Question

1. Với giá trị nào của x thì các căn thức sau đây có nghĩa

a) \(\sqrt{\frac{x-1}{x+3}}\)

b) \(\sqrt{\frac{x-1}{4-x}}\)

c) \(\sqrt{\frac{a^3}{b^2}}\)

2. Biến đoi biểu thức trong dấu căn:

a) \(\sqrt{64+6\sqrt{7}}\)

b) \(\sqrt{16+8\sqrt{3}}\)

c) \(\sqrt{9-2\sqrt{14}}\)

Answer 1

Câu 1:

\(\left\{{}\begin{matrix}\frac{x-1}{x+3}\ge0\\x+3\ne0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x\ge1\\x< -3\end{matrix}\right.\)

b/

\(\left\{{}\begin{matrix}\frac{x-1}{4-x}\ge0\\4-x\ne0\end{matrix}\right.\) \(\Rightarrow1\le x< 4\)

c/

\(\left\{{}\begin{matrix}\frac{a^3}{b^2}\ge0\\b^2\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a^3\ge0\\b\ne0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a\ge0\\b\ne0\end{matrix}\right.\)

Câu 2:

\(\sqrt{64+6\sqrt{7}}=\sqrt{63+2\sqrt{63}+1}=\sqrt{\left(\sqrt{63}+1\right)^2}=1+\sqrt{63}=1+3\sqrt{7}\)

\(\sqrt{16+8\sqrt{3}}=\sqrt{12+2\sqrt{12.4}+4}=\sqrt{\left(\sqrt{12}+\sqrt{4}\right)^2}=\sqrt{12}+\sqrt{4}=2+2\sqrt{3}\)

\(\sqrt{9-2\sqrt{14}}=\sqrt{7-2\sqrt{7.2}+2}=\sqrt{\left(\sqrt{7}-\sqrt{2}\right)^2}=\sqrt{7}-\sqrt{2}\)