Question

Помогите срочно надо

Answer 1

$\left ( \sqrt{7}+3 \right )\left ( 3\sqrt{7}-1 \right )=3\sqrt{7}\cdot \sqrt{7}-\sqrt{7}+3\cdot 3\sqrt{7}-3=3\cdot 7-\sqrt{7}+9\sqrt{7}-3=\\=21-3+8\sqrt{7}=18+8\sqrt{7}\\\\\left ( 4\sqrt{2}-\sqrt{3} \right )\left ( 2\sqrt{2}+5\sqrt{3} \right )=4\sqrt{2}\cdot 2\sqrt{2}+4\sqrt{2}\cdot 5\sqrt{3}-2\sqrt{2}\cdot \sqrt{3}-5\sqrt{3}\cdot \sqrt{3}=\\=8\cdot 2+20\sqrt{2}\cdot \sqrt{3}-2\sqrt{2}\sqrt{3}-5\cdot 3=16+18\sqrt{2}\cdot \sqrt{3}-15=1+18\sqrt{6}$

$\left ( \sqrt{p} -q\right )\left ( \sqrt{p}+1 \right )=\left ( \sqrt{p} \right )^2-q^2=p-q^2\\\\\left ( 6-\sqrt{13} \right )\left ( 6+\sqrt{13} \right )=6^2-\left ( \sqrt{13} \right )^2=36-13=23$

Answer 2

1)

$( \sqrt{7} + 3)(3 \sqrt{7} - 1) = 3 { \sqrt{7} }^{2} - \sqrt{7} + 9 \sqrt{7} - 3 = 3 \times 7 \ - \sqrt{7} + 9 \sqrt{7} = 21 - \sqrt{7} - 9 \sqrt{7} = 21 - 10 \sqrt{7}$

2)

$(4 \sqrt{2} - \sqrt{3})(2 \sqrt{2} + 5 \sqrt{3}) = 8 { \sqrt{2} }^{2} + 20 \sqrt{6} - 2 \sqrt{6} - 5 { \sqrt{3} }^{2} = 8 \times 2 + 18 \sqrt{6} - 5 \times 3 = 16 + 18 \sqrt{6} - 15 = 1 + 18 \sqrt{6}$

3)

$( \sqrt{p} - q)( \sqrt{p} + q) = { \sqrt{p} }^{2} - {q}^{2} = p - {q}^{2}$

4)

$(6 - \sqrt{13} )(6 + \sqrt{13}) = {6}^{2} - { \sqrt{13} }^{2} = 36 - 13 = 23$

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