21 tháng 7 2021 lúc 6:14
\(A=\sqrt{\left(\sqrt{5}-2\right)^2}-\sqrt{5}=\left|\sqrt{5}-2\right|-\sqrt{5}=\sqrt{5}-2-\sqrt{5}=-2\)
\(B=\dfrac{\sqrt{x}\left(\sqrt{x}-1\right)}{\sqrt{x}}+\dfrac{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}{\sqrt{x}-1}=\sqrt{x}+\sqrt{x}+1=2\sqrt{x}+1\)
b.
Để \(3A+B=0\Leftrightarrow3.\left(-2\right)+2\sqrt{x}+1=0\)
\(\Leftrightarrow2\sqrt{x}=5\)
\(\Leftrightarrow\sqrt{x}=\dfrac{5}{2}\)
\(\Leftrightarrow x=\dfrac{25}{4}\)
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