a, \(B=\dfrac{1}{13}\left(\dfrac{-65}{x-7}+\dfrac{26}{x-7}\right)\)
\(=\dfrac{1}{13}\left(\dfrac{-65+26}{x-7}\right)=\dfrac{1}{13}.\dfrac{-39}{x-7}=\dfrac{-3}{x-7}=-\dfrac{3}{x-7}\)
b, Để B đạt giá trị nguyên thì \(\dfrac{3}{x-7}\) phải đạt giá trị nguyên.
\(\Rightarrow x-7\inƯ\left(3\right)\Rightarrow x-7\in\left\{-3;-1;1;3\right\}\)
\(\Rightarrow x\in\left\{4;6;8;10\right\}\)
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\(B=\dfrac{1}{13}\left(\dfrac{-65}{x-7}+\dfrac{26}{x-7}\right)\)
\(B=\dfrac{-65}{13\left(x-7\right)}+\dfrac{26}{13\left(x-7\right)}\)
\(B=\dfrac{-65}{13x-91}+\dfrac{26}{13x-91}\)
\(B=\dfrac{-39}{13x-91}\)
\(B\in Z\Rightarrow\left\{{}\begin{matrix}-65⋮x-7\\26⋮x-7\end{matrix}\right.\)
\(\Rightarrow x-7\in UC\left(-65;26\right)\)
\(UCLN\left(-65;26\right)=13\)
\(\Rightarrow\left[{}\begin{matrix}x-7=13\Rightarrow x=20\\x-7=-13\Rightarrow x=-6\\x-7=1\Rightarrow x=8\\x-7=-1\Rightarrow x=6\end{matrix}\right.\)
a, B=113(−65x−7+26x−7)
=113(−65+26x−7)=113.−39x−7=−3x−7=−3x−7
b, Để B đạt giá trị nguyên thì 3x−7 phải đạt giá trị nguyên.
⇒x−7∈Ư(3)⇒x−7∈{−3;−1;1;3}
⇒x∈{4;6;8;10}
Vậy...........
