Đáp án:
Bài 1:
a.$\sqrt5$
b.$2\sqrt{2}$
Bài 2:
a.$(\sqrt{x}+2)(x-\sqrt{x}+2^2)$
b.$\sqrt{x}(1-\sqrt{x})(1+\sqrt{x}+x)$
c.$(\sqrt{x}-2)(\sqrt{x}-3)$
Giải thích các bước giải:
Bài 1:
a.Ta có:
$\sqrt{7-2\sqrt{10}}+\sqrt{2}$
$=\sqrt{5-2\sqrt{5}\cdot \sqrt{2}+2}+\sqrt{2}$
$=\sqrt{(\sqrt{5}-\sqrt{2})^2}+\sqrt{2}$
$=(\sqrt{5}-\sqrt{2})+\sqrt{2}$
$=\sqrt5$
b.Ta có:
$(\sqrt{3+\sqrt{5}}-\sqrt{3-\sqrt{5}})2$
$=(\sqrt{6+2\sqrt{5}}-\sqrt{6-2\sqrt{5}})\sqrt{2}$
$=(\sqrt{5+2\sqrt{5}+1}-\sqrt{5-2\sqrt{5}+1})\sqrt{2}$
$=(\sqrt{(\sqrt{5}+1)^2}-\sqrt{(\sqrt{5}-1)^2})\sqrt{2}$
$=((\sqrt{5}+1)-(\sqrt{5}-1))\sqrt{2}$
$=2\sqrt{2}$
Bài 2:
a.Ta có:
$x\sqrt{x}+8$
$=(\sqrt{x})^3+2^3$
$=(\sqrt{x}+2)((\sqrt{x})^2-\sqrt{x}+2^2)$
$=(\sqrt{x}+2)(x-\sqrt{x}+2^2)$
b.Ta có:
$\sqrt{x}-x^2$
$=\sqrt{x}(1-x\sqrt{x})$
$=\sqrt{x}(1-(\sqrt{x})^3)$
$=\sqrt{x}(1-\sqrt{x})(1+\sqrt{x}+(\sqrt{x})^2)$
$=\sqrt{x}(1-\sqrt{x})(1+\sqrt{x}+x)$
c.Ta có:
$x-5\sqrt{x}+6$
$=(x-2\sqrt{x})-(3\sqrt{x}-6)$
$=(\sqrt{x}-2)\sqrt{x}-3(\sqrt{x}-2)$
$=(\sqrt{x}-2)(\sqrt{x}-3)$
