Question

3(4a-b)^2-2(a-b)(a+b)+4(a+3b)^2 при a=-0.2, b=-1

Answer 1

$3(4a-b)^{2} -2(a-b)(a+b)+4(a+3b)^{2}= \\ =3(16a^{2} -8ab+b^{2} )-2(a^{2} -b^{2} )+4(a^{2} +6ab+9b^{2} )=\\ =48a^{2} -24ab+3b^{2} -2a^{2} +2b^{2} +4a^{2} +24ab+36b^{2} =\\ =50a^{2} +41b^{2}$

$a=-0,2; a^{2} =0,04\\ b=-1; b^{2} =1$

$50*0,04+41*1=2+41=43$