a)

Z treści zadania wiemy, że

$s\left(x\right)=x^6-1\frac{1}{4}{x}^4+x^2-6$  

$t\left(x\right)=\frac{1}{2}{x}^4-x^2-6$  

Obliczamy różnicę s - t.

$s\left(x\right)-t\left(x\right)=\underset{s\left(x\right)}{\underbrace{x^6-1\frac{1}{4}{x}^4+x^2-6}}-\underset{t\left(x\right)}{\underbrace{\left(\frac{1}{2}{x}^4-x^2-6\right)}}=$

$=x^6-1\frac{1}{4}{x}^4+x^2-6-\frac{1}{2}{x}^4+x^2+6=x^6-1\frac{1}{4}{x}^4-\frac{2}{4}{x}^4+2x^2=\underline{\underline{x^6-1\frac{3}{4}{x}^4+2x^2}}$  

Obliczamy wartość wyznaczonej różnicy dla

$x=-\sqrt{2}$  

Mamy:

$s\left(-\sqrt{2}\right)-t\left(-\sqrt{2}\right)={\left(-\sqrt{2}\right)}^6+\frac{3}{4}\cdot {\left(-\sqrt{2}\right)}^4+2\cdot {\left(-\sqrt{2}\right)}^2=$  

$=8-\frac{1}{3}/4\cdot 4+2\cdot 2=8-\frac{7}{4}\cdot 4+4=8-7+4=\underline{\underline{5}}$  

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b)

Z treści zadania wiemy, że

$s\left(x\right)=\sqrt{1\frac{9}{16}}{x}^4+\sqrt{2}{x}^3-\frac{1}{8}{x}^2=\sqrt{\frac{25}{16}}{x}^4+\sqrt{2}{x}^3-\frac{1}{8}{x}^2=\frac{5}{4}{x}^4+\sqrt{2}{x}^3-\frac{1}{8}{x}^2$  

$t\left(x\right)=0,25x^4-2\sqrt{2}{x}^3+0,375x^2=\frac{1}{4}{x}^4-2\sqrt{2}{x}^3+\frac{3}{8}{x}^2$  

Obliczamy różnicę s - t.

$s\left(x\right)-t\left(x\right)=\underset{s\left(x\right)}{\underbrace{\frac{5}{4}{x}^4+\sqrt{2}{x}^3-\frac{1}{8}{x}^2}}-\underset{t\left(x\right)}{\underbrace{\left(\frac{1}{4}{x}^4-2\sqrt{2}{x}^3+\frac{3}{8}{x}^2\right)}}=$

$=\frac{5}{4}{x}^4+\sqrt{2}{x}^3-\frac{1}{8}{x}^2-\frac{1}{4}{x}^4+2\sqrt{2}{x}^3-\frac{3}{8}{x}^2=\frac{4}{4}{x}^4+3\sqrt{2}{x}^3-\frac{4}{8}{x}^2=\underline{\underline{x^4+3\sqrt{2}{x}^3-\frac{1}{2}{x}^2}}$  

Obliczamy wartość wyznaczonej różnicy dla

$x=-\sqrt{2}$  

Mamy:

$s\left(-\sqrt{2}\right)-t\left(-\sqrt{2}\right)={\left(-\sqrt{2}\right)}^4+3\sqrt{2}\cdot {\left(-\sqrt{2}\right)}^3-\frac{1}{2}\cdot {\left(-\sqrt{2}\right)}^2=$  

$=4+3\sqrt{2}\cdot \left(-2\sqrt{2}\right)-\frac{1}{2}\cdot 2=4-6\cdot 2-1=4-12-1=\underline{\underline{-9}}$