Question

Answer 1

a: Để 2x-3/x-1 là số nguyên thì \(2x-3⋮x-1\)

\(\Leftrightarrow x-1\in\left\{1;-1\right\}\)

hay \(x\in\left\{2;0\right\}\)

b: \(\left|x+6\right|+\left|x-4\right|=\left|x+6\right|+\left|4-x\right|\ge\left|x+6+4-x\right|=\left|10\right|=10\)

Answer 2

a. ta có : \(A=\dfrac{2x-3}{x-1}\)

\(2x-3⋮x-1\\ \Rightarrow\left(2x-2\right)-1⋮x-1\\ m\text{à}2x-2⋮x-1\left(x\in Z\right)\\ \Rightarrow1⋮\left(x-1\right)\\ \Rightarrow x-1\in\text{Ư}\left(1\right)=\left\{1;-1\right\}\)

ta có bảng sau : 

x-1	1	-1
x	2	0

vậy......