Question

10ав-(-5а+в)кв при а =√9 в=√5

Answer 1

$10ab-(-5a+b)^{2} =$
$=10ab-(b-5a)^{2}=$
$=10ab-(b^{2}-2*b*5a+(5a)^{2})=$
$=10ab-(b^{2}-10ab+25a^{2})=$
$=10ab-b^{2}+10ab-25a^{2}=$
$=-25a^{2}-b^{2}+20ab=$
$=-25* (sqrt{9})^{2} -(sqrt{5})^{2}+ sqrt{9}* sqrt{5}=$
$=-25*9-5+ sqrt{45}=$
$= 225-5+3 sqrt{5}=$
$=220+3 sqrt{5}$

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