giúp em với ạ em cảm ơn trước ạ
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(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) \(D=\dfrac{2x+10}{x+3}=\dfrac{2\left(x+3\right)+4}{x+3}=2+\dfrac{4}{x+3}\in Z\Rightarrow\dfrac{4}{x+3}\in Z\)
\(\Rightarrow\left(x+3\right)\inƯ\left(4\right)=\left\{\pm1;\pm2\pm4\right\}\)
\(\Rightarrow x\in\left\{-2;-4;-1;-5;1;-7\right\}\)
1. What size shoes do you take?
2. What newspaper do you read?
3. What color are your eyes?
4. What time did you arrive this morning?
5. What kind of film do you like?
6. How tall is your teacher?
7. How far is it from your house to the office?
8. How much did you pay for your new shirt?
9. How often do you take an English test in class?
10. How long have you been studying English?
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) \(B=\dfrac{1}{99}-\dfrac{1}{99.98}-\dfrac{1}{98.97}-\dfrac{1}{97.96}-...-\dfrac{1}{3.2}-\dfrac{1}{2.1}\)
\(B=\dfrac{1}{99}-\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{96.97}+\dfrac{1}{97.98}+\dfrac{1}{98.99}\right)\)
\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{96}-\dfrac{1}{97}+\dfrac{1}{97}-\dfrac{1}{98}+\dfrac{1}{98}-\dfrac{1}{99}\right)\)
\(B=\dfrac{1}{99}-\left(1-\dfrac{1}{99}\right)\)
\(B=\dfrac{1}{99}-\left(\dfrac{99}{99}-\dfrac{1}{99}\right)\)
\(B=\dfrac{1}{99}-\dfrac{98}{99}\)
\(B=-\dfrac{97}{99}\)