Tính \(\sqrt{3}+\sqrt{12}+3\sqrt{2}.\sqrt{24}\) \(15\sqrt{3}\) \(4\sqrt{3}\) \(6\sqrt{3}\) \(2\sqrt{3}\)
Rút gọn biểu thức \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\) \(\dfrac{3-\sqrt{3}}{3}\) \(\dfrac{3+\sqrt{3}}{3}\) \(2\sqrt{3}\) \(3-\sqrt{3}\) Hướng dẫn giải: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{2}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\) \(=\dfrac{1}{2+\sqrt{3}}+\dfrac{1}{\sqrt{3}}-\dfrac{2}{\sqrt{3}\left(\sqrt{3}+1\right)}\) \(=\dfrac{\sqrt{3}\left(\sqrt{3}+1\right)+\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)-2\left(2+\sqrt{3}\right)}{\sqrt{3}.\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}\) \(=\dfrac{2\sqrt{3}+4}{\sqrt{3}.\left(2+\sqrt{3}\right).\left(\sqrt{3}+1\right)}\) \(=\dfrac{2\left(\sqrt{3}+2\right)}{\sqrt{3}\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}\) \(=\dfrac{2}{\sqrt{3}.\left(\sqrt{3}+1\right)}\) \(=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\) \(=\dfrac{2\sqrt{3}\left(\sqrt{3}-1\right)}{3.2}\) \(=\dfrac{\sqrt{3}\left(\sqrt{3}-1\right)}{3}=\dfrac{3-\sqrt{3}}{3}\)
Rút gọn biểu thức: \(\sqrt{15+6\sqrt{6}}+\sqrt{15-6\sqrt{6}}\) \(6\) \(7\) \(6+2\sqrt{6}\) \(3\sqrt{6}\) Hướng dẫn giải: \(\sqrt{15+6\sqrt{6}}+\sqrt{15-6\sqrt{6}}\) \(=\sqrt{3^2+2.3\sqrt{6}+6}+\sqrt{3^2-2.3.\sqrt{6}+6}\) \(=\sqrt{\left(3+\sqrt{6}\right)^2}+\sqrt{\left(3-\sqrt{6}\right)^2}\) \(=\left|3+\sqrt{6}\right|+\left|3-\sqrt{6}\right|\) \(=3+\sqrt{6}+3-\sqrt{6}\) \(=6\)
Rút gọn biểu thức \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\) 0 2 1 \(2\sqrt{7}\) Hướng dẫn giải: \(\sqrt{4-\sqrt{7}}-\sqrt{4+\sqrt{7}}+\sqrt{2}\) \(=\dfrac{\sqrt{8-2\sqrt{7}}-\sqrt{8+2\sqrt{7}}+2}{\sqrt{2}}\) \(=\dfrac{\sqrt{\left(\sqrt{7}-1\right)^2}-\sqrt{\left(\sqrt{7}+1\right)^2}+2}{\sqrt{2}}\) \(=\dfrac{\sqrt{7}-1-\left(\sqrt{7}+1\right)+2}{\sqrt{2}}\) \(=0\)
Rút gọn biểu thức: \(\sqrt{m+2+2\sqrt{m+1}}-\sqrt{m+2-2\sqrt{m+1}}\) (giả sử các biểu thức đều có nghĩa) \(\sqrt{m+1}+1-\left|\sqrt{m+1}-1\right|\) \(2\sqrt{m+1}\) \(2\) \(2\sqrt{m+1}-2\) Hướng dẫn giải: \(\sqrt{m+2+2\sqrt{m+1}}-\sqrt{m+2-2\sqrt{m+1}}\) \(=\sqrt{\left(\sqrt{m+1}+1\right)^2}-\sqrt{\left(\sqrt{m+1}-1\right)^2}\) \(=\left|\sqrt{m+1}+1\right|-\left|\sqrt{m+1}-1\right|\) \(=\sqrt{m+1}+1-\left|\sqrt{m+1}-1\right|\)
Tính giá trị của biểu thức \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\) tại \(a=\sqrt{3};b=\sqrt{5}\) . \(\sqrt{3}-\sqrt{5}\) \(2\sqrt{15}\) \(\sqrt{3}+\sqrt{5}\) \(\dfrac{\sqrt{3}-\sqrt{5}}{2}\) Hướng dẫn giải: \(\dfrac{a\sqrt{b}+b\sqrt{a}}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\) \(=\dfrac{\sqrt{ab}\left(\sqrt{a}+\sqrt{b}\right)}{\sqrt{ab}}:\dfrac{1}{\sqrt{a}-\sqrt{b}}\) \(=\left(\sqrt{a}+\sqrt{b}\right)\left(\sqrt{a}-\sqrt{b}\right)\) \(=a-b\) Thay \(a=\sqrt{3};b=\sqrt{5}\) thì giá trị biểu thức là: \(a-b=\sqrt{3}-\sqrt{5}\).
Tìm x biết \(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)=-1\). \(x=\dfrac{9}{16}\) \(x=\dfrac{3}{4}\) \(x=\dfrac{3}{4};x=-\dfrac{3}{4}\) \(x=-\dfrac{9}{16}\) Hướng dẫn giải: \(\left(\dfrac{4\sqrt{x}}{2+\sqrt{x}}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{x-2\sqrt{x}}-\dfrac{2}{\sqrt{x}}\right)\) \(=\left(\dfrac{4\sqrt{x}\left(2-\sqrt{x}\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}+\dfrac{8x}{4-x}\right):\left(\dfrac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}-2\right)}-\dfrac{2\left(\sqrt{x}-2\right)}{\sqrt{x}\left(\sqrt{x}-2\right)}\right)\) \(=\dfrac{4x+8\sqrt{x}}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{\sqrt{x}-1-2\sqrt{x}+4}{\sqrt{x}\left(\sqrt{x}-2\right)}\) \(=\dfrac{4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(2+\sqrt{x}\right)\left(2-\sqrt{x}\right)}:\dfrac{-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-2\right)}\) \(=\dfrac{4\sqrt{x}}{2-\sqrt{x}}.\dfrac{\sqrt{x}\left(\sqrt{x}-2\right)}{-\sqrt{x}+3}\) \(=\dfrac{-4x}{-\sqrt{x}+3}\) \(\dfrac{-4x}{-\sqrt{x}+3}=-1\) \(\Leftrightarrow4x=-\sqrt{x}+3\) \(\Leftrightarrow4x+\sqrt{x}-3=0\) \(\Leftrightarrow\left(\sqrt{x}+1\right)\left(4\sqrt{x}-3\right)=0\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}+1=0\\4\sqrt{x}-3=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}\sqrt{x}=-1\left(l\right)\\\sqrt{x}=\dfrac{3}{4}\end{matrix}\right.\) \(\Leftrightarrow x=\dfrac{9}{16}\)
