bài gì

đề bài đâu r

a) \(A=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{19.21}\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{19}-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\left(1-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\left(\dfrac{21}{21}-\dfrac{1}{21}\right)\)

\(A=\dfrac{1}{2}.\dfrac{20}{21}\)

\(A=\dfrac{10}{21}\)

b) \(B=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)

\(B=\dfrac{1}{99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{96.97}+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)

\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{96}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)

\(B=\dfrac{1}{99}-\dfrac{98}{99}\)

\(B=-\dfrac{97}{99}\)

Ta có: \(\int\dfrac{xdx}{x^2+3}\)

Đặt \(u=x^2+3\left(u>0\right)\) 

Có \(du=2xdx\)

\(\Rightarrow\int\dfrac{xdx}{x^2+3}=\)\(\int\dfrac{du}{2u}=\dfrac{1}{2}ln\left(u\right)=\dfrac{1}{2}ln\left(x^2+3\right)\)

(a) \(A=\dfrac{3}{x-2}\in Z\)

\(\Rightarrow\left(x-2\right)\inƯ\left(3\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\x-1=3\\x-1=-3\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\x=4\\x=-2\end{matrix}\right.\)

Vậy: \(x\in\left\{-2;0;2;4\right\}.\)

(b) \(B=-\dfrac{11}{2x-3}\in Z\)

\(\Rightarrow\left(2x-3\right)\inƯ\left(11\right)=\left\{\pm1;\pm3\right\}\)

\(\Rightarrow\left[{}\begin{matrix}2x-3=1\\2x-3=-1\\2x-3=11\\2x-3=-11\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=7\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{-4;1;2;7\right\}.\)

(c) \(C=\dfrac{x+3}{x+1}=\dfrac{\left(x+1\right)+2}{x+1}=1+\dfrac{2}{x+1}\in Z\Rightarrow\dfrac{2}{x+1}\in Z\)

\(\Rightarrow\left(x+1\right)\inƯ\left(2\right)=\left\{\pm1;\pm2\right\}\)

\(\Rightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\\x+1=2\\x+1=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\\x=1\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{-3;-2;0;1\right\}.\)

(d) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)

\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)

\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)