Bài 1:
\(a,B=\dfrac{3-3}{3-2}=0\\ b,S=A:\left(B+1\right)\\ S=\dfrac{x+2\sqrt{x}-x+\sqrt{x}+2-\sqrt{x}-4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\dfrac{3-\sqrt{x}+\sqrt{x}-2}{\sqrt{x}-2}\\ S=\dfrac{2\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\left(\sqrt{x}+2\right)=\dfrac{2\sqrt{x}-2}{\sqrt{x}+2}\\ c,S=\dfrac{2\left(\sqrt{x}+2\right)-6}{\sqrt{x}+2}=2-\dfrac{6}{\sqrt{x}+2}\in Z\\ \Leftrightarrow\sqrt{x}+2\inƯ\left(6\right)=\left\{2;3;6\right\}\left(\sqrt{x}+2\ge2\right)\\ \Leftrightarrow\sqrt{x}\in\left\{0;1;4\right\}\\ \Leftrightarrow x\in\left\{0;1;16\right\}\)
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