2 szkoły ponadpodstawowej

II liceum

II technikum

Matematyka 1. Poziom podstawowy i rozszerzony. Po gimnazjum

Przedstawiamy liczbę&nbsp; 0,(72) &nbsp;w postaci ułamka zwykłego. 0,727272…=d |⋅100 &nbsp; 72,727272…=100d &nbsp; 72+d<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
 53.3-86.7 123.7-153 211-199 66.7-36 137.3-56.3 212-62h199568v120H200432c-178.3
 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>0,727272…​​=100d &nbsp; 72+d=100d &nbsp; 72=99d ∣:99 &nbsp; d=9972​=118​ &nbsp; Zatem:&nbsp;

a) 2, &nbsp;ponieważ&nbsp; 2=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>​&lt;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>​&lt;9<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>​=3. &nbsp;

a) {10, 11} &nbsp; b) R−{10, 11} &nbsp;

a) &nbsp;Wypisujemy dzielniki liczby&nbsp; 8: 1, 2, 4, 8. &nbsp; Sprawdzamy, który z dzielników spełnia nierówność: 1−3⋅(1)2=1−3=−2&gt;−14 &nbsp; Liczba&nbsp; 1 &nbsp;spełnia tę formę zdaniową. <hr> b) &nbsp;Równanie&nbsp; x1​=x &nbsp;spełniają tylko dwie liczby:&nbsp; 1 &nbsp;i&nbsp; −1. &nbsp; Liczba&nbsp; 1 &nbsp;nie spełnia nierówności&nbsp; 1&lt;1, &nbsp;ale liczba&nbsp; −1 &nbsp;ją spełnia:&nbsp; −1&lt;1. &nbsp; Liczba&nbsp; −1 &nbsp;spełnia tę formę zdaniową.&nbsp;

a) A∪B=B &nbsp; b) (A∪B)′=B′={x: x∈C ∧ x≤−3} &nbsp;

Warto zacząć od narysowania danych zbiorów: <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/71597/Wv/d4MKL3i7wmZP2Q0Duw%3D%3D_f2fe7f3433097a215935b54185cbb4e77da91b7f0d6ecdabfa275d28f1872086.jpg" alt="Thumb zad1.86str26"> <hr> a) A−B=⟨8,+∞) &nbsp; <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/71598/oKVJRhX3AHdyO/3GgSWrJg%3D%3D_7f6d0c465946c82438170a4d88e401852ebf11b8b79ff4822176da9488bede45.jpg" alt="Thumb zad1.86astr26">

a) A′=(4,+∞) &nbsp; b) B′=(−∞, 1)∪⟨7,+∞) &nbsp;

a) A=⟨−1, 1⟩, B=(−7, 3) &nbsp; b) A=(−4, 3), B=⟨−2, 0) &nbsp;

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