Đáp án:

$\displaystyle \begin{array}{{>{\displaystyle}l}} 102.\ B=1\\ 103.\ b.\ S=2011-\frac{1}{2011} \end{array}$

Giải thích các bước giải:

$\displaystyle \begin{array}{{>{\displaystyle}l}} 95.\ \\ Số\ hữu\ tỉ\ là:\ \sqrt{144} =12;-\sqrt{\frac{75}{192}} =-\frac{5}{8} ;\sqrt{\frac{225}{36}} =\frac{5}{2} ;\\ \sqrt{16} =4;\sqrt{0,16} =0,4\\ Số\ vô\ tỉ\ là:\ \sqrt{40} ;\sqrt{1,6}\\ 102.\\ B=\sqrt{( x+1)^{2}} -\sqrt{x^{2}} \ với\ x\geqslant -1\\ Ta\ có:\ x\geqslant -1\Rightarrow x^{2} \geqslant 1\ và( x+1)^{2} \geqslant 0\\ \Rightarrow B=x+1-x=1\\ 103.\ a.\ Với\ n >0\ ta\ có:\\ A=1+\frac{1}{n^{2}} +\frac{1}{( n+1)^{2}} =\frac{n^{2}( n+1)^{2} +( n+1)^{2} +n^{2}}{n^{2}( n+1)^{2}}\\ =\frac{n^{4} +2n^{3} +3n^{2} +2n+1}{n^{2}( n+1)^{2}}\\ Ta\ có:\ n^{4} +2n^{3} +3n^{2} +2n+1\\ =n^{4} +n^{3} +n^{2} +n^{3} +n^{2} +n+n^{2} +n+1\\ =n^{2}\left( n^{2} +n+1\right) +n\left( n^{2} +n+1\right) +1\left( n^{2} +n+1\right)\\ =\left( n^{2} +n+1\right)\left( n^{2} +n+1\right) =\left( n^{2} +n+1\right)^{2}\\ \Rightarrow A=\frac{\left( n^{2} +n+1\right)^{2}}{n^{2}( n+1)^{2}} \Rightarrow đpcm\\ b.\ S=\sqrt{1+\frac{1}{1^{2}} +\frac{1}{2^{2}}} +\sqrt{1+\frac{1}{2^{2}} +\frac{1}{3^{2}}} +...+\sqrt{1+\frac{1}{2010^{2}} +\frac{1}{2011^{2}}}\\ Áp\ dụng\ câu\ a\ ta\ có:\\ \sqrt{1+\frac{1}{1^{2}} +\frac{1}{2^{2}}} =\sqrt{\frac{\left( 1^{2} +1+1\right)^{2}}{1^{2}( 1+1)^{2}}} =\frac{3}{2} =1+\frac{1}{1.2} =1+1-\frac{1}{2}\\ \sqrt{1+\frac{1}{2^{2}} +\frac{1}{3^{2}}} =\sqrt{\frac{\left( 2^{2} +2+1\right)^{2}}{2^{2}( 2+1)^{2}}} =\frac{7}{6} =1+\frac{1}{2.3} =1+\frac{1}{2} -\frac{1}{3}\\ ........\\ \sqrt{1+\frac{1}{2010^{2}} +\frac{1}{2011^{2}}} =\sqrt{\frac{\left( 2010^{2} +2010+1\right)^{2}}{2010^{2}( 2010+1)^{2}}} =1+\frac{1}{2010.2011} =1+\frac{1}{2010} -\frac{1}{2011}\\ \Rightarrow S=1+\frac{1}{1} -\frac{1}{2} +1+\frac{1}{2} -\frac{1}{3} +........+1+\frac{1}{2010} -\frac{1}{2011}\\ =2010+1-\frac{1}{2} +\frac{1}{2} -\frac{1}{3} +.........+\frac{1}{2010} -\frac{1}{2011}\\ =2010+1-\frac{1}{2011} =2011-\frac{1}{2011} \end{array}$