Question

Пж помогите........... ​

Answer 1

Ответ:

1)

$\displaystyle (2\sqrt{6} -\sqrt{3)} (\sqrt{3}+2\sqrt{6} )=(2\sqrt{6} -\sqrt{3)} (2\sqrt{6}+\sqrt{3} ) = \\\\(2\sqrt{6} )^{2} - (\sqrt{3} )^{2} =4*6-3=24-3=21$

2)

$\displaystyle (12-\sqrt{7} )} (\sqrt{7}+12)= (12-\sqrt{7} )} (12+\sqrt{7}) = \\\\12^{2} - (\sqrt{7} )^{2} =144-7=137$

3)

$\displaystyle (7\sqrt{a}+2\sqrt{b} )} (2\sqrt{b}-7\sqrt{a} )= (2\sqrt{b}+7\sqrt{a} )} (2\sqrt{b}-7\sqrt{a} )=\\\\(2\sqrt{b} )^{2} - (7\sqrt{a} )^{2} =4b - 49a$

4)

$\displaystyle (\sqrt{10}-2\sqrt{3} )} (2\sqrt{3}+\sqrt{10} )= (\sqrt{10}-2\sqrt{3} )} (\sqrt{10}+2\sqrt{3} )=\\\\(\sqrt{10} )^{2} - (2\sqrt{3} )^{2} =10-4*3=10-12=-2$

По формуле разности квадратов:

$\displaystyle (a-b)(a+b)=a^{2} -b^{2}$