Ta có  4A=\(1+\frac{1}{2^2}+\frac{1}{2^4}+...+\frac{1}{2^{98}}\)

Trừ 4A cho A ta được 

3A = \(1-\frac{1}{2^{100}}\)=> 3A <1 => A<1/3 (đpcm)

Chúc bạn học tốt 

Ta có :\(A=\frac{1}{2^2}+...+\frac{1}{2^{100}}\)

\(2A=\frac{1}{2}+...+\frac{1}{2^{99}}\)

\(2A-A=\left(\frac{1}{2}+...+\frac{1}{2^{99}}\right)-\left(\frac{1}{2^2}+...+\frac{1}{2^{100}}\right)\)

\(A=\frac{1}{2}-\frac{1}{2^{100}}\)

Lại có :

\(\frac{1}{3}=\frac{1}{2}-\frac{1}{6}\)

Vì \(\frac{1}{2^{100}}< \frac{1}{6}\)

\(\Rightarrow\frac{1}{2}-\frac{1}{2^{100}}>\frac{1}{2}-\frac{1}{6}\)

\(\Rightarrow A>\frac{1}{3}\)

Vậy \(A>\frac{1}{3}\)(ĐPCM)

Ta có :

\(\frac{1}{5^2}>\frac{1}{5.6}\)

\(\frac{1}{6^2}>\frac{1}{6.7}\)

\(..............\)

\(\frac{1}{100^2}>\frac{1}{100.101}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}>\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{100.101}\)

\(\Rightarrow A>\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{100}-\frac{1}{101}\)

\(\Rightarrow A>\frac{1}{5}-\frac{1}{101}=\frac{96}{505}>\frac{96}{576}=\frac{1}{6}\left(1\right)\)

Lại có :

\(\frac{1}{5^2}< \frac{1}{4.5}\)

\(\frac{1}{6^2}< \frac{1}{5.6}\)

\(...............\)

\(\frac{1}{100^2}< \frac{1}{99.100}\)

\(\Rightarrow A=\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}< \frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}\)

\(\Rightarrow A< \frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)

\(\Rightarrow A< \frac{1}{4}-\frac{1}{100}< \frac{1}{4}\left(2\right)\)

Từ (1) và (2) => Điều phải chứng minh

Đặt \(A=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{90}\)

         \(=\left(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\right)+\left(\frac{1}{46}+\frac{1}{47}+...+\frac{1}{90}\right)\)

Đặt \(B=\frac{1}{31}+\frac{1}{32}+...+\frac{1}{45}\)

 Ta có: \(\frac{1}{31}>\frac{1}{45}\)

           \(\frac{1}{32}>\frac{1}{45}\)

           ....................

          \(\frac{1}{45}=\frac{1}{45}\)

\(\Rightarrow B>\frac{1}{45}.15\)

\(\Rightarrow B>\frac{1}{3}\)

Đặt \(C=\frac{1}{46}+\frac{1}{47}+...+\frac{1}{90}\)

Ta có: \(\frac{1}{46}>\frac{1}{90}\)

           \(\frac{1}{47}>\frac{1}{90}\)

          .....................

         \(\frac{1}{90}=\frac{1}{90}\)

\(\Rightarrow C>\frac{1}{90}.45\)

\(\Rightarrow C>\frac{1}{2}\)

\(\Rightarrow B+C>\frac{1}{3}+\frac{1}{2}\)

Hay \(A>\frac{5}{6}\left(1\right)\)

Lại có: \(A=\left(\frac{1}{31}+...+\frac{1}{59}\right)+\left(\frac{1}{60}+...+\frac{1}{90}\right)\)

Đặt \(D=\frac{1}{31}+...+\frac{1}{59}\)

Ta có: \(\frac{1}{31}< \frac{1}{30}\)

          . ...................

           \(\frac{1}{59}< \frac{1}{30}\)

\(\Rightarrow D< \frac{1}{30}.60\)

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

Đăt \(E=\frac{1}{60}+...+\frac{1}{90}\)

Ta có: \(\frac{1}{60}=\frac{1}{60}\)

             .................

          \(\frac{1}{90}< \frac{1}{60}\)

\(\Rightarrow E< \frac{1}{60}.31\)

\(\Rightarrow E< \frac{31}{60}< 1\)

\(\Rightarrow E< 1\)

\(\Rightarrow E+D< 1+\frac{1}{2}\)

Hay \(A< \frac{3}{2}\left(2\right)\)

Từ (1) và (2) \(\Rightarrow\frac{5}{6}< A< \frac{3}{2}\)

Mình làm hơi ngáo có gì thì cứ nói 

\(\frac{1}{3^2}<\frac{1}{3.4}\)

\(\frac{1}{4^2}<\frac{1}{4.5}\)

\(\frac{1}{5^2}<\frac{1}{5.6}\)

\(...\)

\(\frac{1}{100^2}<\frac{1}{100.101}\)

\(\Rightarrow\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{100.101}\)

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{100}-\frac{1}{101}\)

\(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{3}-\frac{1}{101}\)

Mà \(\frac{1}{3}<\frac{1}{2}\) nên \(\frac{1}{3}-\frac{1}{101}<\frac{1}{2}\)

hay \(\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+...+\frac{1}{100^2}<\frac{1}{2}\)

Đặt A=1/3^2+1/4^2+1/5^2+...+1/100^2

Suy raA<1/2*3+1/3*4+1/4*5+..+1/99*100

A<1/2-1/100<1/2

Ta có điều phải chứng minh.

Đặt   \(A=\frac{1}{3}-\frac{2}{3^2}+...+\frac{99}{3^{99}}-\frac{100}{3^{100}}\)

\(\Rightarrow3A=1-\frac{2}{3}+...+\frac{99}{3^{98}}-\frac{100}{3^{99}}\)

\(4A=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)

Đặt    \(B=1+\frac{1}{3}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3B=3+1+...+\frac{3}{3^{98}}\)

\(2B=3-\frac{1}{3^{99}}\)

\(B=\frac{3}{2}-\frac{1}{3^{99}.2}\)

Thay B vào 4A ta có:

\(4A=\frac{3}{2}-\frac{1}{3^{99}.2}\)

\(A=\frac{3}{2.4}-\frac{1}{3^{99}.2.4}\)

\(A=\frac{3}{8}-\frac{1}{3^{99}.8}\)

Vì \(\frac{3}{8}>\frac{3}{16}\)

\(\Rightarrow\frac{3}{8}-\frac{1}{3^{99}.8}< \frac{3}{16}\)

Vậy \(A< \frac{3}{16}\)

a)\(-\frac{2}{5}+\frac{2}{3}x+\frac{1}{6}x=-\frac{4}{5}\Leftrightarrow\frac{5}{6}x=-\frac{2}{5}\Leftrightarrow x=-\frac{12}{25}\)
Vậy nghiệm là x = -12/25

b)\(\frac{3}{2}x-\frac{2}{5}-\frac{2}{3}x=-\frac{4}{15}\Leftrightarrow\frac{5}{6}x=\frac{2}{15}\Leftrightarrow x=\frac{4}{25}\)
Vậy nghiệm là x = 4/25

c)\(\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}=\frac{x+1}{13}+\frac{x+1}{14}\)\(\Leftrightarrow\left(x+1\right)\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\right)=0\)
\(\Leftrightarrow x+1=0\left(\frac{1}{10}+\frac{1}{11}+\frac{1}{12}-\frac{1}{13}-\frac{1}{14}\ne0\right)\)\(\Leftrightarrow x=-1\)
Vậy nghiệm là x = -1
 

Cảm ơn bạnh nha. Chúc bạn buổi tối ấm =)))) <3

Đề gõ sai, xin sửa lại:
Chứng minh:

\({1 \over {11}^2} + {1 \over {12}^2} + {1 \over {13}^2} + {1 \over {14}^2} + ... + {1 \over {100}^2}<{1 \over {10}}\)

Cảm ơn

Đặt biểu thức là A     ta có:

1/11^2 < 1/10.11 = 1/10 - 1/11

1/12^2 < 1/11.12 = 1/11 - 1/12

1/13^2 < 1/12.13 = 1/12 - 1/13

. . . . . . . . . 

1/100^2 < 1/99.100 = 1/99 - 1/100

 => A < 1/10 - 1/11 + 1/11 - 1/12 + 1/12 - 1/13 + . . . .+ 1/99 - 1/100

  => A < 1/10 -  1/100

 => A < 1/10

                Bạn nhớ k cho mình nha

Có\(\frac{1}{11^2}\)<\(\frac{1}{10.11}\);...;\(\frac{1}{100}\)<\(\frac{1}{99.100}\)\(\Rightarrow\)\(\frac{1}{11^2}\)+...+\(\frac{1}{100^2}\)<\(\frac{1}{10}\)-\(\frac{1}{11}\)+\(\frac{1}{11}\)-\(\frac{1}{12}\)+.......+\(\frac{1}{99}\)-\(\frac{1}{100}\)<\(\frac{1}{10}\)-\(\frac{1}{100}\)