Question

[loan]

tìm xez sao cho 

a,2x+3chia hết cho x

b,8x+4chia hết cho 2x -1

c.x2-x+7chia hết cho x-1

Answer 1

[Nguyễn acc 2]

\(a,\) Vì \(2x⋮x\Rightarrow3⋮x\Rightarrow x\inƯ\left(3\right)=\left\{-3;-1;1;3\right\}\)

\(b,\left(8x+4\right)⋮\left(2x-1\right)\\ \Rightarrow\left[\left(8x-4\right)+8\right]⋮\left(2x-1\right)\\ \Rightarrow\left[4\left(2x-1\right)+8\right]⋮\left(2x-1\right)\)

\(Vì.4\left(2x-1\right)⋮\left(2x-1\right)\Rightarrow8⋮\left(2x-1\right)\Rightarrow\left(2x-1\right)\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

Ta có bảng:

2x-1	-8	-4	-2	-1	1	2	4	8
x	-3,5(loại)	-1,5(loại)	-0,5(loại)	0	1	1,5(loại)	2,5(loại)	4,5(loại)

Vậy \(x\in\left\{0;1\right\}\)

\(c,\left(x^2-x+7\right)⋮\left(x-1\right)\\ \Rightarrow\left[x\left(x-1\right)+7\right]⋮\left(x-1\right)\)

\(Vì.x\left(x-1\right)⋮\left(x-1\right)\Rightarrow7⋮\left(x-1\right)\Rightarrow x-1\inƯ\left(7\right)=\left\{-7;-1;1;7\right\}\)

Ta có bảng:

x-1	-7	-1	1	7
x	-6	0	2	8

Vậy \(x\in\left\{-6;0;2;8\right\}\)