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\(\left(3\sqrt{7}\right)^2=63>28=\left(\sqrt{28}\right)^2\) hoặc \(3\sqrt{7}>2\sqrt{7}=\sqrt{28}\)
\(\dfrac{x-2\sqrt{x}}{x-4}=\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}=\dfrac{\sqrt{x}}{\sqrt{x}+2}\)
\(T=\dfrac{\sqrt{27}+3}{\sqrt{3}}=\dfrac{3\sqrt{3}+3}{\sqrt{3}}=\dfrac{3\left(\sqrt{3}+1\right)}{\sqrt{3}}=\sqrt{3}\left(\sqrt{3}+1\right)=3+\sqrt{3}\)
Dạ √27 + 3 / √3 ạ 3 với căn 3 là chung với nhau ý ạ chứ ko phải như trên ý ạ
\(B=\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\) ĐK : \(x\ge0;x\ne1\)
\(=\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}-\frac{2}{\sqrt{x}+3}\)
\(=\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-1}\)
\(=\frac{3\sqrt{x}+1}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}-\frac{2}{\sqrt{x}+3}\)
\(=\frac{3\sqrt{x}+1-2\cdot\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)
\(=\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)
\(=\frac{\sqrt{x}+3}{\left(\sqrt{x}+3\right)\cdot\left(\sqrt{x}-1\right)}\)
\(=\frac{1}{\sqrt{x}-1}\)
\(B=\frac{3\sqrt{x}+1}{x+2\sqrt{x}-3}-\frac{2}{\sqrt{x}+3}\)
\(=\frac{3\sqrt{x}+1-2\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{3\sqrt{x}+1-2\sqrt{x}+2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}\)
\(=\frac{\sqrt{x}+3}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+3\right)}=\frac{1}{\sqrt{x}-1}\)