Câu 2:
a) ĐKXĐ: \(n\ne7\)
b) Thay n=20 vào A, ta được:
\(A=\dfrac{8}{20-7}=\dfrac{8}{13}\)
Thay n=3 vào A, ta được:
\(A=\dfrac{8}{3-7}=\dfrac{8}{-4}=-2\)
Câu 3:
a) Ta có: \(3x=6\cdot9\)
nên 3x=54
hay x=18
b) Ta có: \(3\cdot18=6\cdot9\)
nên \(\dfrac{3}{6}=\dfrac{9}{18};\dfrac{3}{9}=\dfrac{6}{18};\dfrac{6}{3}=\dfrac{18}{9};\dfrac{9}{3}=\dfrac{18}{6}\)
5.
\(\dfrac{1}{9}>\dfrac{3}{x}>\dfrac{1}{10}\Leftrightarrow\dfrac{3}{27}>\dfrac{3}{x}>\dfrac{3}{30}\)
\(\Rightarrow27< x< 30\Rightarrow\left[{}\begin{matrix}x=28\\x=29\end{matrix}\right.\)
Tổng: \(28+29=...\)
6.
\(\dfrac{1}{9}>\dfrac{3}{\left|x\right|}>\dfrac{1}{10}\Leftrightarrow\dfrac{3}{27}>\dfrac{3}{\left|x\right|}>\dfrac{3}{30}\)
\(\Leftrightarrow27< \left|x\right|< 30\)
\(\Leftrightarrow\left[{}\begin{matrix}\left|x\right|=28\\\left|x\right|=29\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=28\\x=-28\\x=29\\x=-29\end{matrix}\right.\)
Tổng: \(28+\left(-28\right)+29+\left(-29\right)=...\)
