Gọi i là đại diện cho các số từ 1 đến 2011

ĐKXĐ:  \(a_i\ne0\left(i=1,2,3,..,2011\right)\)  

Xét \(a_i=1\)  Ta có: \(\frac{1}{a^{11}_i}=1>\frac{2011}{2048}\Rightarrow\frac{1}{x^{11}_1}+\frac{1}{x^{11}_2}+...+\frac{1}{x^{11}_{2011}}>\frac{2011}{2048}\left(loai\right)\) 

Xét \(a_i\ge2\) Ta có: \(\frac{1}{a^{11}_i}\le\frac{1}{2048}\Rightarrow\frac{1}{x^{11}_1}+\frac{1}{x^{11}_2}+...+\frac{1}{x^{11}_{2011}}\le\frac{2011}{2048}\)

Dấu "=" xảy ra khi \(a_i=2\) 

Thay vào ta có: 

\(M=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2011}}\) 

\(\Rightarrow2M-M=\left(1+\frac{1}{2}+...+\frac{1}{2^{2010}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2011}}\right)\) 

\(\Rightarrow M=1-\frac{1}{2^{2011}}\)

M=a+b=c+d=e+f.M=a+b=c+d=e+f.

⇒⎧⎪ ⎪ ⎪ ⎪⎨⎪ ⎪ ⎪ ⎪⎩a7=b11=a+b7+11=M18(1)c11=d13=c+d11+13=M24(2)e13=f17=e+f13+17=M30(3)⇒{a7=b11=a+b7+11=M18(1)c11=d13=c+d11+13=M24(2)e13=f17=e+f13+17=M30(3)

Kết hợp (1),(2)và(3)(1),(2)và(3)

⇒M∈BCNN(18;24;30).⇒M∈BCNN(18;24;30).

⇒M∈{0;360;720;1080;...}⇒M∈{0;360;720;1080;...}

Mà MM là số tự nhiên nhỏ nhất có 4 chữ số.

⇒M=1080.⇒M=1080.

Vậy M=1080.

nhớ cho mình 1 k nhé chúc bạn học tốt

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

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

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

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

=> A > B

Vậy A > B

Phương thảo nhé

có ai kb với mik ko

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

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

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

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

=> A > B

Vậy A > B

\(3,\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)^2-\frac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\frac{2}{xy}:\left[\left(\frac{1}{x}\right)^2-2.\frac{1}{x}.\frac{1}{y}+\left(\frac{1}{y}\right)^2\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\frac{2}{xy}:\left[\frac{1}{x^2}-\frac{2}{xy}+\frac{1}{y^2}\right]-\frac{x^2+y^2}{x^2-2xy+y^2}\)

\(=\frac{2}{xy}:\left[\frac{y^2-2.xy+x^2}{x^2y^2}\right]-\frac{x^2+y^2}{\left(x-y\right)^2}\)

\(=\frac{2}{xy}.\frac{x^2y^2}{x^2-2xy+y^2}-\frac{x^2+y^2}{x^2-2xy+y^2}\)

\(=\frac{2xy}{x^2-2xy+y^2}+\frac{-x^2-y^2}{x^2-2xy-y^2}\)

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

\(\frac{2011^3+11^3}{2011^3+2000^3}\)

\(=\frac{\left(2011+11\right)\left(2011^2-2011.11+11^2\right)}{\left(2011+2000\right)\left(2011^2-2011.2000+2000^2\right)}\)

\(=\frac{\left(2011+11\right)\left[2011^2-11\left(2011-11\right)\right]}{\left(2011+2000\right)\left[2011^2-2000\left(2011-2000\right)\right]}\)

\(=\frac{\left(2011+11\right)\left(2011^2-11.2000\right)}{\left(2011+2000\right)\left(2011^2-2000.11\right)}\)

\(=\frac{2011+11}{2011+2000}\left(2011^2-11.2000\ne0\right)\)

                                          đpcm

\(A=\left(\frac{a+1}{ab+1}+\frac{ab+a}{ab-1}-1\right):\left(\frac{a+1}{ab+1}-\frac{ab+a}{ab-1}+1\right)\)

\(A=\left[\frac{\left(a+1\right)\left(ab-1\right)+\left(ab+a\right)\left(ab+1\right)-\left(ab+1\right)\left(ab-1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{\left(a+1\right)\left(ab-1\right)-\left(ab+a\right)\left(ab+1\right)+\left(ab+1\right)\left(ab-1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]\)\(A=\left[\frac{a^2b-a+ab-1+a^2b^2+ab+a^2b+a-a^2b^2+1}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{a^2b-a+ab-1-a^2b^2-ab-a^2b-a+a^2b^2-1}{\left(ab+1\right)\left(ab-1\right)}\right]\)\(A=\left[\frac{2a^2b+2ab}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{2a^2b-2a}{\left(ab+1\right)\left(ab-1\right)}\right]\)

\(A=\left[\frac{2ab\left(a+1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{2a\left(ab-1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]\)

\(A=\left[\frac{2ab\left(a+1\right)}{\left(ab+1\right)\left(ab-1\right)}\right]:\left[\frac{2a}{\left(ab+1\right)}\right]\left(ab-1\ne0\right)\)

\(A=\frac{b\left(a+1\right)}{ab-1}\left(ab+1\ne0;2a\ne0\right)\)

1)

\(\frac{a}{b}=\frac{a\left(b+c\right)}{b\left(b+c\right)}=\frac{ab+ac}{b\left(b+c\right)}\)

\(\frac{a+c}{b+c}=\frac{b\left(a+c\right)}{b\left(b+c\right)}=\frac{ab+bc}{b\left(b+c\right)}\)

mà ab = ab; ac > bc ( vì a > b )

=> \(\frac{a}{b}>\frac{a+c}{b+c}\left(đpcm\right)\)