Rút gọn các biểu thức sau: (căn 8 - 3 căn 2 + căn 10) . căn 2 - căn 5

Giải Toán 9 Ôn tập chương 1

Video Giải Bài 71 trang 40 Toán lớp 9 Tập 1

Bài 71 trang 40 Toán lớp 9 Tập 1: 

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

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

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

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

Lời giải:

a) 

$\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right)\sqrt{2}-\sqrt{5}=\left(\sqrt{2^2.2}-3\sqrt{2}+\sqrt{5.2}\right)\sqrt{2}-\sqrt{5}=\left(2\sqrt{2}-3\sqrt{2}+\sqrt{2}.\sqrt{5}\right).\sqrt{2}-\sqrt{5}=\left(2-3+\sqrt{5}\right).\sqrt{2}.\sqrt{2}-\sqrt{5}=\left(\sqrt{5}-1\right).2-\sqrt{5}=2\sqrt{5}-2-\sqrt{5}=\sqrt{5}-2$

b)

$0,2\sqrt{{\left(-10\right)}^2.3}+2\sqrt{{\left(\sqrt{3}-\sqrt{5}\right)}^2}=0,\mathrm{2.10.}\sqrt{3}+2.\left.\sqrt{3}-\sqrt{5}\right|=2\sqrt{3}+2\left(\sqrt{5}-\sqrt{3}\right)=2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}$

c)

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

d)

$2\sqrt{{\left(\sqrt{2}-3\right)}^2}+\sqrt{2.{\left(-3\right)}^2}-5\sqrt{{\left(-1\right)}^4}=2\left.\sqrt{2}-3\right|+3\sqrt{2}-5\sqrt{1}=2.\left(3-\sqrt{2}\right)+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}$