Giải câu 4 trang 33 toán VNEN 9 tập 1

a) $\left ( \sqrt{\frac{9}{2}} + \sqrt{\frac{1}{2}} - \sqrt{2} \right )$.$\sqrt{2}$

= $\left ( \frac{\sqrt{9}}{\sqrt{2}}+ \frac{\sqrt{1}}{\sqrt{2}} - \sqrt{2} \right )$.$\sqrt{2}$ 

= 3 + 1 - 2 = 2

b) ($\sqrt{3}$ - $\sqrt{2}$ + 1)($\sqrt{3}$ - 1)

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

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

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

c) $(\sqrt{2} + \sqrt{5})^{2}$ = $(\sqrt{2})^{2}$ + $(\sqrt{5})^{2}$ + 2$\sqrt{2}$$\sqrt{5}$ = 2 + 5 + 2$\sqrt{2}$$\sqrt{5}$  = 7 + 2$\sqrt{10}$

d) ($\sqrt{8}$ - 5$\sqrt{2}$ + $\sqrt{20}$).$\sqrt{5}$ - $\left ( 3\sqrt{\frac{1}{10}} + 10 \right )$

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

= (- 3$\sqrt{2}$ + 2$\sqrt{5}$).$\sqrt{5}$ - $\left ( 3\frac{\sqrt{10}}{10} + 10 \right )$

= - 3$\sqrt{10}$ + 10 - 3$\sqrt{\frac{\sqrt{10}}{10}}$ - 10 

= - 33$\frac{\sqrt{10}}{10}$.