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\(a,A=\sqrt{27}+\frac{2}{\sqrt{3}-2}-\sqrt{\left(1-\sqrt{3}\right)^2}\)
\(=3\sqrt{3}+\frac{2\left(\sqrt{3}+2\right)}{\left(\sqrt{3}-2\right)\left(\sqrt{3}+2\right)}-\left(\sqrt{3}-1\right)\)
\(=3\sqrt{3}+\frac{2\sqrt{3}+4}{3-4}-\sqrt{3}+1\)
\(=3\sqrt{3}-2\sqrt{3}-4-\sqrt{3}+1\)
\(=-3\)
\(B=\left(\frac{1}{x-\sqrt{x}}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{x-2\sqrt{x}+1}\)
\(=\left(\frac{1}{\sqrt{x}\left(\sqrt{x}-1\right)}+\frac{1}{\sqrt{x}-1}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}.\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}+1}\)
\(=\frac{\sqrt{x}-1}{\sqrt{x}}\)
b, Ta có \(B< A\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}< -3\)
\(\Leftrightarrow\frac{\sqrt{x}-1}{\sqrt{x}}+3< 0\)
\(\Leftrightarrow\frac{\sqrt{x}-1+3\sqrt{x}}{\sqrt{x}}< 0\)
\(\Leftrightarrow\frac{4\sqrt{x}-1}{\sqrt{x}}< 0\)
\(\Leftrightarrow4\sqrt{x}-1< 0\left(Do\sqrt{x}>0\right)\)
\(\Leftrightarrow\sqrt{x}< \frac{1}{4}\)
\(\Leftrightarrow0< x< \frac{1}{2}\)(Kết hợp ĐKXĐ)
Vậy ...
a/ \(Q=\left(\frac{\sqrt{x}-2+7}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right):\left(\frac{\sqrt{x}-1-\sqrt{x}+2}{\sqrt{x}-2}\right)\)
\(Q=\left(\frac{\sqrt{x}+5}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\right).\left(\sqrt{x}-2\right)\)
\(Q=\frac{\sqrt{x}+5}{\sqrt{x}+2}\)
b/ i, x= \(\sqrt{\left(5+\sqrt{2}\right)^2}-\sqrt{\left(4+\sqrt{2}\right)^2}=5+\sqrt{2}-4-\sqrt{2}=1\)
\(\Rightarrow Q=\frac{5+1}{2+1}=2\)
ii, x= \(\frac{\sqrt{2\left(2-\sqrt{3}\right)}}{2-\sqrt{3}}-\frac{\sqrt{2\left(2+\sqrt{3}\right)}}{2+\sqrt{3}}\)\(=\frac{\sqrt{4-2\sqrt{3}}}{2-\sqrt{3}}-\frac{\sqrt{4+2\sqrt{3}}}{2+\sqrt{3}}=\frac{\left(\sqrt{3}-1\right)\left(2+\sqrt{3}\right)-\left(\sqrt{3}+1\right)\left(2-\sqrt{3}\right)}{4-3}\)
\(=2\sqrt{3}+3-2-\sqrt{3}-2\sqrt{3}+3-2+3=5-\sqrt{3}\)
\(Q=\frac{\sqrt{5-\sqrt{3}}+5}{\sqrt{5-\sqrt{3}}+2}\)
Đến đây chưa nghĩ ra :D
Sửa chút đoạn sau cho bạn trên.
ii, \(x=\sqrt{\frac{2}{2-\sqrt{3}}}-\sqrt{\frac{2}{2+\sqrt{3}}}\)
\(=\sqrt{2}.\sqrt{2-\sqrt{3}}\left(2+\sqrt{3}\right)-\sqrt{2}.\sqrt{2+\sqrt{3}}\left(2-\sqrt{3}\right)\)
\(=2\sqrt{3}-\sqrt{3}-2+3-\left(2\sqrt{3}+2-3-\sqrt{3}\right)\)\(=2\)
\(\Rightarrow Q=\frac{\sqrt{2}+5}{\sqrt{2}+2}=\frac{8-3\sqrt{2}}{2}\) (Trục căn thức ở mẫu, lấy \(2-\sqrt{2}\) )