Question

Tím x 

a,\(\left(x+3\right).\left(x-4\right)< 0\left(x\in Z\right)\)

b, \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2012}{2014}\)

Answer 1

a, \(\left(x+3\right)\left(x-4\right)< 0\)

\(\Rightarrow x^2-x-12< 0\)

\(\Rightarrow\left(x-0,5\right)^2< 12,25\)

\(\Rightarrow3,5>x-0,5>-3,5\)

\(\Rightarrow4>x>-3\)

b,\(\Rightarrow\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+...+\frac{2}{x\left(x+1\right)}=\frac{2012}{2014}\)

\(\Rightarrow2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2012}{2014}\)

\(\Rightarrow2.\frac{x-1}{2x+2}=\frac{2012}{2014}\)

\(\Rightarrow\frac{x-1}{x+1}=\frac{2012}{2014}\Rightarrow x=2013\)

chúc bạn học tốt ^^

Answer 2

\(\left(x+3\right)\left(x-4\right)< 0\)

Ta có 2 trường hợp

Trường hợp 1:

\(PT\Leftrightarrow\hept{\begin{cases}x+3>0\\x-4< 0\end{cases}\Leftrightarrow\hept{\begin{cases}x>-3\\x< 4\end{cases}}}\)

\(\Rightarrow x< 4\left(1\right)\)

Trường hợp 2:

\(PT\Leftrightarrow\hept{\begin{cases}x+3< 0\\x-4>0\end{cases}\Leftrightarrow\hept{\begin{cases}x< -3\\x>4\end{cases}}}\)

\(\Rightarrow x< -3\left(2\right)\)

Từ (1) và (2)

\(\Rightarrow4>x< -3\)

Vậy \(x\in\){-4;-5;-6;-7-;-8;.....}