III gimnazjum

Matematyka wokół nas 3

a) &nbsp; (5−1+5)2−(251​)−1:5−2−5−2= 5−2+2+52−25:(251​)−5−2

a) &nbsp; {[(−2)2]−3−(24)−2}⋅2131​ ={2−6−2−8}⋅2131​=

{[(34)−8]−4⋅(31​)17⋅6−5⋅189⋅(271​)40} &nbsp; :⎩<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.016em' style='width:0.8889em' viewBox='0 0 888.89 16' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V16 H384z M384 0 H504 V16 H384z'/></svg>⎨<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.016em' style='width:0.8889em' viewBox='0 0 888.89 16' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V16 H384z M384 0 H504 V16 H384z'/></svg>⎧​(0,0625)−3⋅12−5⋅[(−31​)−2]4⎭<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.016em' style='width:0.8889em' viewBox='0 0 888.89 16' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V16 H384z M384 0 H504 V16 H384z'/></svg>⎬<svg xmlns="http://www.w3.org/2000/svg" width='0.8889em' height='0.016em' style='width:0.8889em' viewBox='0 0 888.89 16' preserveAspectRatio='xMinYMin'><path d='M384 0 H504 V16 H384z M384 0 H504 V16 H384z'/></svg>⎫​ &nbsp; =

[(21​)−3]2=(2x+1)3⋅(−21​)2 (21​)−6=(23x+3)⋅(−2)−2

a) &nbsp; 16⋅4−1=42⋅4−1=41=4 b)

a) &nbsp; x−3⋅(x1​)2⋅x6=x−3⋅x−2⋅x6=x

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