$\text{a)}{\left(x^2-2x\right)}^2+5\left(x^2-2x\right)+4=0$  

Podstawiamy x²-2x=t.

$t^2+5t+4=0$  

$\Delta =5^2-4\cdot 1\cdot 4=25-16=9,\sqrt{\Delta }=\sqrt{9}=3$  

$t=\frac{-5-3}{2}=\frac{-8}{2}=-4\text{lub}t=\frac{-5+3}{2}=\frac{-2}{2}=-1$  

Wracamy z podstawieniem do zmiennej x.

$x^2-2x=-4\text{lub}{x}^2-2x=-1\mid +1$  

$x^2-2x+1=-3\text{lub}{x}^2-2x+1=0$  

$\underset{\text{r. sprzeczne}}{{\left(x-1\right)}^2=-3}\text{lub}{\left(x-1\right)}^2=0$  

${\left(x-1\right)}^2=0$  

$x-1=0$  

$x=1$  

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$\text{b)}{\left(x^2+4x\right)}^2+7\left(x^2+4x\right)+12=0$  

Podstawiamy x²+4x=t.

$t^2+7t+12=0$  

$\Delta =7^2-4\cdot 1\cdot 12=49-48=1,\sqrt{\Delta }=\sqrt{1}=1$  

$t=\frac{-7-1}{2}=\frac{-8}{2}=-4\text{lub}t=\frac{-7+1}{2}=\frac{-6}{2}=-3$  

Wracamy z podstawieniem do zmiennej x.

$x^2+4x=-4\text{lub}{x}^2+4x=-3\mid +4$  

$x^2+4x+4=0\text{lub}{x}^2+4x+4=1$  

${\left(x+2\right)}^2=0\text{lub}{\left(x+2\right)}^2=1$  

$x+2=0\text{lub}x+2=-1\text{lub}x+2=1$  

$x=-2\text{lub}x=-3\text{lub}x=-1$  

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$\text{c)}\left(x^2-5x\right)\left(x^2-5x+2\right)=24$  

Podstawiamy x²-5x=t.

$t\left(t+2\right)=24$  

$t^2+2t=24\mid +1$  

$t^2+2t+1=25$  

${\left(t+1\right)}^2=25$  

$t+1=-5\text{lub}t+1=5$  

$t=-6\text{lub}t=4$  

Wracamy z podstawieniem do zmiennej x.

$x^2-5x=-6\text{lub}{x}^2-5x=4$  

$\underset{\Delta =1}{x^2-5x+6=0}\text{lub}\underset{\Delta =41}{x^2-5x-4=0}$  

$x=\frac{5-1}{2}=\frac{4}{2}=2\text{lub}x=\frac{5+1}{2}=\frac{6}{2}=3\text{lub}x=\frac{5-\sqrt{41}}{2}\text{lub}x=\frac{5+\sqrt{41}}{2}$  

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$\text{d)}\left(x^2+2x\right)\left(x^2+2x-1\right)=2$  

Podstawiamy x²+2x=t.

$t\left(t-1\right)=2$  

$t^2-t-2=0$  

$\Delta ={\left(-1\right)}^2-4\cdot 1\cdot \left(-2\right)=1+8=9,\sqrt{\Delta }=\sqrt{9}=3$  

$t=\frac{1-3}{2}=\frac{-2}{2}=-1\text{lub}t=\frac{1+3}{2}=\frac{4}{2}=2$  

Wracamy z podstawieniem do zmiennej x.

$x^2+2x=-1\text{lub}{x}^2+2x=2\mid +1$  

$x^2+2x+1=0\text{lub}{x}^2+2x+1=3$  

${\left(x+1\right)}^2=0\text{lub}{\left(x+1\right)}^2=3$  

$x+1=0\text{lub}x+1=-\sqrt{3}\text{lub}x+1=\sqrt{3}$  

$x=-1\text{lub}x=-1-\sqrt{3}\text{lub}x=-1+\sqrt{3}$