Ta có:

\(\dfrac{1}{101}+\dfrac{1}{102}+...+\dfrac{1}{299}+\dfrac{1}{300}>\dfrac{1}{300}.200=\dfrac{200}{300}=\dfrac{2}{3}\)

\(\Rightarrow\) biểu thức trên lớn hơn \(\dfrac{2}{3}\).

A=\(\dfrac{2}{7}+\dfrac{-3}{8}+\dfrac{11}{7}+\dfrac{1}{3}+\dfrac{1}{7}+\dfrac{5}{-3}\)

A=\(\left(\dfrac{2}{7}+\dfrac{11}{7}+\dfrac{1}{7}\right)+\left(\dfrac{1}{3}+\dfrac{5}{-3}\right)+\dfrac{-3}{8}\)

A=\(2+\dfrac{-4}{3}+\dfrac{-3}{8}\)

A=\(\dfrac{7}{24}\)

B=\(\left(\dfrac{3}{17}+\dfrac{14}{17}\right)+\left(\dfrac{-18}{35}+\dfrac{17}{-35}\right)+\left(\dfrac{-5}{13}+\dfrac{-8}{13}\right)\)

B=\(\dfrac{17}{17}+\dfrac{-35}{35}+\dfrac{-13}{13}\)

B=\(1+\left(-1\right)+\left(-1\right)=-1\)

C=\(\dfrac{-3}{17}+\left(\dfrac{2}{3}+\dfrac{3}{17}\right)\)

C=\(\dfrac{-3}{17}+\dfrac{2}{3}+\dfrac{3}{17}=\left(\dfrac{-3}{17}+\dfrac{3}{17}\right)+\dfrac{2}{3}\)

C=0+\(\dfrac{2}{3}=\dfrac{2}{3}\)

D=\(\left(\dfrac{-1}{6}+\dfrac{5}{-12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-1}{6}+\dfrac{5}{-12}+\dfrac{7}{12}\)

D=\(\dfrac{-2}{12}+\dfrac{-5}{12}+\dfrac{7}{12}=\left(\dfrac{-2}{12}+\dfrac{-5}{12}\right)+\dfrac{7}{12}\)

D=\(\dfrac{-7}{12}+\dfrac{7}{12}=0\)

\(A=\dfrac{3}{4}\cdot\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot...\cdot\dfrac{899}{900}\)

\(A=\dfrac{1\cdot3}{2\cdot2}\cdot\dfrac{2\cdot4}{3\cdot3}\cdot\dfrac{3\cdot5}{4\cdot4}\cdot...\cdot\dfrac{29\cdot31}{30\cdot30}\)

\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot...\cdot30\right)^2}\)

\(A=\dfrac{1\cdot\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot31}{\left(2\cdot3\cdot4\cdot5\cdot...\cdot29\right)^2\cdot30\cdot30}\)

\(A=\dfrac{1\cdot31}{30}=\dfrac{31}{30}\)

Ta có : \(\dfrac{1}{101}>\dfrac{1}{300}\)

...

\(\dfrac{1}{299}>\dfrac{1}{300}\)

Do đó :

\(\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{1}{300}+\dfrac{1}{300}..+\dfrac{1}{300}\)

\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+..+\dfrac{1}{300}>\dfrac{200}{300}=\dfrac{2}{3}\)

Vậy...

1) \(\dfrac{2}{x+1}=\dfrac{x+1}{8}\Leftrightarrow\left(x+1\right)\left(x+1\right)=2.8\Leftrightarrow\left(x+1\right)^2=16\)

\(\Leftrightarrow\left\{{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\) vậy \(x=3;x=-5\)

2) thiếu quế phải nha

3) \(\dfrac{x-4}{x-7}=\left(\dfrac{-3}{5}\right)^2\Leftrightarrow\dfrac{x-4}{x-7}=\dfrac{9}{25}\Leftrightarrow9.\left(x-7\right)=25.\left(x-4\right)\)

\(\Leftrightarrow9x-63=25x-100\Leftrightarrow25x-9x=-63+100\)

\(\Leftrightarrow16x=37\Leftrightarrow x=\dfrac{37}{16}\) vậy \(x=\dfrac{37}{16}\)

4) ta có : \(x+y=20\Leftrightarrow y=20-x\)

\(\dfrac{3+x}{7+y}=\dfrac{3}{7}\Leftrightarrow7\left(3+x\right)=3\left(7+y\right)\Leftrightarrow21+7x=21+3y\)

\(\Leftrightarrow7x=3y\Leftrightarrow7x-3y=0\Leftrightarrow7x-3\left(20-x\right)=0\)

\(\Leftrightarrow7x-60+3x=0\Leftrightarrow10x=60\Leftrightarrow x=6\)

\(\Rightarrow6+y=20\Leftrightarrow y=14\) vậy \(x=6;y=14\)

a.\(\dfrac{2}{x+1}=\dfrac{x+1}{8}\)

\(\Rightarrow\left(x+1\right)\left(x+1\right)=2.8\)

\(\Rightarrow\left(x+1\right)^2=16\)

\(\Rightarrow\left(x+1\right)^2=(\pm4)^2\)

\(\rightarrow x+1=\pm4\)

\(+)x+1=-4\)

\(x=\left(-4\right)-1\)

\(x=-5\)

\(+)x+1=4\)

\(x=4-1\)

\(x=3\)

Vậy \(x\in\left\{-5;3\right\}\)

Câu còn lại tương tự nhé. Trừ câu 2 bạn xem lại nhé

a) Giải

Đặt \(M=\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\)

\(\Rightarrow A< A.M\)

hay \(A< \left(\dfrac{1}{2}.\dfrac{3}{4}.\dfrac{5}{6}...\dfrac{99}{100}\right).\left(\dfrac{2}{3}.\dfrac{4}{5}.\dfrac{6}{7}...\dfrac{98}{99}\right)\)

\(\Rightarrow A< \dfrac{1}{2}.\dfrac{2}{3}.\dfrac{3}{4}.\dfrac{4}{5}.\dfrac{5}{6}.\dfrac{6}{7}...\dfrac{98}{99}.\dfrac{99}{100}\)

\(\Leftrightarrow A< \dfrac{1.2.3.4.5.6...98.99}{2.3.4.5.6.7...99.100}\)

\(\Rightarrow A< \dfrac{1}{100}< \dfrac{1}{10}\)

Vậy \(A< \dfrac{1}{10}\)

Ace Legona, Hoàng Thị Ngọc Anh, ... giúp mình câu này với!

Đoàn Đức Hiếu, ... giúp với!

Đặt \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)

Ta có:

\(A=\dfrac{1}{2^2}+\dfrac{1}{3^2}+...+\dfrac{1}{2008^2}< \)\(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2007\cdot2008}\left(1\right)\)

Lại có: \(B=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+...+\dfrac{1}{2007\cdot2008}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)

\(=1-\dfrac{1}{2008}< 1\left(2\right)\)

Từ \(\left(1\right);\left(2\right)\) ta có \(A< B< 1\Rightarrow A< 1\)

A=\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+....+\dfrac{1}{2008^2}\)

A<\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2007.2008}\)

A<\(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2007}-\dfrac{1}{2008}\)

A<\(1-\dfrac{1}{2008}\)

A<\(\dfrac{2007}{2008}< 1\)

=> A<1

Vậy A<1