a: Thay x=16 vào A, ta được:
\(A=\dfrac{4+2}{4}=\dfrac{6}{4}=\dfrac{3}{2}\)
b: \(B=\dfrac{x-\sqrt{x}+4}{x-2\sqrt{x}}-\dfrac{3}{\sqrt{x}-2}\)
\(=\dfrac{x-\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{3}{\sqrt{x}-2}\)
\(=\dfrac{x-4\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\left(\sqrt{x}-2\right)^2}{\sqrt{x}\left(\sqrt{x}-2\right)}=\dfrac{\sqrt{x}-2}{\sqrt{x}}\)
c: C=B:A
\(=\dfrac{\sqrt{x}-2}{\sqrt{x}}:\dfrac{\sqrt{x}+2}{\sqrt{x}}=\dfrac{\sqrt{x}-2}{\sqrt{x}+2}\)
\(C-1=\dfrac{\sqrt{x}-2-\left(\sqrt{x}+2\right)}{\sqrt{x}+2}=\dfrac{-4}{\sqrt{x}+2}< 0\)
=>C<1
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