a)
[latex]frac{25^{x^2}}{ (sqrt[4]{5})^{8x}} leq frac{625}{5^{4x}}[/latex]
dziedzina: [latex]x in R[/latex]
[latex]frac{(5^2)^{x^2}}{ (5^{ frac{1}{4} })^{8x}} leq frac{5^4}{5^{4x}}[/latex]
[latex]frac{5^{2x^2}}{5^{2x}} leq 5^{4-4x}[/latex]
[latex]5^{2x^2-2x} leq 5^{4-4x}[/latex]
[latex]2x^2-2x le 4-4x[/latex]
[latex]2x^2-2x -4+4xle 0[/latex]
[latex]2x^2+2x -4le 0 /:2[/latex]
[latex]x^2+x -2le 0[/latex]
ciąg dalszy w załączniku 1
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b)
[latex]frac{4}{3}left(frac{2}{3}
ight) ^{x+2} + frac{4}{9}left(frac{2}{3}
ight) ^{x+1} geq frac{8}{9} / cdot frac{9}{4}[/latex]
dziedzina: [latex]x in R[/latex]
[latex]3left(frac{2}{3}
ight) ^{x+2} + left(frac{2}{3}
ight) ^{x+1} geq 2[/latex]
[latex]3 cdot frac{2}{3} cdot left(frac{2}{3}
ight) ^{x+1} + left(frac{2}{3}
ight) ^{x+1} geq 2[/latex]
[latex]2left(frac{2}{3}
ight) ^{x+1} + left(frac{2}{3}
ight) ^{x+1} geq 2[/latex]
[latex]3left(frac{2}{3}
ight) ^{x+1}geq 2 /:3[/latex]
[latex]left(frac{2}{3}
ight) ^{x+1}geq frac{2}{3}[/latex]
[latex]left(frac{2}{3}
ight) ^{x+1}geq left(frac{2}{3}
ight)^1[/latex]
Podstawa jest mniejsza od 1 więc zmieniamy znak nierówności
[latex]x+1 le 1[/latex]
[latex]x le 1-1[/latex]
[latex]x le 0[/latex]
