Question

1.Cum se calculeaza (x+y) la puterea 2011 + (-x-y) la puterea 2011 dar (6x-4y+2) la puterea 2012 - 16 la puterea 503 x ( 3x -2y+1) la puterea 2012?
2.Consideram un romb ABCD si un punct E ∉ (ABC). Fie M,N,P,Q mijloacele segmentelor AE,EC,BC, respectiv CD. Aratati ca MN perpendicular pe PQ.

Answer 1

Binomul lui Newton la cazul general este:              
$(x+y)^{n}= \Sigma_{k = 0}^{n}C_{n}^{k}* x^{n-k}* y^{k} \\ \\ (x-y)^{n}= \Sigma_{k = 0}^{n} (-1)^{k}* C_{n}^{k}* x^{n-k}* y^{k}$

Rezolvarea exercitiilor de algebra:

$(x+y)^{2011} + (-x-y)^{2011} = \\ =(x+y)^{2011} + [-(x+y)]^{2011}= \\=(x+y)^{2011} + [(-1)(x+y)]^{2011}= \\ =(x+y)^{2011} + (-1)^{2011}*(x+y)^{2011}= \\=(x+y)^{2011} + (-1)*(x+y)^{2011}= \\=(x+y)^{2011} - (x+y)^{2011}= 0$

   

$\,\,\,\,\,\,(6x-4y+2)^{2012}-16^{503}*(3x-2y+1)^{2012}= \\= (6x-4y+2)^{2012}-( 2^{4} )^{503}*(3x-2y+1)^{2012}= \\= (6x-4y+2)^{2012}-2^{4*503} *(3x-2y+1)^{2012}= \\ =(6x-4y+2)^{2012}-2^{2012} *(3x-2y+1)^{2012}= \\=(6x-4y+2)^{2012}-[2 *(3x-2y+1)]^{2012}= \\=(6x-4y+2)^{2012}-(2*3x-2*2y+2*1)^{2012}= \\ =(6x-4y+2)^{2012}-(6x-4y+2)^{2012}= 0$