Question

Cho biểu thức \(A=\dfrac{\sqrt{x}-5}{\sqrt{x}+3}\)

Tìm x thuộc Z để A có giá trị nguyên

Answer 1

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

=> Để A nguyên thì:\(\dfrac{8}{\sqrt{x}+3}\in Z\)

<=> \(\sqrt{x}+3\inƯ\left(8\right)\)

<=> \(\sqrt{x}+3=\left\{-8;-4;-2;-1;1;2;4;8\right\}\)

<=> \(\sqrt{x}=\left\{-11;-7;-5;-4;-2;-1;1;5\right\}\)

<=> x = {1 ; 25}

Vậy.............