a) 

Zał: 

$x\ne 0$  

 

${\left(\frac{5}{9}\right)}^{\frac{1}{x}}\ge \frac{5}{9}$  

${\left(\frac{5}{9}\right)}^{\frac{1}{x}}\ge {\left(\frac{5}{9}\right)}^1$  

Zmieniamy znak nierówności na przeciwny: 

$\frac{1}{x}\le 1\mid \cdot x^2$  

$x\le x^2\mid -x$  

$0\le x^2-x$  

$0\le x\left(x-1\right)$  

$x\in (-\infty ,0⟩\cup ⟨1,+\infty )\wedge x\ne 0$  

$x\in \left(-\infty ,0\right)\cup ⟨1,+\infty )$  

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b)

Zał:

$x\ne 0$   

 

${\left(\frac{2}{3}\right)}^{\frac{3}{x}}<\sqrt[3]{\frac{27}{8}}$  

${\left(\frac{2}{3}\right)}^{\frac{3}{x}}<\frac{3}{2}$  

${\left(\frac{2}{3}\right)}^{\frac{3}{x}}<{\left(\frac{2}{3}\right)}^{-1}$  

Zmieniamy znak nierówności na przeciwny:

$\frac{3}{x}>-1\mid \cdot x^2$  

$3x>-x^2\mid +x^2$  

$x^2+3x>0$  

$x\left(x+3\right)>0$  

$x\in \left(-\infty ,-3\right)\cup \left(0,+\infty \right)$  

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c)

Zał:

$x+2\ne 0$  

$x\ne -2$  

 

${\left(\frac{1}{9}\right)}^{\frac{2x-3}{x+2}}<81$  

${\left(\frac{1}{9}\right)}^{\frac{2x-3}{x+2}}<{\left(\frac{1}{9}\right)}^{-2}$  

Zmieniamy znak nierówności na przeciwny:

$\frac{2x-3}{x+2}>-2\mid \cdot {\left(x+2\right)}^2$  

$\left(2x-3\right)\left(x+2\right)>-2{\left(x+2\right)}^2\mid +2{\left(x+2\right)}^2$  

$\left(2x-3\right)\left(x+2\right)+2{\left(x+2\right)}^2>0$  

$\left(x+2\right)\cdot \left(\left(2x-3\right)+2\left(x+2\right)\right)>0$  

$\left(x+2\right)\left(2x-3+2x+4\right)>0$  

$\left(x+2\right)\left(4x+1\right)>0$  

$4\left(x+2\right)\left(x+\frac{1}{4}\right)>0\mid :4$  

$\frac{x+2}{x+\frac{1}{4}}>0$  

$x\in \left(-\infty ,-2\right)\cup \left(-\frac{1}{4},+\infty \right)$  

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d)

Zał:

$x-1\ne 0$  

$x\ne 1$  

 

$\frac{4}{25}\ge {\left(2,5\right)}^{\frac{x+1}{x-1}}$  

${\left(\frac{2}{5}\right)}^2\ge {\left(\frac{5}{2}\right)}^{\frac{x+1}{x-1}}$  

${\left(\frac{5}{2}\right)}^{-2}\ge {\left(\frac{5}{2}\right)}^{\frac{x+1}{x-1}}$  

$-2\ge \frac{x+1}{x-1}\mid \cdot {\left(x-1\right)}^2$  

$-2{\left(x-1\right)}^2\ge \left(x+1\right)\left(x-1\right)\mid +2{\left(x-1\right)}^2$  

$0\ge \left(x+1\right)\left(x-1\right)+2{\left(x-1\right)}^2$  

$0\ge \left(x-1\right)\left(\left(x+1\right)+2\left(x-1\right)\right)$  

$0\ge \left(x-1\right)\left(x+1+2x-2\right)$  

$0\ge \left(x-1\right)\left(3x-1\right)$  

$0\ge 3\left(x-1\right)\left(x-\frac{1}{3}\right)\mid :3$  

$0\ge \left(x-1\right)\left(x-\frac{1}{3}\right)$  

$x\in ⟨\frac{1}{3},1⟩\wedge x\ne 1$  

$x\in ⟨\frac{1}{3},1)$  

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e)

Zał:

$x-1\ne 0$  

$x\ne 1$  

 

$4^{\frac{x+2}{x-1}}\le 2^8$  

${\left(2^2\right)}^{\frac{x+2}{x-1}}\le 2^8$  

$2^{\frac{2x+4}{x-1}}\le 2^8$  

$\frac{2x+4}{x-1}\le 8\mid \cdot {\left(x-1\right)}^2$  

$2\left(x+2\right)\left(x-1\right)\le 8{\left(x-1\right)}^2\mid :2$  

$\left(x+2\right)\left(x-1\right)\le 4{\left(x-1\right)}^2\mid -4{\left(x-1\right)}^2$  

$\left(x+2\right)\left(x-1\right)-4{\left(x-1\right)}^2\le 0$  

$\left(x-1\right)\left(\left(x+2\right)-4\left(x-1\right)\right)\le 0$  

$\left(x-1\right)\left(x+2-4x+4\right)\le 0$  

$\left(x-1\right)\left(-3x+6\right)\le 0$  

$-3\left(x-1\right)\left(x-2\right)\le 0\mid :\left(-3\right)$  

$\left(x-1\right)\left(x-2\right)\ge 0$  

$x\in (-\infty ,1⟩\cup ⟨2,+\infty )\wedge x\ne 1$  

$x\in \left(-\infty ,1\right)\cup ⟨2,+\infty )$  

 

Uwaga !!!

W zbiorze zadań podano błędną odpowiedź.

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f)

Zał:

$x+2\ne 0$  

$x\ne -2$  

 

${\left(\frac{3}{7}\right)}^{\frac{4x-3}{x+2}}\ge \sqrt[4]{\frac{49}{9}}$  

${\left(\frac{3}{7}\right)}^{\frac{4x-3}{x+2}}\ge {\left(\frac{49}{9}\right)}^{\frac{1}{4}}$  

${\left(\frac{3}{7}\right)}^{\frac{4x-3}{x+2}}\ge {\left({\left(\frac{3}{7}\right)}^{-2}\right)}^{\frac{1}{4}}$  

${\left(\frac{3}{7}\right)}^{\frac{4x-3}{x+2}}\ge {\left(\frac{3}{7}\right)}^{-\frac{1}{2}}$  

Zmieniamy znak nierówności na przeciwny:

$\frac{4x-3}{x+2}\le -\frac{1}{2}\mid \cdot \left(-2{\left(x+2\right)}^2\right)$  

$-2\left(4x-3\right)\left(x+2\right)\ge {\left(x+2\right)}^2\mid +2\left(4x-3\right)\left(x+2\right)$  

$0\ge {\left(x+2\right)}^2+2\left(4x-3\right)\left(x+2\right)$  

$0\ge \left(x+2\right)\left(\left(x+2\right)+2\left(4x-3\right)\right)$  

$0\ge \left(x+2\right)\left(x+2+8x-6\right)$  

$0\ge \left(x+2\right)\left(9x-4\right)$  

$0\ge 9\left(x+2\right)\left(x-\frac{4}{9}\right)\mid :9$  

$0\ge \left(x+2\right)\left(x-\frac{4}{9}\right)$  

$x\in ⟨-2,\frac{4}{9}⟩\wedge x\ne -2$  

$x\in (-2,\frac{4}{9}⟩$