a) Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{10}}\)=> \(2.A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\)

=> \(2.A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)

\(A=1-\frac{1}{2^{10}}\)=> \(1-A=1-\left(1-\frac{1}{2^{10}}\right)=\frac{1}{2^{10}}>\frac{1}{2^{11}}\)=> đpcm

b) Đặt B = \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\)

Vì \(\frac{1}{2^2}

Ta có : \(\frac{1}{2^2}< \frac{1}{1\cdot2};\frac{1}{3^2}< \frac{1}{2\cdot3};\frac{1}{4^2}< \frac{1}{3\cdot4};...;\frac{1}{11^2}< \frac{1}{10\cdot11}\)

\(=>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{10\cdot11}\)

\(=>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}\)

\(=>\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< 1-\frac{1}{11}=\frac{10}{11}\)(đpcm)

Nếu bạn chưa hiểu thì bạn hỏi lại mình nhé! Chúc bạn học tốt!

Ta có:

\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}\)

\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{10.11}=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{10}-\frac{1}{11}=\frac{1}{2}-\frac{1}{11}=\frac{9}{22}\)

Mà \(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{11^2}< \frac{9}{22}< \frac{10}{11}\) nên \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{11^2}< \frac{10}{11}\)

bạn cho mình hỏi đpcm là j vậy

\(1-\frac{1}{2}-\frac{1}{2^2}-...-\frac{1}{2^{10}}\)

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

Đặt \(A=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\)

\(\Rightarrow2A=1+\frac{1}{2}+...+\frac{1}{2^9}\)

\(\Rightarrow2A-A=\left(1+\frac{1}{2}+...+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{10}}\right)\)

\(\Rightarrow A=1-\frac{1}{2^{10}}\)

Thay A vào (1)

\(\Rightarrow1-\left(1-\frac{1}{2^{10}}\right)\)

\(=1-1+\frac{1}{2^{10}}=\frac{1}{2^{10}}\)

Ta có: 2¹⁰ < 2¹¹

\(\Rightarrow\frac{1}{2^{10}}>\frac{1}{2^{11}}\)(đpcm)

\(N=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)

\(N< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)

\(N< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(N< 1-\frac{1}{100}\)

\(N< \frac{99}{100}< \frac{75}{100}=\frac{3}{4}\)

\(a,\)

Để A là phân số thì \(n-2\ne0\Rightarrow n\ne2\)

b, Ta có :

\(A=\frac{n+1}{n-2}=\frac{n-2+3}{n-2}=1+\frac{3}{n-2}\)

Mà \(3⋮n+2\Rightarrow n+2\inƯ(3)=\left\{\pm1;\pm3\right\}\)

Tự xét bảng

Dễ thấy : \(P< 1\)

Áp dụng kết quả bài trên nếu \(\frac{a}{b}< 1\)thì \(\frac{a+m}{b+m}>\frac{a}{b}\)với \((m>0)\)

Vậy \(P=\frac{10^{11}-1}{10^{12}-1}=\frac{(10^{11}-1)+11}{(10^{12}-1)+11}=\frac{10^{11}+10}{10^{12}+10}\)

\(P< \frac{10\cdot(10^{10}+1)}{10\cdot(10^{11}+1)}=\frac{10^{10}+1}{10^{11}+1}=Q\)

Do đó P < Q

đỡ hơn chưa??? mong các bn giúp mình vs

Vê trái: 

\(=\frac{2}{\left(x-1\right)\left(x+1\right)}+\frac{4}{\left(x-2\right)\left(x+2\right)}+...+\frac{20}{\left(x-10\right)\left(x+10\right)}\)

\(=\frac{\left(x+1\right)-\left(x-1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-10\right)}{\left(x+10\right)\left(x-10\right)}\)

\(=\frac{1}{x-1}-\frac{1}{x+1}+\frac{1}{x-2}-\frac{1}{x+2}+...+\frac{1}{x-10}-\frac{1}{x+10}\)

\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\)

Vế phải:

\(=\frac{\left(x+1\right)-\left(x-10\right)}{\left(x-10\right)\left(x+1\right)}+\frac{\left(x+2\right)-\left(x-9\right)}{\left(x-9\right)\left(x+2\right)}+...+\frac{\left(x+10\right)-\left(x-1\right)}{\left(x-1\right)\left(x+10\right)}\)

\(=\frac{1}{x-10}-\frac{1}{x+1}+\frac{1}{x-9}-\frac{1}{x+2}+...+\frac{1}{x-1}-\frac{1}{x+10}\)

\(=\left(\frac{1}{x-1}+\frac{1}{x-2}+...+\frac{1}{x-10}\right)-\left(\frac{1}{x+1}+\frac{1}{x+2}+...+\frac{1}{x+10}\right)\) = vế phải

=> đpcm

\(M=\frac{1}{3^2}+\frac{2}{3^3}+\frac{3}{3^4}+...+\frac{10}{3^{11}}\)

\(\Rightarrow3M=\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\)

\(\Rightarrow3M-M=\left(\frac{1}{3}+\frac{2}{3^2}+...+\frac{10}{3^{10}}\right)-\left(\frac{1}{3^2}+\frac{2}{3^3}+...+\frac{10}{3^{11}}\right)\)

\(\Rightarrow2M=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}-\frac{10}{3^{11}}\)

Đặt \(A=\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\)

\(\Rightarrow3A=1+\frac{1}{3}+...+\frac{1}{3^9}\)

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+...+\frac{1}{3^9}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{10}}\right)\)

\(\Rightarrow2A=1-\frac{1}{3^{10}}< 1\)

\(\Rightarrow2A< 1\)

\(\Rightarrow A< \frac{1}{2}\)

\(\Rightarrow2M< \frac{1}{2}-\frac{10}{3^{11}}\)

\(\Rightarrow M< \frac{\frac{1}{2}-\frac{10}{3^{11}}}{2}\)

\(\Rightarrow M< \frac{1}{4}-\frac{1}{2.3^{11}}< \frac{1}{4}\)

\(\Rightarrow M< \frac{1}{4}\left(đpcm\right)\)