CMR:
a/ $\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}=}\sqrt{3}+\sqrt{5}-\sqrt{2}$
b/ $\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}=\sqrt{2}+1$
CMR: a/ $\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}=}\sqrt{3}+\sqrt{5}-\sqrt{2}$ b/ $\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}=\sqrt{2}+1$
By Amara
Đáp án:
Giải thích các bước giải:
a/ $\sqrt{10+\sqrt{60}-\sqrt{24}-\sqrt{40}}$
$=\sqrt{(\sqrt{2})^2+(\sqrt{3})^2+(\sqrt{5})^2+2\sqrt{3}.\sqrt{5}-2\sqrt{2}.\sqrt{3}-2\sqrt{2}.\sqrt{5}}$
$=\sqrt{(\sqrt{3}+\sqrt{5}-\sqrt{2})^2}$
$=\sqrt{3}+\sqrt{5}-\sqrt{2}$
b/ $\sqrt{6+\sqrt{24}+\sqrt{12}+\sqrt{8}}-\sqrt{3}$
$=\sqrt{(\sqrt{2})^2+(\sqrt{3})^2+1+2\sqrt{2}.\sqrt{3}+2.1\sqrt{3}+2.1\sqrt{2}}-\sqrt{3}$
$=\sqrt{(\sqrt{2}+\sqrt{3}+1)^2}-\sqrt{3}$
$=\sqrt{2}+\sqrt{3}+1-\sqrt{3}$
$=\sqrt{2}+1$
a) Ta có:
$10 + \sqrt{60} -\sqrt{24} – \sqrt{40}$
$= 5 + 3 + 2 + 2\sqrt{15} – 2\sqrt{6} – 2\sqrt{10}$
$= (\sqrt5)^2 + (\sqrt3)^2 + (\sqrt2)^2 + 2\sqrt3.\sqrt5 – 2\sqrt2.\sqrt3 – 2\sqrt2.\sqrt5$
$= (\sqrt2 -\sqrt5 – \sqrt3)^2$
$= (\sqrt3 + \sqrt5 – \sqrt2)^2$
Do đó:
$\sqrt{10 + \sqrt{60} -\sqrt{24} – \sqrt{40}} = \sqrt3 + \sqrt5 -\sqrt2$
b) Ta có:
$6 + \sqrt{24} + \sqrt{12} + \sqrt8$
$= 1 + 2 + 3 + 2\sqrt6 + 2\sqrt3 + 2\sqrt2$
$= 1^2 + (\sqrt2)^2 + (\sqrt3)^2 + 2\sqrt2.\sqrt3 + 2\sqrt3.1 + 2\sqrt2.1$
$= (1 + \sqrt2 + \sqrt3)^2$
Do đó:
$\sqrt{6 + \sqrt{24} + \sqrt{12} + \sqrt8} = 1 + \sqrt2 + \sqrt3$
Hay $\sqrt{6 + \sqrt{24} + \sqrt{12} + \sqrt8} -\sqrt3 = \sqrt2 + 1$