\(\sqrt[3]{27}=3\)

\(\sqrt[3]{27}=3\) nha bạn

a) \(A=\sqrt{7-2\sqrt{6}}-2\sqrt{6}-1=\sqrt{6-2\sqrt{6}.1+1}-2\sqrt{6}-1\)

\(=\sqrt{\left(\sqrt{6}-1\right)^2}-2\sqrt{6}-1=\sqrt{6}-1-2\sqrt{6}-1=-\sqrt{6}-2\)

b) \(B=\sqrt{7+4\sqrt{3}}-2\sqrt{3}+1=\sqrt{4+2.2\sqrt{3}+3}-2\sqrt{3}+1\)

\(\sqrt{\left(2+\sqrt{3}\right)^2}-2\sqrt{3}+1=2+\sqrt{3}-2\sqrt{3}+1=-\sqrt{3}+3\)

c) \(C=\sqrt{13-4\sqrt{3}}+2\sqrt{3}=\sqrt{12-2.2\sqrt{3}.1+1}+2\sqrt{3}\)

\(=\sqrt{\left(2\sqrt{3}\right)^2-2.2\sqrt{3}.1+1}+2\sqrt{3}=\sqrt{\left(2\sqrt{3}-1\right)^2}+2\sqrt{3}\)

\(=2\sqrt{3}-1+2\sqrt{3}=4\sqrt{3}-1\)

sao giải mà bạn ko cho đúng zạy 2 bài rùi đó

\(\sqrt{\frac{4a^2}{25}}=\frac{\sqrt{4a^2}}{\sqrt{25}}=\frac{\sqrt{4}.\sqrt{a^2}}{5}\) hiểu chưa

a) ĐKXĐ :

\(6-2x\ne0\Leftrightarrow x\ne3\)

 \(A=\frac{x^2-6x+9}{6-2x}=\frac{\left(x-3\right)^2}{-2.\left(x-3\right)}=\frac{x-3}{-2}\)

b) ĐKXĐ:

\(\sqrt{x}-7\ne0\text{ và }x\ge0\)

\(\Leftrightarrow\sqrt{x}\ne7\text{ và }x\ge0\)

\(\Leftrightarrow x\ne49\text{ và }x\ge0\)

\(B=\frac{x-49}{\sqrt{x}-7}=\frac{\left(\sqrt{x}-7\right)\left(\sqrt{x}+7\right)}{\sqrt{x}-7}=\sqrt{x}+7\)

c) 

\(C=\frac{2-\sqrt{2}}{\sqrt{2}}=\frac{\left(\sqrt{2}\right)^2-\sqrt{2}}{\sqrt{2}}=\frac{\sqrt{2}.\left(\sqrt{2}-1\right)}{\sqrt{2}}=\sqrt{2}-1\)

dễ