Câu 9:
a: \(A=\left(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(x+1\right)}+\dfrac{1}{x+1}\right)\cdot\dfrac{x+1}{\sqrt{x}-1}\)
\(=\dfrac{\sqrt{x}+1}{x+1}\cdot\dfrac{x+1}{\sqrt{x}-1}=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}\)
b: Thay \(x=4+2\sqrt{3}\) vào A, ta đc:
\(A=\dfrac{\sqrt{3}+1+1}{\sqrt{3}+1-1}=\dfrac{2+\sqrt{3}}{\sqrt{3}}\)
c: Để A nguyên thì \(\sqrt{x}-1+2⋮\sqrt{x}-1\)
=>\(\sqrt{x}-1\in\left\{1;-1;2;-2\right\}\)
=>\(x\in\left\{4;0;9\right\}\)
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