Đáp án:
Giải thích các bước giải:
a). $\sqrt[]{5-2√6}$ - $\sqrt[]{5+2√6}$
= $\sqrt[]{3-2.√3.√2+2}$ - $\sqrt[]{3+2.√3.√2+2}$
= $\sqrt[]{(√3-√2)^2}$ - $\sqrt[]{(√3 + √2)^2}$
= √3 - √2 - √3 - √2
= -2$\sqrt[]{2}$
b). (5 + 4√2) (3 + 2 $\sqrt[]{1+√2}$ ) (3 - 2 $\sqrt[]{1+√2}$ )
= (5 + 4$\sqrt[]{2}$ ) [9 - (2$\sqrt[]{1+√2}$ )² ]
= (5 + 4$\sqrt[]{2}$ ) [9 - 4 (1 + $\sqrt[]{2}$ )
= (5 + 4$\sqrt[]{2}$ ) (9 - 4 - 4$\sqrt[]{2}$ )
= (5 + 4$\sqrt[]{2}$ ) (5 - 4$\sqrt[]{2}$ )
= 25 - (4$\sqrt[]{2}$)²
= 25 - 32
= -7
c). $\sqrt[]{√2 - 1}$ . $\sqrt[]{2 - \sqrt[]{3-√2} }$ . $\sqrt[]{2 + \sqrt[]{3-√2}}$
= $\sqrt[]{√2 - 1}$ . $\sqrt[]{4 - (\sqrt[]{3-√2})^2 }$
= $\sqrt[]{√2 - 1}$ . $\sqrt[]{4 - 3+√2 }$
= $\sqrt[]{√2 - 1}$ . $\sqrt[]{1 +√2 }$
= $\sqrt[]{2 - 1}$
= $\sqrt[]{1}$
= 1
d). (2 - $\sqrt[]{3}$ ) ( $\sqrt[]{6}$ + $\sqrt[]{2}$ ) $\sqrt[]{2 + \sqrt[]{3}}$
= (2 - $\sqrt[]{3}$ ) (1 + $\sqrt[]{3}$ ) $\sqrt[]{4 + 2\sqrt[]{3} }$
= (2 - $\sqrt[]{3}$ ) (1 + $\sqrt[]{3}$ ) $\sqrt[]{3 + 1 + 2\sqrt[]{3}}$
= (2 - $\sqrt[]{3}$ ) (1 + $\sqrt[]{3}$ ) $\sqrt[]{(\sqrt[]{3} + 1)^2}$
= (2 - $\sqrt[]{3}$ ) (1 + $\sqrt[]{3}$ ) ( $\sqrt[]{3}$ + 1)
= (2 - $\sqrt[]{3}$ ) (1 + $\sqrt[]{3}$ )²
= (2 - $\sqrt[]{3}$ ) ( 1 + 2$\sqrt[]{3}$ + 3)
= (2 - $\sqrt[]{3}$ ) ( 4 + 2$\sqrt[]{3}$ )
= 2 (2 - $\sqrt[]{3}$ ) ( 2 + $\sqrt[]{3}$ )
= 2 (4 - $\sqrt[]{3}$ )
= 2 . 1
= 2
