a. Với \(x\ge0;x\ne1;x\ne9\):
\(B=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
\(=\left[\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\right]\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+3+2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{3\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{3\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-1\right)}=\dfrac{3}{\sqrt{x}+3}\)
b. Để \(B\) nhận giá trị nguyên thì: \(\dfrac{3}{\sqrt{x}+3}\) có giá trị nguyên
\(\Rightarrow3⋮\sqrt{x}+3\)
\(\Rightarrow\sqrt{x}+3\inƯ\left(3\right)\)
\(\Rightarrow\sqrt{x}+3\in\left\{1;3;-1;-3\right\}\)
\(\Rightarrow\sqrt{x}\in\left\{-2;0;-4;-6\right\}\)
Mà \(\sqrt{x}\ge0\forall x\ge0\) nên \(\sqrt{x}=0\Leftrightarrow x=0\) (tm ĐKXĐ)
\(\text{#}Toru\)
a: \(B=\left(\dfrac{1}{\sqrt{x}-3}+\dfrac{2}{\sqrt{x}+3}\right):\dfrac{\sqrt{x}-1}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}+3+2\sqrt{x}-6}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\cdot\dfrac{\sqrt{x}-3}{\sqrt{x}-1}\)
\(=\dfrac{3\sqrt{x}-3}{\sqrt{x}-1}\cdot\dfrac{1}{\sqrt{x}+3}=\dfrac{3}{\sqrt{x}+3}\)
b Để B nguyên thì \(3⋮\sqrt{x}+3\)
mà \(\sqrt{x}+3>=3\forall x\) thỏa mãn ĐKXĐ
nên \(\sqrt{x}+3=3\)
=>x=0
