\(A=\dfrac{\sqrt{x}+1}{\sqrt{x}-2}\left(dk:x\ge0;x\ne4;x\ne9\right)\)
thay \(x=\dfrac{1}{4}\left(tm\right)\) vào A có
\(A=\dfrac{\sqrt{\dfrac{1}{4}}+1}{\sqrt{\dfrac{1}{4}}-2}==\dfrac{\dfrac{1}{2}+1}{\dfrac{1}{2}-2}-1\)
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b) \(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{x-5\sqrt{x}+6}\)
\(B=\dfrac{\sqrt{x}+2}{\sqrt{x}-3}+\dfrac{\sqrt{x}-8}{\left(\sqrt{x}-2\right)\left(\sqrt{x}-3\right)}\)
\(B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x-4+\sqrt{x}-8}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{x+\sqrt{x}-12}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+4\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}-2\right)}\)
\(B=\dfrac{\sqrt{x}+4}{\sqrt{x}-2}\)
vậy ...
