.

A=(1/2^2-1) (1/3^2-1) (1/4^2-1) .... (1/100^2-1)

A=(1/2^2-2^2/2^2) (1/3^2-3^2/3^2) ...... (1/100^2-100^2/100^2)

A=1-2^2/2^2 . 1-3^2/3^2 .... 1-100^2/100^2

A=-(2^2-1/2^2 . 3^2-1/3^2 ..... 100^2-1/100^2)

A=-(1.3/2^2 x 2.4/3^2 ..... 99.101/100^2)

A=-(1.3.2.4.....99.101/2.2.3.3.....100.100)

A=-[(1.2.3....99).(3.4.5.....101) / (2.3.4...100) . (2.3.4...100) ]

A=-101/200 < -1/2

A   = \(\dfrac{1}{2^2}\) + \(\dfrac{1}{4^2}\) + \(\dfrac{1}{6^2}\)+.......+\(\dfrac{1}{\left(2.n\right)^2}\)

A  =  \(\dfrac{1}{2^2}\) + \(\dfrac{1}{\left(2.2\right)^2}\)+ \(\dfrac{1}{\left(2.3\right)^2}\) +....+\(\dfrac{1}{\left(2.n\right)^2}\)

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

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

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

     4A =  1 + \(\dfrac{1}{2^2}\) + \(\dfrac{1}{3^2}\) +......+ \(\dfrac{1}{n^2}\)

     4A = 1 + \(\dfrac{1}{2.2}\) + \(\dfrac{1}{3.3}\) + ...+\(\dfrac{1}{n.n}\)

       1   = 1

     \(\dfrac{1}{2.2}\) < \(\dfrac{1}{1.2}\)

      \(\dfrac{1}{3.3}\) < \(\dfrac{1}{2.3}\)

     ...................

 \(\dfrac{1}{n.n}\) < \(\dfrac{1}{\left(n-1\right).n}\)

Cộng vế với vế ta có: 

4A = 1+\(\dfrac{1}{2.2}\)+\(\dfrac{1}{3.3}\)+....+\(\dfrac{1}{n.n}\) <1+ \(\dfrac{1}{1.2}\)+ \(\dfrac{1}{2.3}\)+ ......+ \(\dfrac{1}{\left(n-1\right).n}\)

4A < 1+ \(\dfrac{1}{1}\) - \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\) - \(\dfrac{1}{3}\)+....+\(\dfrac{1}{n-1}\) - \(\dfrac{1}{n}\) = 2 - \(\dfrac{1}{n}\)

A < ( 2 - \(\dfrac{1}{n}\)): 4 

A < 2 : 4 - \(\dfrac{1}{n}\) : 4

A < \(\dfrac{1}{2}\) - \(\dfrac{1}{4n}\) < \(\dfrac{1}{2}\)

Vậy A < \(\dfrac{1}{2}\) 

Ta có :2²A=1+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{4^2}\)+...+\(\dfrac{1}{n^2}\)

            2²A-A=1-\(\dfrac{1}{\left(2n\right)^2}\)

            3A=\(\dfrac{\left(2n\right)^2-1}{\left(2n\right)^2}\) <\(\dfrac{n^2}{\left(2n\right)^2}\)=\(\dfrac{1}{2}\)

          3A<\(\dfrac{1}{2}\) suy ra A<\(\dfrac{1}{2}\)

1/3²< 1/2.3

1/4²< 1/3.4

...

1/100²< 1/99.100

=> 1/2² + 1/3² + 1/4² + ... + 1/100²< 1/2² + 1/2.3 + 1/3.4 + ... + 1/99.100

A < 1/4 + 1/2 -1/3 + 1/3 - 1/4 +... + 1/99 - 1/100

A < 1/4 + 1/2 -1/100 < 1/4 + 1/2 = 3/4

=> A < 3/4

a) Trước hết ta chứng minh \(a^2-1=\left(a-1\right)\left(a+1\right)\text{tự chứng minh }\)

Áp dụng bổ đề trên ta có:

\(-A=\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)\cdot...\cdot\left(1-\dfrac{1}{100^2}\right) =\dfrac{2^2-1}{2^2}\cdot\dfrac{3^2-1}{3^2}\cdot...\cdot\dfrac{100^2-1}{100^2}=\dfrac{1\cdot3}{2^2}\cdot\dfrac{2\cdot4}{3^2}\cdot...\cdot\dfrac{99\cdot101}{100^2}=\dfrac{1\cdot2\cdot3^2\cdot...\cdot99^2\cdot100\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\dfrac{1\cdot101}{2\cdot100}>\dfrac{1}{2}\\ \Rightarrow A< -\dfrac{1}{2}\)

b)

TH1: x chẵn  mà x là số nguyên tố => x=2

=> y^2 = 117+4=121 => y=11 (thỏa mãn)

TH2:  x lẻ => x^2 lẻ  . Mà 117 lẻ

=> x^2+117 chẵn => y^2 chẵn => y chẵn mà y là số nguyên tố

=> y=2 

=>x^2+117= 4=> x^2 = -113 (vô lý)

Vậy x=2;y=11

\(M=1-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\right)\)

Đặt \(N=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{10}}\)

\(2N=1+\dfrac{1}{2}+...+\dfrac{1}{2^9}\)

\(\Rightarrow2N-N=1-\dfrac{1}{2^{10}}\)

\(\Rightarrow N=1-\dfrac{1}{2^{10}}\)

\(\Rightarrow M=1-\left(1-\dfrac{1}{2^{10}}\right)=\dfrac{1}{2^{10}}>\dfrac{1}{2^{11}}\)

Vậy \(M>\dfrac{1}{2^{11}}\)

Ta có:

\(A=\left(\dfrac{1}{2}-1\right)\cdot\left(\dfrac{1}{3}-1\right)\cdot\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{10}-1\right)\)

\(A=-\dfrac{1}{2}\cdot-\dfrac{2}{3}-\dfrac{3}{4}\cdot...\cdot-\dfrac{9}{10}\)

\(A=\dfrac{-1\cdot-2\cdot-3\cdot...\cdot-9}{2\cdot3\cdot4\cdot...\cdot10}\)

\(A=-\dfrac{1}{10}\)

Mà: \(10>9\)

\(\Rightarrow\dfrac{1}{10}< \dfrac{1}{9}\)

\(\Rightarrow-\dfrac{1}{10}>-\dfrac{1}{9}\)

\(\Rightarrow A>-\dfrac{1}{9}\)

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{n\left(n-1\right)}\\ A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}=1-\dfrac{1}{n}< 1\left(\dfrac{1}{n}>0\right)\)

\(A=\left(\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}}{x+\sqrt{x}+1}+\dfrac{1}{1-\sqrt{x}}\right):\dfrac{\sqrt{x}-1}{2}\left(dkxd:x\ge0;x\ne1\right)\)

\(=\left[\dfrac{x+2}{\left(\sqrt{x}\right)^3-1}+\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1}{\sqrt{x}-1}\right]\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\left[\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{x-\sqrt{x}}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{x+\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\right]\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{x+2+x-\sqrt{x}-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\cdot\dfrac{2}{\sqrt{x}-1}\)

\(=\dfrac{\left(x-2\sqrt{x}+1\right)\cdot2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)\cdot\left(\sqrt{x}-1\right)}\)

\(=\dfrac{\left(\sqrt{x}-1\right)^2\cdot2}{\left(\sqrt{x}-1\right)^2\cdot\left(x+\sqrt{x}+1\right)}\)

\(=\dfrac{2}{x+\sqrt{x}+1}\)

Xét: \(A-2=\dfrac{2}{x+\sqrt{x}+1}-2\)

\(=\dfrac{2}{x+\sqrt{x}+1}-\dfrac{2\left(x+\sqrt{x}+1\right)}{x+\sqrt{x}+1}\)

\(=\dfrac{2-2x-2\sqrt{x}-2}{x+\sqrt{x}+1}\)

\(=\dfrac{-2x-2\sqrt{x}}{x+\sqrt{x}+1}\)

\(=\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\)

Với \(x\ge0;x\ne1\Leftrightarrow\left\{{}\begin{matrix}x+\sqrt{x}\ge0\\x+\sqrt{x}+1>0\end{matrix}\right.\)

\(\Leftrightarrow\dfrac{x+\sqrt{x}}{x+\sqrt{x}+1}\ge0\)

\(\Leftrightarrow\dfrac{-2\left(x+\sqrt{x}\right)}{x+\sqrt{x}+1}\le0\)

\(\Rightarrow A-2\le0\Leftrightarrow A\le2\)

Vậy: \(A\le2\).