Question

va rog repede ca nu am timp dau coroana​

Answer 1

Explicație pas cu pas:

1.

$7\sqrt3-4\sqrt3+5\sqrt3-6\sqrt3=\sqrt3(7-4+5-6)=2\sqrt3\\\boxed{d) \ 2\sqrt3}$

2.

$\sqrt{(-2)^6\cdot (-5)^2}-\sqrt{(-2)^4\cdot3^4}-\sqrt{4^3}=\\\sqrt{64\cdot25}-\sqrt{16\cdot81}-\sqrt{64}=\\8\cdot5-4\cdot9-8=\\40-36-8=-4$

$\boxed{a) \ -4}$

3.

$|\sqrt2-\sqrt5|-|\sqrt3-\sqrt5|+|\sqrt3-2|-|\sqrt2-3|=\\\sqrt5-\sqrt2-(\sqrt5-\sqrt3)+2-\sqrt3-(3-\sqrt2)=\\\sqrt5-\sqrt2-\sqrt5+\sqrt3+2-\sqrt3-3+\sqrt2=\\2-3=-1\\\boxed{b) \ -1}$

4.

$3\sqrt{50}-4\sqrt{32}+2\sqrt{18}-\sqrt8=3\cdot5\sqrt2-4\cdot4\sqrt2+2\cdot3\sqrt2-2\sqrt2=\sqrt2(15-16+6-2)=3\sqrt2\\\boxed{c) \ 3\sqrt2}$

5.

$\sqrt{3^2+4^2+12^2}-\sqrt{15^2-12^2}=\sqrt{9+16+144}-\sqrt{225-144}=\sqrt{169}-\sqrt{81}=13-9=4\\\boxed{c) \ 4}$

6.

$2\sqrt2\cdot\sqrt3+3\sqrt3\cdot \sqrt2+4\sqrt3\cdot3\sqrt2-5\sqrt3\cdot4\sqrt2=\\2\sqrt6+3\sqrt6+12\sqrt6-20\sqrt6=\sqrt6(2+3+12-20)=-3\sqrt6\\\boxed{ b) \ -3\sqrt6}$