Bài 71 (Trang 40 SGK Toán 9, Tập 1):

Rút gọn các biểu thức sau:

$\mathrm{a}) \left(\sqrt{8}-3\sqrt{2} +\sqrt{10}\right)\sqrt{2}-\sqrt{5};$

$\mathrm{b}) 0,2\sqrt{{\left(-10\right)}^2.3} +2\sqrt{{\left(\sqrt{3}-\sqrt{5}\right)}^2};$

$\mathrm{c}) \left(\frac{1}{2}.\sqrt{\frac{1}{2}} -\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right):\frac{1}{8};$

$\mathrm{d}) \sqrt{{\left(\sqrt{2}-3\right)}^2} +\sqrt{2.{\left(-3\right)}^2} -5\sqrt{{\left(-1\right)}^4}.$

Hướng dẫn giải:

$$\begin{gathered} \mathrm{a}) \left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5} = \left(\sqrt{2^2.2}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}- \sqrt{5} \\ = \left(2\sqrt{2} - 3\sqrt{2} + \sqrt{10}\right)\sqrt{2} - \sqrt{5} \\ = 2.{\left(\sqrt{2}\right)}^2 - 3{\left(\sqrt{2}\right)}^2 + \sqrt{10}.\sqrt{2} - \sqrt{5} = 4 - 6 + \sqrt{20} -\sqrt{5} = -2+\sqrt{4.5}-\sqrt{5 }= -2 + 2\sqrt{5} -\sqrt{5} \\ = -2 + \sqrt{5}. \end{gathered}$$

$$\begin{gathered} \mathrm{b}) 0,2\sqrt{\left({\left(-10\right)}^2.3\right)} + 2\sqrt{{\left(\sqrt{3} - \sqrt{5}\right)}^2} \\ = 0,2.10\sqrt{3} + 2\left|\sqrt{3} - \sqrt{5}\right| \\ = 2\sqrt{3} +\hspace{0.17em}2\left(\sqrt{5} - \sqrt{3}\right) \\ = 2\sqrt{3} + 2\sqrt{5} - 2\sqrt{3} \\ = 2\sqrt{5} \end{gathered}$$

$$\begin{gathered} \mathrm{c}) \left(\frac{1}{2}.\sqrt{\frac{1}{2}}-\frac{3}{2}\sqrt{2}+\frac{4}{5}\sqrt{200}\right):\frac{1}{8} = \left(\frac{1}{2}.\sqrt{\frac{1}{2}}-3\sqrt{\frac{2}{2^2}}+ \frac{4}{5}\sqrt{\frac{400}{2}}\right):\frac{1}{8} \\ = \left(\frac{1}{2}.\sqrt{\frac{1}{2}}- 3\sqrt{\frac{1}{2}}+\frac{4}{5}.20\sqrt{\frac{1}{2}}\right):\frac{1}{8} \\ = \left(\frac{1}{2}.\sqrt{\frac{1}{2}}-3\sqrt{\frac{1}{2}}+16\sqrt{\frac{1}{2}}\right).8 = \frac{27}{2}\sqrt{\frac{1}{2}}.8= 27.4\sqrt{\frac{1}{2}} = 27.2\sqrt{2^2\frac{1}{2}} =54\sqrt{2} \end{gathered}$$

$$\begin{gathered} d) 2\sqrt{{\left(\sqrt{2}-3\right)}^2} +\sqrt{2{\left(-3\right)}^2} -5\sqrt{{\left(-1\right)}^2} \\ = 2\left|\sqrt{2} -3\right|+3\sqrt{2} -5\sqrt{1} =2\left(3-\sqrt{2}\right)+ 3\sqrt{2}-5 \\ = 6 - 2\sqrt{2}+3\sqrt{2}-5 =1+\sqrt{2} \end{gathered}$$