bạn kia giải sai rồi . làm kiểu vay sẽ ngc dau . 

quy luật là \(\frac{1}{a}+\frac{1}{a+3}+\frac{1}{a+9}+...+\frac{1}{a+3k}\)à bạn?

\(\left|x\right|< 5\\ \Rightarrow-5< x< 5\\ \Rightarrow x\in\left\{-4;-3;-2;-1;0;1;2;3;4\right\}\)

toán lớp 5 thì đúng hơn 

\(x\left(x+3\right)^2-3x=\left(x+2\right)^3+1\\ \Rightarrow x\left(x^2+6x+9\right)-3x=x^3+9+6x\left(x+2\right)\\ =x^3+6x^2+6x=x^3+6x^2+12x+9\\ \Rightarrow12x+9=6x\\ \Rightarrow6x=-9\\ \Rightarrow x=-\dfrac{3}{2}\)

\(A=-\dfrac{4}{x^2-4x+10}\\ =-\dfrac{4}{\left(x^2-2.x.2+4+6\right)}\\ =-\dfrac{4}{\left(x-2\right)^2+6}\)

\(\left(x-2\right)^2\ge0\\ \Rightarrow\left(x-2\right)^2+6\ge6\\ \Rightarrow\dfrac{4}{\left(x-2\right)^2+6}\le\dfrac{2}{3}\\ \Rightarrow A=-\dfrac{4}{\left(x-2\right)^2+6}\ge-\dfrac{2}{3}\)

Min A=-2/3 khi x=2

cac ban tra loi chi tiet dum minh nha

\(A=\dfrac{2010}{2}+\dfrac{2010}{6}+...+\dfrac{2010}{9900}\\ =2010\left(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\right)\\ =2010\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\right)\\ =2010\left(\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\right)\\ =2010.\dfrac{99}{100}=\dfrac{19899}{10}\)