Giải câu 4 trang 33 toán VNEN 9 tập 1

Giải câu 4 trang 33 toán VNEN 9 tập 1.

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

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

= 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 $\sqrt{2}$ + $\sqrt{20}$ ). $\sqrt{5}$ - $(3 \sqrt{\frac{1}{10}} + 10)$

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

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

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

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