Question

147. Вычислите:
а) (5 – 125);
6) (/7 + 28);
в) (/13 – 12) - (13 + 12); е) (Wз – (12).
r) (5/3 +6,2)-(573 – 6/2);
д) (2 – (8) ;
помогите с 147

Answer 1

Ответ:

Объяснение:

$(\sqrt{5}-\sqrt{125})^2=\sqrt{5}^2-2\sqrt{125}\sqrt{5}+\sqrt{125}^2=5-2\sqrt{5*5*5*5}+125=130-2*25=130-50=80$

$(\sqrt{7}+ \sqrt{28})^2=7+2\sqrt{7} \sqrt{28}+28=35+2\sqrt{7*7*4} =35+2*14=35+28=63$

$(\sqrt{13}- \sqrt{12})( \sqrt{13} +\sqrt{12})=\sqrt{13}^2- \sqrt{12}^2=13-12=1$

$(5\sqrt{3}+6 \sqrt{2})( 5\sqrt{3}-6\sqrt{2})=25\sqrt{3}^2-36 \sqrt{2}^2=25*3-36*2=75-72=3$

$(\sqrt{2} -\sqrt{8})^3= \sqrt{2}^3-6\sqrt{8}+24\sqrt{2}-\sqrt{8}^3=2\sqrt{2}-16\sqrt{2}-12\sqrt{2}+24\sqrt{2}=-2\sqrt{2}$

$(\sqrt{3} -\sqrt{12})^3= \sqrt{3}^3-9\sqrt{12}+36\sqrt{3}-\sqrt{12}^3=3\sqrt{3}-24\sqrt{3}-18\sqrt{3}+26\sqrt{3}=-13\sqrt{3}$