tìm X để A nguyên :3x/x-3

Question

trả lời giúp

 

Answer 1

\(\dfrac{3x}{x-3} (\text{ĐK:}x \not =3)\)

\(\text{Ta có: }3x=3(x-3)+9\)

\(\text{Ta có: }3(x-3)\ \vdots\ (x-3)\)

\(\text{Để }3x\ \vdots\ (x-3)\text{ thì }9\ \vdots\ (x-3)\)

\(\Rightarrow (x-3)\in Ư(9)=\{\pm1\ ;\pm3\ ;\pm9\}\)

\(\text{Lập bảng}\)

\(x-3\)	\(-9 \)	\(-3\)	\(-1\)	\(1 \)	\(3 \)	\(9\)
\(x\)	\(-6\)	\(0\)	\(2\)	\(4\)	\(6\)	\(12\)

\(\text{Vậy }x\in\{-6;0;2;4;6;12\}\)

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Answer 2

\(\dfrac{3x}{x-3} (\text{ĐK:}x \not =3)\)

\(\text{Ta có: }3x=3(x-3)+9\)

\(\text{Ta có: }3(x-3)\ \vdots\ (x-3)\)

\(\text{Để }3x\ \vdots\ (x-3)\text{ thì }9\ \vdots\ (x-3)\)

\(\Rightarrow (x-3)\in Ư(9)=\{\pm1\ ;\pm3\ ;\pm9\}\)

\(\text{Lập bảng}\)

\(x-3\)	\(-9 \)	\(-3\)	\(-1\)	\(1 \)	\(3 \)	\(9\)
\(x\)	\(-6\)	\(0\)	\(2\)	\(4\)	\(6\)	\(12\)

\(\text{Vậy }x\in\{-6;0;2;4;6;12\}\)

Tick mình nha