a: Khi x=3+2căn 2 thì \(A=\dfrac{3+2\sqrt{2}-\sqrt{2}-1+2}{\sqrt{2}+1+3}=\dfrac{4+\sqrt{2}}{4+\sqrt{2}}=1\)
b: \(B=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)+2\left(\sqrt{x}+3\right)-3\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-4+2\sqrt{x}+6-3\sqrt{x}-4}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}\)
\(=\dfrac{x-\sqrt{x}-2}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}+1}{\sqrt{x}+3}\)
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