a)

Obliczamy:

${\left(\sqrt{4-\sqrt{15}}-\sqrt{4+\sqrt{15}}\right)}^2=$

Korzystamy ze wzoru skróconego mnożenia na kwadrat różnicy ${\left(a-b\right)}^2=a^2-2ab+b^2$:

$={\left(\sqrt{4-\sqrt{15}}\right)}^2-2\cdot \sqrt{4-\sqrt{15}}\cdot \sqrt{4+\sqrt{15}}+{\left(\sqrt{4+\sqrt{15}}\right)}^2=$

$=4-\sqrt{15}-2\cdot \sqrt{\left(4-\sqrt{15}\right)\left(4+\sqrt{15}\right)}+4+\sqrt{15}=$

Korzystamy ze wzoru skróconego mnożenia na różnicę kwadratów $a^2-b^2=\left(a-b\right)\left(a+b\right)$:

$=8-2\cdot \sqrt{4^2-{\left(\sqrt{15}\right)}^2}=8-2\cdot \sqrt{16-15}=8-2\cdot \sqrt{1}=8-2\cdot 1=8-2=\boxed{6}$

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b)

Obliczamy:

$\sqrt{2{\left(2-2\sqrt{2}\right)}^2}+\sqrt{2{\left(2+2\sqrt{2}\right)}^2}=$

$=\sqrt{2}\cdot \sqrt{{\left(2-2\sqrt{2}\right)}^2}+\sqrt{2}\cdot \sqrt{{\left(2+2\sqrt{2}\right)}^2}=$

Korzystamy z tego, że $\sqrt{a^2}=\left|a\right|$ i otrzymujemy:

$=\sqrt{2}\cdot \left|2-2\sqrt{2}\right|+\sqrt{2}\cdot \left|2+2\sqrt{2}\right|=$

Zauważmy, że:

$2-2\sqrt{2}<0$, więc $\left|2-2\sqrt{2}\right|=-\left(2-2\sqrt{2}\right)=-2+2\sqrt{2}=2\sqrt{2}-2$

$2+2\sqrt{2}>0$, więc $\left|2+2\sqrt{2}\right|=2+2\sqrt{2}$

Zatem:

$=\sqrt{2}\cdot \left(2\sqrt{2}-2\right)+\sqrt{2}\left(2+2\sqrt{2}\right)=2\cdot 2-2\sqrt{2}+2\sqrt{2}+2\cdot 2=$

$=4-2\sqrt{2}+2\sqrt{2}+4=\boxed{8}$

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c)

Obliczamy:

${\left[{\left(2\sqrt{2}-\sqrt{7}\right)}^{\frac{1}{2}}+{\left(2\sqrt{2}+\sqrt{7}\right)}^{\frac{1}{2}}\right]}^2=$

 Korzystamy ze wzoru skróconego mnożenia na kwadrat sumy ${\left(a+b\right)}^2=a^2+2ab+b^2$:

$={\left[{\left(2\sqrt{2}-\sqrt{7}\right)}^{\frac{1}{2}}\right]}^2+2\cdot {\left(2\sqrt{2}-\sqrt{7}\right)}^{\frac{1}{2}}\cdot {\left(2\sqrt{2}+\sqrt{7}\right)}^{\frac{1}{2}}+{\left[{\left(2\sqrt{2}+\sqrt{7}\right)}^{\frac{1}{2}}\right]}^2=$

$={\left(2\sqrt{2}-\sqrt{7}\right)}^{\frac{1}{2}\cdot 2}+2\cdot {\left[\left(2\sqrt{2}-\sqrt{7}\right)\left(2\sqrt{2}+\sqrt{7}\right)\right]}^{\frac{1}{2}}+{\left(2\sqrt{2}+\sqrt{7}\right)}^{\frac{1}{2}\cdot 2}=$

Korzystamy ze wzoru skróconego mnożenia na różnicę kwadratów $a^2-b^2=\left(a-b\right)\left(a+b\right)$:

$=2\sqrt{2}-\sqrt{7}+2\cdot {\left[{\left(2\sqrt{2}\right)}^2-{\left(\sqrt{7}\right)}^2\right]}^{\frac{1}{2}}+2\sqrt{2}+\sqrt{7}=$

$=4\sqrt{2}+2\cdot {\left(4\cdot 2-7\right)}^{\frac{1}{2}}=4\sqrt{2}+2\cdot {\left(8-7\right)}^{\frac{1}{2}}=4\sqrt{2}+2\cdot 1^{\frac{1}{2}}=$

$=4\sqrt{2}+2\cdot 1=\boxed{4\sqrt{2}+2}$

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d)

Obliczamy:

$\frac{\sqrt{2}-1}{\sqrt{2}+1}-\sqrt[3]{\frac{5\sqrt{2}-7}{5\sqrt{2}+7}}=\frac{\sqrt{2}-1}{\sqrt{2}+1}-\frac{\sqrt[3]{5\sqrt{2}-7}}{\sqrt[3]{5\sqrt{2}+7}}=$

Możemy na przykład sprowadzić ułamki do wspólnego mianownika:

$=\frac{\sqrt{2}-1}{\sqrt{2}+1}\cdot \frac{\sqrt[3]{5\sqrt{2}+7}}{\sqrt[3]{5\sqrt{2}+7}}-\frac{\sqrt[3]{5\sqrt{2}-7}}{\sqrt[3]{5\sqrt{2}+7}}\cdot \frac{\sqrt{2}+1}{\sqrt{2}+1}=$

$=\frac{\left(\sqrt{2}-1\right)\cdot \sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\cdot \left(\sqrt{2}+1\right)}{\left(\sqrt{2}+1\right)\cdot \sqrt[3]{5\sqrt{2}+7}}=$

$=\frac{\sqrt[3]{{\left(\sqrt{2}-1\right)}^3}\cdot \sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\cdot \sqrt[3]{{\left(\sqrt{2}+1\right)}^3}}{\left(\sqrt{2}+1\right)\cdot \sqrt[3]{5\sqrt{2}+7}}=$

Korzystamy ze wzoru skróconego mnożenia:

• na sześcian różnicy ${\left(a-b\right)}^3=a^3-3a^{2}b+3a{b}^2-b^3$ i zauważamy, że:

${\left(\sqrt{2}-1\right)}^3={\left(\sqrt{2}\right)}^3-3\cdot {\left(\sqrt{2}\right)}^2\cdot 1+3\cdot \sqrt{2}\cdot 1^2-1^3=2\sqrt{2}-6+3\sqrt{2}-1=5\sqrt{2}-7$

• na sześcian sumy ${\left(a+b\right)}^3=a^3+3a^{2}b+3a{b}^2+b^3$ i zauważamy, że:

${\left(\sqrt{2}+1\right)}^3={\left(\sqrt{2}\right)}^3+3\cdot {\left(\sqrt{2}\right)}^2\cdot 1+3\cdot \sqrt{2}\cdot 1^2+1^3=2\sqrt{2}+6+3\sqrt{2}+1=5\sqrt{2}+7$

Zatem:

$=\frac{\sqrt[3]{5\sqrt{2}-7}\cdot \sqrt[3]{5\sqrt{2}+7}-\sqrt[3]{5\sqrt{2}-7}\cdot \sqrt[3]{5\sqrt{2}+7}}{\left(\sqrt{2}+1\right)\cdot \sqrt[3]{5\sqrt{2}+7}}=$

$=\frac{0}{\left(\sqrt{2}+1\right)\cdot \sqrt[3]{5\sqrt{2}+7}}=0$