Question

1) √(27 + 10 × √2) + √(11 – 6 × √2)
2) 2 × √(7 – 4 × √3) + √(13 – 4 × √3)
по действиям пожалуйста и с обьяснением

Answer 1

1)

√(27+10√2)+√(11-6√2)=√(25+2*5*√2+2)+√(9-2*3*√2+2)=

=√(5²+2*5*√2+(√2)√+√(3²-2*3*√2+(√2)²)=√(5+√2)²+√(3-√2)²=5+√2+3-√2=8.

2)

2*√(7-4*√3)+√(11-4√3)=2*√(4-2*2*√3+3)+√(1-4*√3+12)=

=2*√(2²-2*2*√3+(√3)²)+√(1²-2*1*2√3+(2√3)²=2*√(2-√3)²+√(1-2√3)²=

2*(2-√3)+|1-2√3|=4-2√3+2√3-1=3.

Answer 2

$displaystyle tt 1) sqrt{27+10sqrt{2}} +sqrt{11-6sqrt{2}} = \\ =sqrt{25+10sqrt{2}+2} +sqrt{9-6sqrt{2}+2} = sqrt{(5+sqrt{2})^2} +sqrt{(3-sqrt{2})^2} = \\ = 5+sqrt{2}+3-sqrt{2}=8$

$displaystyle tt 2) 2sqrt{7-4sqrt{3}} +sqrt{13-4sqrt{3}} = \\ 2sqrt{4-4sqrt{3}+3} +sqrt{1-4sqrt{3}+12}=2sqrt{(2-sqrt{3})^2} +sqrt{(1-2sqrt{3})^2}= \\=2(2-sqrt{3})+|1-2sqrt{3}| =4-2sqrt{3}+2sqrt{3}-1 = 3$