Question

Tính

Answer 1

a) \(\sqrt{12}-\sqrt{27}+\sqrt{3}=\sqrt{4.3}-\sqrt{9.3}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}=0\)

b) \(\left(\sqrt{12}-2\sqrt{75}\right)\sqrt{3}=\sqrt{36}-2\sqrt{225}=6-30=-24\)

c) \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}=\sqrt{36.7}-\sqrt{100.7}+\sqrt{144.7}-\sqrt{64.7}\)

\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}=0\)

d) \(3\left(\sqrt{12}+\sqrt{27}-\sqrt{3}\right)=3\left(\sqrt{4.3}+\sqrt{9.3}-\sqrt{3}\right)\)

\(=3\left(2\sqrt{3}+3\sqrt{3}-\sqrt{3}\right)=3.4\sqrt{3}=12\sqrt{3}\)

Answer 2

e, \(\sqrt{12}-\sqrt{27}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}=0\)

b, \(\left(\sqrt{12}-2\sqrt{75}\right)\sqrt{2}=\left(2\sqrt{3}-2.5\sqrt{3}\right)\sqrt{2}\)

\(=4\sqrt{3}-10\sqrt{6}\)

tương tự 

Answer 3

Lời giải:

a.

$\sqrt{12}-\sqrt{27}+\sqrt{3}=2\sqrt{3}-3\sqrt{3}+\sqrt{3}=(2-3+1)\sqrt{3}=0.\sqrt{3}=0$

b.

$(\sqrt{12}-2\sqrt{75})\sqrt{3}=(2\sqrt{3}-10\sqrt{3}).\sqrt{3}$

$=-8\sqrt{3}.\sqrt{3}=-8.3=-24$

c.

$\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}$

$=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}$

$=\sqrt{7}(6-10+12-8)=0\sqrt{7}=0$

d.

$3(\sqrt{12}+\sqrt{27}-\sqrt{3})=3(2\sqrt{3}+3\sqrt{3}-\sqrt{3})$

$=3.4\sqrt{3}=12\sqrt{3}$

Answer 4

a) \(\sqrt{12}-\sqrt{27}+\sqrt{3}\)

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

=0

b) \(\left(\sqrt{12}-2\sqrt{75}\right)\cdot\sqrt{3}\)

\(=\left(2\sqrt{3}-10\sqrt{3}\right)\cdot\sqrt{3}\)

=-24

c) \(\sqrt{252}-\sqrt{700}+\sqrt{1008}-\sqrt{448}\)

\(=6\sqrt{7}-10\sqrt{7}+12\sqrt{7}-8\sqrt{7}\)

\(=0\)