\(\text{A = }\frac{\text{-1}}{\text{2011}}-\frac{\text{3}}{\text{11}^2}-\frac{\text{5}}{\text{11}^2.\text{11}}-\frac{\text{7}}{\text{11}^2.\text{11}^2}=\text{ }\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)\)

\(\text{B = }\frac{\text{-1}}{\text{2011}}-\frac{7}{\text{11}^2}-\frac{5}{\text{11}^2.\text{11}}-\frac{3}{\text{11}^2.\text{11}^2}=\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

\(\text{Vì }3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}< 7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\)

\(\Rightarrow\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(3-\frac{\text{5}}{\text{11}}-\frac{\text{7}}{\text{11}^2}\right)>\frac{\text{-1}}{\text{2011}}-\frac{\text{1}}{\text{11}^2}.\left(7-\frac{5}{\text{11}}-\frac{3}{\text{11}^2}\right)\)

=> A > B

Vậy A > B

\(A=\frac{x}{x-1}+\frac{x}{x+1}+\frac{2x^2}{1-x^2}\)

\(A=\frac{x}{x-1}+\frac{x}{x+1}+\frac{-2x^2}{x^2-1}\)

\(A=\frac{x\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{x\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{-2x^2}{\left(x+1\right)\left(x-1\right)}\)

\(A=\frac{x^2+x+x^2-x-2x^2}{\left(x+1\right)\left(x-1\right)}=\frac{1}{\left(x+1\right)\left(x-1\right)}\)

đề s ý 

Đề đúng mà bạn

1212;

1212;

1212.

k cho mình nhé.

1212

tk nhe@@@@@@@@@@@!!

aitk minh minh tk lai

bye

Ta so sánh từng số hạng : 

\(\frac{\sqrt{2}-\sqrt{1}}{1+2}=\frac{\left(\sqrt{2}-\sqrt{1}\right)\left(\sqrt{2}+\sqrt{1}\right)}{\left(1+2\right)\left(\sqrt{2}+\sqrt{1}\right)}=\frac{1}{\left(1+2\right)\left(\sqrt{2}+\sqrt{1}\right)}< \frac{1}{2}\)

\(\frac{\sqrt{3}-\sqrt{2}}{2+3}=\frac{\left(\sqrt{3}-\sqrt{2}\right)\left(\sqrt{3}+\sqrt{2}\right)}{\left(2+3\right)\left(\sqrt{3}+\sqrt{2}\right)}=\frac{1}{\left(2+3\right)\left(\sqrt{2}+\sqrt{3}\right)}< \frac{1}{2}\)

..........................................................................................................................................................................................

\(\frac{\sqrt{2015}-\sqrt{2014}}{2014+2015}=\frac{\left(\sqrt{2015}-\sqrt{2014}\right)\left(\sqrt{2015}+\sqrt{2014}\right)}{\left(2014+2015\right)\left(\sqrt{2015}+\sqrt{2014}\right)}=\frac{1}{\left(2014+2015\right)\left(\sqrt{2015}+\sqrt{2015}\right)}< \frac{1}{2}\)

Vì mỗi số hạng của M đều nhỏ hơn 1/2 nên M < 1/2

Bài này mình làm chưa đúng nhé :) Để lát mình làm cách khác.

Ta có:

\(\frac{\sqrt{n+1}-\sqrt{n}}{n\left(n+1\right)}=\frac{\sqrt{n+1}-\sqrt{n}}{2n+1}\)\(=\frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n+1}}\)

\(< \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{4n^2+4n}}=\frac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n\left(n+1\right)}}\)\(=\frac{1}{2}\left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)\)

Áp dụng vào bài toán ta có:

\(M< \frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}\right)\)\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{2015}}\right)\)

cách làm thì ko biết, nhưng kq thì CÓ LẼ là : M < 0,5

do M= 0,4783935314...  < 0,5