1}{2^2.501^2}\)

A = \(\dfrac{1}{2^2}\) \(\times\)( \(1\) + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+.......+ \(\dfrac{1}{501^2}\))

ta có : \(\dfrac{1}{2^2}\)   < \(\dfrac{1}{1.2}\)

           \(\dfrac{1}{3^2}\)   < \(\dfrac{1}{2.3}\)

          ................

         \(\dfrac{1}{501^2}\) < \(\dfrac{1}{500.501}\)

Cộng vế với vế ta được

           \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +.....+ \(\dfrac{1}{501^2}\) < \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+.......+\dfrac{1}{500.501}\)

           \(\dfrac{1}{2^2}\) +  \(\dfrac{1}{3^2}\) +.....+ \(\dfrac{1}{501^2}\) < \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\) + \(\dfrac{1}{2}-\dfrac{1}{3}\)+.....+ \(\dfrac{1}{500}-\dfrac{1}{501}\)

            \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\)+......+ \(\dfrac{1}{501^2}\) < 1 - \(\dfrac{1}{501}\) < 1 

   =>A = \(\dfrac{1}{4}\) \(\times\) ( 1 + \(\dfrac{1}{2^2}\)+ \(\dfrac{1}{3^2}\)+.....+\(\dfrac{1}{501^2}\)) < \(\dfrac{1}{4}\) \(\times\)(1 + 1)

    A <  \(\dfrac{1}{4}\)  \(\times\) 2

    A < \(\dfrac{1}{2}\)

Chứng minh bằng phép biến đổi tương đương:

1.

\(\Leftrightarrow4+x+y\ge4\sqrt{x+y}\)

\(\Leftrightarrow x+y-4\sqrt{x+y}+4\ge0\)

\(\Leftrightarrow\left(\sqrt{x+y}-2\right)^2\ge0\) (luôn đúng)

Vậy BĐT đã cho đúng

2.

\(\Leftrightarrow\dfrac{y+z}{xyz}\ge\dfrac{4}{x^2+yz}\)

\(\Leftrightarrow\left(y+z\right)\left(x^2+yz\right)\ge4xyz\)

\(\Leftrightarrow x^2y+x^2z+y^2z+z^2y-4xyz\ge0\)

\(\Leftrightarrow y\left(x^2+z^2-2xz\right)+z\left(x^2+y^2-2xy\right)\ge0\)

\(\Leftrightarrow y\left(x-z\right)^2+z\left(x-y\right)^2\ge0\) (đúng)

\(\frac{1}{3^2}<\frac{1}{2.3}\)

\(\frac{1}{4^2}<\frac{1}{3.4}\)

...

\(\frac{1}{100^2}<\frac{1}{99.100}\)

===>\(\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}<\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}=\frac{1}{2}-\frac{1}{100}=\frac{49}{100}<\frac{50}{100}=\frac{1}{2}\)