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.
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Binomul lui Newton la cazul general este:

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 (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](https://tex.z-dn.net/?f=+%28x%2By%29%5E%7B2011%7D+%2B+%28-x-y%29%5E%7B2011%7D+%3D++%5C%5C+%3D%28x%2By%29%5E%7B2011%7D+%2B+%5B-%28x%2By%29%5D%5E%7B2011%7D%3D+%5C%5C%3D%28x%2By%29%5E%7B2011%7D+%2B+%5B%28-1%29%28x%2By%29%5D%5E%7B2011%7D%3D++%5C%5C+%3D%28x%2By%29%5E%7B2011%7D+%2B+%28-1%29%5E%7B2011%7D%2A%28x%2By%29%5E%7B2011%7D%3D+%5C%5C%3D%28x%2By%29%5E%7B2011%7D+%2B+%28-1%29%2A%28x%2By%29%5E%7B2011%7D%3D++%5C%5C%3D%28x%2By%29%5E%7B2011%7D+-+%28x%2By%29%5E%7B2011%7D%3D+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 \,\,\,\,\,\,(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](https://tex.z-dn.net/?f=+%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%5C%2C%286x-4y%2B2%29%5E%7B2012%7D-16%5E%7B503%7D%2A%283x-2y%2B1%29%5E%7B2012%7D%3D++%5C%5C%3D+%286x-4y%2B2%29%5E%7B2012%7D-%28+2%5E%7B4%7D+%29%5E%7B503%7D%2A%283x-2y%2B1%29%5E%7B2012%7D%3D+%5C%5C%3D+%286x-4y%2B2%29%5E%7B2012%7D-2%5E%7B4%2A503%7D+%2A%283x-2y%2B1%29%5E%7B2012%7D%3D+%5C%5C++%3D%286x-4y%2B2%29%5E%7B2012%7D-2%5E%7B2012%7D+%2A%283x-2y%2B1%29%5E%7B2012%7D%3D++%5C%5C%3D%286x-4y%2B2%29%5E%7B2012%7D-%5B2+%2A%283x-2y%2B1%29%5D%5E%7B2012%7D%3D+%5C%5C%3D%286x-4y%2B2%29%5E%7B2012%7D-%282%2A3x-2%2A2y%2B2%2A1%29%5E%7B2012%7D%3D+%5C%5C+%3D%286x-4y%2B2%29%5E%7B2012%7D-%286x-4y%2B2%29%5E%7B2012%7D%3D++0+)
Rezolvarea exercitiilor de algebra:
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