Question

Расписать 3 примера:
1)Вычислите
1/(√7-√6)-3/(√6-√3)-4/(√7+√3)
2)Вычислите 2a²-5ab+2b²
при a=√7+√6 и b=√7-√6
3)Упростите
3√(3*(2/3))-√132+4*√(2*(1/16))

Answer 1

Решение смотрите на фото

Answer 2

1)

1/(√7-√6)-3/(√6-√3)-4/(√7+√3) $frac{1}{sqrt{7}-sqrt{6}} -frac{3}{sqrt{6}-sqrt{3}} -frac{4}{sqrt{7}+sqrt{3}} = frac{1*(sqrt{7}+sqrt{6})}{(sqrt{7}-sqrt{6})(sqrt{7}+sqrt{6})} -frac{3*(sqrt{6}+sqrt{3})}{(sqrt{6}-sqrt{3})*(sqrt{6}+sqrt{3})} -frac{4*(sqrt{7}-sqrt{3})}{(sqrt{7}+sqrt{3})*((sqrt{7}-sqrt{3}))} = frac{(sqrt{7}+sqrt{6})}{1} -frac{3*(sqrt{6}+sqrt{3})}{6-3} -frac{4*(sqrt{7}-sqrt{3})}{7-3} =$$frac{(sqrt{7}+sqrt{6})}{1} -frac{3*(sqrt{6}+sqrt{3})}{3} -frac{4*(sqrt{7}-sqrt{3})}{4} = sqrt{7} +sqrt{6} -sqrt{6} -sqrt{3} -sqrt{7} +sqrt{3} =0$

2) 2a²-5ab+2b² при a=√7+√6 и b=√7-√6;

$2*(sqrt{7}+sqrt{6})^2 -5*(sqrt{7}+sqrt{6})(sqrt{7}-sqrt{6})+2*(sqrt{7}-sqrt{6})^2=2*(7+2sqrt{7*6}+6) -5*(7-6)+2*(7-2sqrt{7*6}+6)=2*(13+2sqrt{42} )-5+2*2*(13+2sqrt{42} )=26+4sqrt{42} -5+26+4sqrt{42} =52$

3) $3sqrt{3*frac{2}{3}} -sqrt{132} +4*sqrt{2*frac{1}{16}} = 3sqrt{2} -sqrt{4*33} +4*frac{1}{4}sqrt{2} = 3sqrt{2} -2sqrt{33} +sqrt{2} = 4sqrt{2} -2sqrt{33} = 2(2sqrt{2} -sqrt{33})$