Question

Answer 1

Bài 5 :

\(A=\dfrac{3x+5}{x-1}=\dfrac{3x-3+8}{x-1}\\ =\dfrac{3\left(x-1\right)+8}{x-1}=3+\dfrac{8}{x-1}\)

Để `A in ZZ` thì `8/(x-1) in ZZ`

`=>8 vdots x-1`

`=>x-1 i Ư(8)={+-1;+-2;+-4;+-8}`

`=>x in {2;0;3;-1;5;-3;9;-7}`

Vậy...

Answer 2

Bài 4 :

\(A=\dfrac{54\cdot107-53}{53\cdot107+54}=\dfrac{\left(53+1\right)\cdot107-53}{53\cdot107+54}\\ =\dfrac{53\cdot107+107-53}{53\cdot107+54}=\dfrac{53\cdot107+54}{53\cdot107+54}=1\left(1\right)\)

\(B=\dfrac{135\cdot269-133}{134\cdot269+135}=\dfrac{\left(134+1\right)\cdot269-133}{134\cdot269+135}\\ =\dfrac{134\cdot269+269-133}{134\cdot269+135}=\dfrac{134\cdot269+136}{134\cdot269+135}\\ \Rightarrow B>1\left(2\right)\)

Từ `(1)` và `(2)=>B>A`

Answer 3

5:

A là số nguyên

=>3x-3+8 chia hết cho x-1

=>x-1 thuộc {1;-1;2;-2;4;-4;8;-8}

=>x thuộc {2;0;3;-1;5;-3;9;-7}

Answer 4

Các bài 1,2,3,4 em làm được chưa?