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MATeMAtyka 1. Zakres podstawowy. Po gimnazjum

Rozwiązanie 1

a) 9−41​x2≥0   ∣⋅(−4) −36+x2≤0 x2−36≤0 (x−6)(x+6)≤0

a)   −2x2+6x−29​&lt;0   Δ=62−4⋅(−2)⋅(−29​)=      =36−36=0   x=−2⋅(−2)6​=   −−46​=46​=23​=121​     <img src="https://prod.odrabiamy-assets.pl/uploads/content_image/image/31584/akHAWvsb1mXjVv9qzh83yQ%3D%3D_d7bff9e9f87373c4d5e29ed6546e803df9abff1b946f8df8e14118b088f9fb67.jpg"> x∈(−∞, 121​)  ∪  (121​, +∞)​​

a) −x2+4x+1&gt;3x2+4x   ∣−3x2−4x −4x2+1&gt;0   ∣⋅(−1) 4x2−1&lt;0 (2x−1)(2x+1)&lt;0

a)   x2−3x+1≥−x2+2x+8   ∣+x2−2x−8   2x2−5x−7≥0

Wyrażenie znajdujące się pod pierwiastkiem kwadratowym musi być nieujemne.      a)   x2−4≥0   (x−2)(x+2)≥0

a)   x2≤8   ∣−8   x2−8≤0   (x−8<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
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M834 80h400000v40h-400000z'/></svg>​)(x+8<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
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M834 80h400000v40h-400000z'/></svg>​)≤0   (x−22<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
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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
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c-6,0,-10,-1,-12,-3s-194,-422,-194,-422s-65,47,-65,47z
M834 80h400000v40h-400000z'/></svg>​)(x+22<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
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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
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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>​)≤0

Wypiszmy zbiór dzielników całkowitych liczby 12: A={−12, −6, −4, −3, −2, −1, 1, 2, 3, 4, 6, 12}        a)   x2−2x+4≤0   Δ=(−2)2−4⋅1⋅4=4−16=−12&lt;0

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