$a)$  

$x^3+x^2-12x=0$  

$x\underset{x_2=\frac{-1+7}{2}=\frac{6}{2}=3}{\underset{x_1=\frac{-1-7}{2}=\frac{-8}{2}=-4}{\underset{\sqrt{\Delta }=7}{\underset{=1+48=49}{\underset{\Delta =1^2-4\cdot 1\cdot \left(-12\right)=}{\underbrace{\left(x^2+x-12\right)}}}}}}=0$  

$x=0\text{lub}x=-4\text{lub}x=3$  

 

 

 

$b)$  

$x^4+6x^3+5x^2=0$  

$x^2\underset{x_2=\frac{-6+4}{2}=\frac{-2}{2}=-1}{\underset{x_1=\frac{-6-4}{2}=\frac{-10}{2}=-5}{\underset{\sqrt{\Delta }=4}{\underset{=36-20=16}{\underset{\Delta =6^2-4\cdot 1\cdot 5=}{\underbrace{\left(x^2+6x+5\right)}}}}}}=0$  

$x=0\text{lub}x=-5\text{lub}x=-1$  

 

 

 

$c)$  

$2x^3+5x^2+2x=0$    

$x\underset{x_2=\frac{-5+3}{2\cdot 2}=\frac{-2}{4}=-\frac{1}{2}}{\underset{x_1=\frac{-5-3}{2\cdot 2}=\frac{-8}{4}=-2}{\underset{\sqrt{\Delta }=3}{\underset{=25-16=9}{\underset{\Delta =5^2-4\cdot 2\cdot 2=}{\underbrace{\left(2x^2+5x+2\right)}}}}}}=0$  

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

 

 

$d)$  

$7x^3+2x=x^2\mid -x^2$  

$7x^3-x^2+2x=0$  

$x\underset{=1-56<0}{\underset{\Delta ={\left(-1\right)}^2-4\cdot 7\cdot 2=}{\underbrace{\left(7x^2-x+2\right)}}}=0$  

Czynnik kwadratowy nie daje rozwiązania, więc równanie ma tylko jedno rozwiązanie.

$x=0$  

 

 

 

$e)$  

$3x^4+6x^3=3x^2\mid :3$  

$x^4+2x^3=x^2\mid -x^2$  

$x^4+2x^3-x^2=0$  

$x^2\underset{x_2=\frac{-2+2\sqrt{2}}{2}=-1+\sqrt{2}}{\underset{x_1=\frac{-2-2\sqrt{2}}{2}=-1-\sqrt{2}}{\underset{\sqrt{\Delta }=\sqrt{8}=\sqrt{4}\cdot \sqrt{2}=2\sqrt{2}}{\underset{\Delta =2^2-4\cdot 1\cdot \left(-1\right)=4+4=8}{\underbrace{\left(x^2+2x-1\right)}}}}}=0$  

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

 

 

 

$f)$  

$x^5+\frac{1}{4}{x}^3=x^4\mid -x^4$  

$x^5-x^4+\frac{1}{4}{x}^3=0$  

$x^3\left(x^2-x+\frac{1}{4}\right)=0$  

$x^3\left(x^2-2\cdot x\cdot \frac{1}{2}+{\left(\frac{1}{2}\right)}^2\right)=0$  

$x^3{\left(x-\frac{1}{2}\right)}^2=0$  

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

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

$x=0\text{lub}x=\frac{1}{2}$  

 

 

 

$g)$  

$x^3-\sqrt{2}{x}^2-7x=0$  

$x\underset{x_2=\frac{\sqrt{2}+\sqrt{30}}{2}}{\underset{x_1=\frac{\sqrt{2}-\sqrt{30}}{2}}{\underset{\sqrt{\Delta }=\sqrt{30}}{\underset{=2_{28}-30}{\underset{\Delta ={\left(-\sqrt{2}\right)}^2-4\cdot 1\cdot \left(-7\right)=}{\underbrace{\left(x^2-\sqrt{2}x-7\right)}}}}}}=0$  

$x=0\text{lub}x=\frac{\sqrt{2}-\sqrt{30}}{2}\text{lub}x=\frac{\sqrt{2}+\sqrt{30}}{2}$  

 

 

 

$h)$  

$x^5+x^4+2x^3=0$  

$x^3\underset{=1-4<0}{\underset{\Delta =1^2-4\cdot 1\cdot 1=}{\underbrace{\left(x^2+x+2\right)}}}=0$  

 

Czynnik kwadratowy nie daje rozwiązania, więc równanie ma tylko jedno rozwiązanie.

$x^3=0$  

$x=0$  

 

 

 

$i)$  

$\frac{x^4}{2}+x^2=\sqrt{2}{x}^3\mid -\sqrt{2}{x}^3$  

$\frac{1}{2}{x}^4-\sqrt{2}{x}^3+x^2=0$  

$x^2\underset{x_0=\frac{\sqrt{2}}{2\cdot \left(\frac{1}{2}\right)}=\frac{\sqrt{2}}{1}=\sqrt{2}}{\underset{2-2=0}{\underset{\Delta ={\left(-\sqrt{2}\right)}^2-4\cdot \frac{1}{2}\cdot 1=}{\underbrace{\left(\frac{1}{2}{x}^2-\sqrt{2}x+1\right)}}}}=0$  

$x=0\text{lub}x=\sqrt{2}$