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`

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...

\(x+\dfrac{5}{6}=\dfrac{2}{5}-\left(-\dfrac{2}{3}\right)\\ x+\dfrac{5}{6}=\dfrac{2}{5}+\dfrac{2}{3}\\ x+\dfrac{5}{6}=\dfrac{16}{15}\\ x=\dfrac{16}{15}-\dfrac{5}{6}\\ x=\dfrac{7}{30}\)

\(\dfrac{2^7\cdot9^3}{6^5\cdot8^2}=\dfrac{2^7\cdot\left(3^2\right)^3}{\left(2\cdot3\right)^5\cdot\left(2^3\right)^2}\\ =\dfrac{2^7\cdot3^6}{2^5\cdot3^5\cdot2^6}=\dfrac{2^7\cdot3}{2^{11}}\\ =\dfrac{3}{2^4}=\dfrac{3}{16}\)