Proszę o pomoc w matematyce klasa 1 technikum zadanie 5 i 6!!

[tex]5. \\a) \: {(x - 1)}^{2} - (x - 1)(x + 1) = {x}^{2} + 2 \times ( - 1) \times x + { (- 1)}^{2} - ({x}^{2} - {1}^{2} ) = {x}^{2} - 2x + 1 - {x}^{2} + 1 = - 2x + 2 \\ - 2x + 2 = - 2 \times ( \sqrt{3} + 1) + 2 = - 2 \sqrt{3} - 2 + 2 = - 2 \sqrt{3} \\ b) \: {(x + 3)}^{2} - {(x - 3)}^{2} - 12 = {x}^{2} + 2 \times 3 \times x + {3}^{2} - ( {x}^{2} + 2 \times ( - 3) \times x + { (- 3)}^{2} ) - 12 = {x}^{2} + 6x + 9 - ( {x}^{2} - 6x + 9 ) = {x}^{2} + 6x + 9 - {x}^{2} + 6x - 9 = 6x \\ 6x = 6 \times ( \sqrt{3} + 1) = 6 \sqrt{3} + 6[/tex]
[tex]6. \\ a) \: {( \sqrt{2} + 3) }^{2} + {( \sqrt{2} - 3) }^{2} = { \sqrt{2} }^{2} + 2 \times 3 \times \sqrt{2} + {3}^{2} + { \sqrt{2} }^{2} + 2 \times ( - 3) \times \sqrt{2} + {( - 3)}^{2} = 2 + 6 \sqrt{2} + 9 + 2 - 6 \sqrt{2} + 9 = 22 \\ b) \: {(2 + \sqrt{3}) }^{2} -{ (2 - \sqrt{3} ) }^{2} = {2}^{2} + 2 \times 2 \times \sqrt{3} + { \sqrt{3} }^{2} - ( {2}^{2} + 2 \times 2 \times ( - \sqrt{3} ) + {( - \sqrt{3} )}^{2} ) = 4 + 4 \sqrt{3} + 3 - (4 - 4 \sqrt{3} + 3) = 7 + 4 \sqrt{3} - 7 + 4 \sqrt{3} = 8 \sqrt{3} \\ c) \:{(2 \sqrt{3} - 1) }^{2} - {( \sqrt{3} + 2)}^{2} = {(2 \sqrt{3}) }^{2} + 2 \times ( - 1) \times 2 \sqrt{3} + {( - 1)}^{2} - ( { \sqrt{3} }^{2} + 2 \times 2 \times \sqrt{3} + {2}^{2} ) = 12 - 4 \sqrt{3} + 1 - (3 + 4 \sqrt{3} + 4) = 13 - 4 \sqrt{3} - 7 - 4 \sqrt{3} = 6 - 8 \sqrt{3} \\ d) \: {(2 \sqrt{3} - 1) }^{2} + {(2 \sqrt{3} + 1) }^{2} = {(2 \sqrt{3}) }^{2} + 2 \times ( - 1) \times 2 \sqrt{3} + {( - 1)}^{2} + {(2 \sqrt{3} )}^{2} + 2 \times 1 \times 2 \sqrt{3} + {1}^{2} = 12 - 4 \sqrt{3} + 1 + 12 + 4 \sqrt{3} + 1 = 26[/tex]