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Matematyka 1. Zakres rozszerzony

Rozwiązanie 1

x=0,2(5)   ∣⋅10   10x=2,(5)   10x=2,(5)<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMinYMin slice'><path d='M0 6l6-6h17c12.688 0 19.313.3 20 1 4 4 7.313 8.3 10 13
 35.313 51.3 80.813 93.8 136.5 127.5 55.688 33.7 117.188 55.8 184.5 66.5.688
 0 2 .3 4 1 18.688 2.7 76 4.3 172 5h399450v120H429l-6-1c-124.688-8-235-61.7
-331-161C60.687 138.7 32.312 99.3 7 54L0 41V6z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMidYMin slice'><path d='M199572 214
c100.7 8.3 195.3 44 280 108 55.3 42 101.7 93 139 153l9 14c2.7-4 5.7-8.7 9-14
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 11.7-311.7 78.3-403 201-6 8-9.7 12-11 12-.7.7-6.7 1-18 1s-17.3-.3-18-1c-1.3 0
-5-4-11-12-44.7-59.3-101.3-106.3-170-141s-145.3-54.3-229-60H0V214z'/></svg><svg xmlns="http://www.w3.org/2000/svg" width='400em' height='0.548em' viewBox='0 0 400000 548' preserveAspectRatio='xMaxYMin slice'><path d='M399994 0l6 6v35l-6 11c-56 104-135.3 181.3-238 232-57.3
 28.7-117 45-179 50H-300V214h399897c43.3-7 81-15 113-26 100.7-33 179.7-91 237
-174 2.7-5 6-9 10-13 .7-1 7.3-1 20-1h17z'/></svg>2,3+0,2(5)​​

Wyznaczamy część wspólną danych przedziałów: A∩B=⟨1, 4⟩   <hr> Szacujemy przedział, do którego należy liczba  a:   a=412+3<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​​&lt;412+4​=416​=4   a=412+3<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​​&gt;412​=3 Otrzymaliśmy:

a) Zauważmy, że: 5<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​&gt;4<svg xmlns="http://www.w3.org/2000/svg" width='400em' height='1.08em' viewBox='0 0 400000 1080' preserveAspectRatio='xMinYMin slice'><path d='M95,702
c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5,-10,-9.5,-14
c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54
c44.2,-33.3,65.8,-50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10
s173,378,173,378c0.7,0,35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429
c69,-144,104.5,-217.7,106.5,-221
l0 -0
c5.3,-9.3,12,-14,20,-14
H400000v40H845.2724
s-225.272,467,-225.272,467s-235,486,-235,486c-2.7,4.7,-9,7,-19,7
c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​=2      Zatem łatwo zauważyć, że:

Zauważmy, że co 4 lata jest rok przestępny (trwa on 366 dni). W ciągu 100 lat jest 25 lat przestępnych.

Z treści zadania wiemy, że: x=y+1,  x∈Z   x−1=y,  x∈Z   oraz x=6k+4,  k∈Z        Zatem otrzymujemy:

a) x=6k−2=6k−6+4=6(k−1)+4,  r=4

a) 2x2=8   ∣:2   x2=4   x=2  ∨  x=−2   x∈{−2, 2}   <hr> b) (x+1)(2x−6)=0   x+1=0  ∨  2x−6=0   x=−1  ∨  2x=6  ∣:2   x=−1  ∨  x=3   x∈{−1, 3}

a)   3⋅21​(21​−4m)+21​=1,5⋅21​(2⋅21​+6)   23​(21​−4m)+21​=23​⋅21​(1+6)

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