Câu 71 Trang 40: Rút gọn biểu thức

Ta có :

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

=  $\sqrt{16}-6+\sqrt{20}-\sqrt{5}$

=  $4-6+2\sqrt{5}-\sqrt{5}=\sqrt{5}-2$

Vậy  $(\sqrt{8}-3\sqrt{2}+\sqrt{10})\sqrt{2}-\sqrt{5}=\sqrt{5}-2$

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

=  $0,2\left | -10 \right |\sqrt{3}+2\left | \sqrt{3}-\sqrt{5} \right |$

=  $0,2.10.\sqrt{3}+2(\sqrt{5}-\sqrt{3})$

=  $2\sqrt{3}+2\sqrt{5}-2\sqrt{3}=2\sqrt{5}$

Vậy $0,2.\sqrt{(-10^{2}).3}+2\sqrt{(\sqrt{3}-\sqrt{5})^{2}}=2\sqrt{5}$

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

=  $\left ( \frac{1}{2}\sqrt{\frac{2}{2^{2}}}-\frac{3}{2} \sqrt{2}+\frac{4}{5}\sqrt{10^{2}.2}\right ):\frac{1}{8}$\

=  $\left ( \frac{1}{4}\sqrt{2}-\frac{3}{2} \sqrt{2}+8\sqrt{2}\right ):\frac{1}{8}$

=  $\frac{27}{4}\sqrt{2}:\frac{1}{8}=\frac{27}{4}\sqrt{2}.8=54\sqrt{2}$

Vậy $\left ( \frac{1}{2}\sqrt{\frac{1}{2}}-\frac{3}{2} \sqrt{2}+\frac{4}{5}\sqrt{200}\right ):\frac{1}{8}=54\sqrt{2}$

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

=  $2\left | \sqrt{2}-3 \right |+\left | -3 \right |\sqrt{2}-5\left | -1 \right |$

=  $2(3-\sqrt{2})+3\sqrt{2}-5=6-2\sqrt{2}+3\sqrt{2}-5=1+\sqrt{2}$

Vậy $2\sqrt{(\sqrt{2}-3)^{2}}+\sqrt{2(-3)^{2}}-5\sqrt{(-1)^{4}}=1+\sqrt{2}$