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Question

a)6 chia hết ( x-1)

b)15 chia hết (2x+1)

c)x+16chia hết x +1

Nguyễn Thanh Hằng · Accepted Answer

a/ $6⋮x-1$

$\Leftrightarrow x-1\inƯ\left(6\right)$

$\Leftrightarrow\left[{}\begin{matrix}x-1=1\x-1=2\x-1=3\x-1=6\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=2\x=3\x=4\x=7\end{matrix}\right.$

Vậy ...........

b/ $15⋮2x+1$

$\Leftrightarrow2x+1\inƯ\left(15\right)$

$\Leftrightarrow\left[{}\begin{matrix}2x+1=1\2x+1=3\2x+1=5\2x+1=15\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\x=2\x=7\end{matrix}\right.$

Vậy ...

c/ $x+16⋮x+1$

Mà $x+1⋮x+1$

$\Leftrightarrow15⋮x+1$

$\Leftrightarrow x+1\inƯ\left(15\right)$

$\Leftrightarrow\left[{}\begin{matrix}x+1=1\x+1=3\x+1=5\x+1=15\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\x=4\x=14\end{matrix}\right.$

Vậy ..Câu trả lời: a/ $6⋮x-1$

$\Leftrightarrow x-1\inƯ\left(6\right)$

$\Leftrightarrow\left[{}\begin{matrix}x-1=1\x-1=2\x-1=3\x-1=6\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=2\x=3\x=4\x=7\end{matrix}\right.$

Vậy ...........

b/ $15⋮2x+1$

$\Leftrightarrow2x+1\inƯ\left(15\right)$

$\Leftrightarrow\left[{}\begin{matrix}2x+1=1\2x+1=3\2x+1=5\2x+1=15\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=0\x=1\x=2\x=7\end{matrix}\right.$

Vậy ...

c/ $x+16⋮x+1$

Mà $x+1⋮x+1$

$\Leftrightarrow15⋮x+1$

$\Leftrightarrow x+1\inƯ\left(15\right)$

$\Leftrightarrow\left[{}\begin{matrix}x+1=1\x+1=3\x+1=5\x+1=15\end{matrix}\right.$ $\Leftrightarrow\left[{}\begin{matrix}x=0\x=2\x=4\x=14\end{matrix}\right.$

Vậy ..

Siêu sao bóng đá · Answer

a) 6 $⋮$ ( x - 1 )

$\Rightarrow$ x - 1 $\in$Ư(6)

$\Rightarrow$ x - 1 $\in$ { 1;2;3;6 }

$\Rightarrow$ x $\in$ { 2;3;4;7 }

Vậy ....

b) 15 $⋮$ ( 2x + 1 )

$\Rightarrow$ 2x + 1 $\in$ Ư(15)

$\Rightarrow$ 2x + 1 $\in$ { 1;3;5;15 }

$\Rightarrow$ 2x $\in$ { 0;2;4;14 }

$\Rightarrow$ x $\in$ { 0;1;2;7 }

Vậy ....

c) (x + 16) $⋮$ (x + 1)

$\Rightarrow$ (x + 1 + 15) $⋮$ (x+1)

Mà x + 1$⋮$ x+1

$\Rightarrow$ 15 $⋮$ x + 1

$\Rightarrow$ x + 1 $\in$ Ư(15)

$\Rightarrow$ x + 1 $\in$ { 1;3;5;15 }

$\Rightarrow$ x $\in$ { 0;2;4;14}

Vậy ....Câu trả lời: a) 6 $⋮$ ( x - 1 )

$\Rightarrow$ x - 1 $\in$Ư(6)

$\Rightarrow$ x - 1 $\in$ { 1;2;3;6 }

$\Rightarrow$ x $\in$ { 2;3;4;7 }

Vậy ....

b) 15 $⋮$ ( 2x + 1 )

$\Rightarrow$ 2x + 1 $\in$ Ư(15)

$\Rightarrow$ 2x + 1 $\in$ { 1;3;5;15 }

$\Rightarrow$ 2x $\in$ { 0;2;4;14 }

$\Rightarrow$ x $\in$ { 0;1;2;7 }

Vậy ....

c) (x + 16) $⋮$ (x + 1)

$\Rightarrow$ (x + 1 + 15) $⋮$ (x+1)

Mà x + 1$⋮$ x+1

$\Rightarrow$ 15 $⋮$ x + 1

$\Rightarrow$ x + 1 $\in$ Ư(15)

$\Rightarrow$ x + 1 $\in$ { 1;3;5;15 }

$\Rightarrow$ x $\in$ { 0;2;4;14}

Vậy ....

x+6	-9	-3	-1	1	3	9
x	-15	-9	-7	-5	-3	3

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2x+1	1	5
x	0	2

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