a)

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

 

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

 

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

 

$=8-2\cdot \sqrt{16-15}=$  

 

$=8-2\cdot \sqrt{1}=$  

 

$=8-2=$  

 

$=6$  

 

 

b)

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

 

$=\sqrt{2\left(4-8\sqrt{2}+8\right)}+\sqrt{2\left(4+8\sqrt{2}+8\right)}=$                  

 

$=\sqrt{2\left(12-8\sqrt{2}\right)}+\sqrt{2\left(12+8\sqrt{2}\right)=}$     

 

$=\sqrt{24-16\sqrt{2}}+\sqrt{24+16\sqrt{2}}$   

 

$\sqrt{24-16\sqrt{2}}+\sqrt{24+16\sqrt{2}}=a\mid {}^2$  

 

$24-16\sqrt{2}+2\cdot \sqrt{24-16\sqrt{2}}\cdot \sqrt{24+16\sqrt{2}}+24+16\sqrt{2}=a^2$   

 

$48+2\cdot \sqrt{576-512}=a^2$  

 

$48+2\cdot \sqrt{64}=a^2$  

 

$48+2\cdot 8=a^2$  

 

$48+16=a^2$  

 

$64=a^2$  

 

$a=8$  

 

Uwaga!

$\sqrt{x^2}=\left|x\right|$  , więc a nie może być równe -8         

 

$\sqrt{2{\left(2-2\sqrt{2}\right)}^2}+\sqrt{2{\left(2+2\sqrt{2}\right)}^2}=8$   

 

    

 

 

c)

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

 

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

 

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

 

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

 

$=4\sqrt{2}+2\cdot 1^{\frac{1}{2}}=$  

 

$=4\sqrt{2}+2$   

 

 

d)

$\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}=a\mid {}^2$  

 

$11-6\sqrt{2}+2\cdot \sqrt{11-6\sqrt{2}}\cdot \sqrt{11+6\sqrt{2}}+11+6\sqrt{2}=a^2$  

 

$22+2\cdot \sqrt{\left(11-6\sqrt{2}\right)\cdot \left(11+6\sqrt{2}\right)}=a^2$  

 

$22+2\cdot \sqrt{121-72}=a^2$  

 

$22+2\sqrt{49}=a^2$   

 

$22+2\cdot 7=a^2$  

 

$22+14=a^2$  

 

$36=a^2$  

 

$a=6$  

 

Uwaga!

$\sqrt{x^2}=\left|x\right|$                   

 

$\sqrt{11-6\sqrt{2}}+\sqrt{11+6\sqrt{2}}=6$