Giá trị của số \(\sqrt{\left(2-\sqrt{3}\right)^26}\) bằng bao nhiêu? \(\left(2-\sqrt{3}\right).\sqrt{6}\) \(\left(\sqrt{3}-2\right).\sqrt{6}\) \(6\left(2-\sqrt{3}\right)\) \(6\left(\sqrt{3}-2\right)\) Hướng dẫn giải: \(\sqrt{\left(2-\sqrt{3}\right)^26}=\left|2-\sqrt{3}\right|.\sqrt{6}=\left(2-\sqrt{3}\right).\sqrt{6}\)
Tính \(\sqrt{3}+\sqrt{12}+\sqrt{48}\) \(7\sqrt{3}\) \(7+\sqrt{3}\) \(6\sqrt{3}\) \(3\sqrt{7}\) Hướng dẫn giải: \(\sqrt{3}+\sqrt{12}+\sqrt{48}\)\(=\sqrt{3}+\sqrt{4.3}+\sqrt{16.3}=\sqrt{3}+2\sqrt{3}+4\sqrt{3}=\)\(=\sqrt{3}\left(1+2+4\right)=7\sqrt{3}\)
Tính \(\sqrt{6}+\sqrt{24}-\sqrt{54}\) \(\sqrt{6}\) \(2\sqrt{6}\) \(3\sqrt{6}\) \(0\) Hướng dẫn giải: \(\sqrt{6}+\sqrt{24}-\sqrt{54}=\sqrt{6}+\sqrt{4.6}+\sqrt{9.6}\)\(=\sqrt{6}+2\sqrt{6}-3\sqrt{6}=0\)
Rút gọn \(\sqrt{4x}+\sqrt{9x}-\sqrt{16x}+\sqrt{25x}\left(x\ge0\right)\) \(6\sqrt{x}\) \(4\sqrt{x}\) \(5\sqrt{x}\) \(3\sqrt{x}\) Hướng dẫn giải: \(\sqrt{4x}+\sqrt{9x}-\sqrt{16x}+\sqrt{25x}\) \(=2\sqrt{x}+3\sqrt{x}-4\sqrt{x}+5\sqrt{x}\) \(=\sqrt{x}\left(2+3-4+5\right)=6\sqrt{x}\)
Rút gọn biểu thức \(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}\) \(\sqrt{a}\) \(\sqrt{ab}\) \(\sqrt{a}+\sqrt{b}\) \(a\) Hướng dẫn giải: \(\dfrac{a+\sqrt{ab}}{\sqrt{a}+\sqrt{b}}=\dfrac{\sqrt{a}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{a}+\sqrt{b}}=\sqrt{a}\).
Sắp xếp các số \(4\sqrt{3},2\sqrt{5},6\sqrt{2},\sqrt{30}\) theo thứ tự tăng dần. \(2\sqrt{5}< \sqrt{30}< 4\sqrt{3}< 6\sqrt{2}\) \(\sqrt{30}< 2\sqrt{5}< 4\sqrt{3}< 6\sqrt{2}\) \(6\sqrt{2}< 4\sqrt{3}< \sqrt{30}< 2\sqrt{5}\) \(4\sqrt{3}< \sqrt{30}< 6\sqrt{2}< 2\sqrt{5}\) Hướng dẫn giải: \(4\sqrt{3}=\sqrt{48},2\sqrt{5}=\sqrt{20},6\sqrt{2}=\sqrt{72},\sqrt{30}\). Vì \(\sqrt{20}< \sqrt{30}< \sqrt{48}< \sqrt{72}\) nên \(2\sqrt{5}< \sqrt{30}< 4\sqrt{3}< 6\sqrt{2}\).
Biết \(\sqrt{36x}-\sqrt{4x}=3\). Tìm x. \(x=\dfrac{9}{16}\) \(x=\dfrac{3}{4}\) \(x=\dfrac{1}{4}\) \(x=\dfrac{5}{4}\) Hướng dẫn giải: \(\sqrt{36x}-\sqrt{4x}=3\Leftrightarrow6\sqrt{x}-2\sqrt{x}=3\)\(\Leftrightarrow4\sqrt{x}=3\)\(\Leftrightarrow\sqrt{x}=\dfrac{3}{4}\)\(\Leftrightarrow x=\dfrac{9}{16}\)
Rút gọn \(\dfrac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\) \(\sqrt{3}\) \(\sqrt{3}+\sqrt{2}\) \(\sqrt{6}\) \(\sqrt{2}\) Hướng dẫn giải: \(\dfrac{3+\sqrt{6}}{\sqrt{3}+\sqrt{2}}\)\(=\dfrac{\sqrt{3}\left(\sqrt{3}+\sqrt{2}\right)}{\sqrt{3}+\sqrt{2}}=\sqrt{3}\)
Rút gọn \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}\) \(9+4\sqrt{5}\) \(3\) \(1\) \(5-2\sqrt{5}\) Hướng dẫn giải: \(\dfrac{\sqrt{5}+2}{\sqrt{5}-2}=\dfrac{\left(\sqrt{5}+2\right)\left(\sqrt{5}+2\right)}{\left(\sqrt{5}-2\right)\left(\sqrt{5}+2\right)}\)\(=\dfrac{\left(\sqrt{5}+2\right)^2}{1}=5+2.2.\sqrt{5}+2^2=9+4\sqrt{5}\)
Rút gọn \(ab\sqrt{1+\dfrac{1}{a^2b^2}}\left(ab>0\right)\) \(\sqrt{a^2b^2}+1\) \(\sqrt{a^2b^2+1}\) \(\sqrt{ab+1}\) \(ab+1\) Hướng dẫn giải: \(ab\sqrt{1+\dfrac{1}{a^2b^2}}=\sqrt{a^2b^2\left(1+\dfrac{1}{a^2b^2}\right)}=\sqrt{a^2b^2+1}\)
