Question

va rog frumos tot ce este incercuit cu albastru​

Answer 1

Răspuns:

$g)7 \sqrt{3} - 3(2 + \sqrt{3} ) + 1 = \\ 7 \sqrt{3} - 6 - 3 \sqrt{3} + 1 = \\ = 4 \ \sqrt{3} - 5$

$b)4 \sqrt{3} ( - \frac{1}{2} \sqrt{2}) = - 2 \sqrt{6}$

$f) \sqrt{6} \times \sqrt{2} + 2 \sqrt{12} - 3 + \sqrt{12} = \\ \sqrt{12} + 2 \sqrt{12} + \sqrt{12} - 3 = 4 \sqrt{12 } - 3$

$h)4 - 2( \sqrt{3} + 1) + 5 \sqrt{3} = \\ 4 - 2 \sqrt{3} - 2 + 5 \sqrt{3} = 2 + 3 \sqrt{3}$

$i)4( \sqrt{3} + 2) - 2 \sqrt{3} - 8 = \\ 4 \sqrt{3} + 8 - 2 \sqrt{3} - 8 = 2 \sqrt{3}$

$j)10 \sqrt{5} - 5( \sqrt{5} + 2) - 4 + 3 \sqrt{5 } = \\ 10 \sqrt{5 } - 5 \sqrt{5} - 10 - 4 + 3 \sqrt{5} = \\ = 8 \sqrt{3} - 14$

$k)3( \sqrt{6} + 1) - 2(1 + \sqrt{6} ) = \\ 3 \sqrt{6} + 3 - 2 - 2 \sqrt{6} = \sqrt{6 } + 1$

$l)3( \sqrt{7} - 2 ) + 2(3 - \sqrt{7} ) = \\ 3 \sqrt{7 } - 6 + 6 - 2 \sqrt{7} = \sqrt{7}$

Answer 2

Răspuns:

Semnul acesta "^" este radicalul

b)4^3 x -^2 supra 2

 -2^3 ^2= -2^6

g) 7^3-6-3^3+1

=4^3-5

f)^12+4^3-3+2^3

=8^3-3

h)4-2^3-2+5^3

=2+3^3

i)4^3+8-2^3-8

=2^3

j) 10^5-5^5-10-4+3^5

=8^5-14

k) 3^6+3-2-2^6

=^6+1

l) 3^7-6+6-2^7

=^7

Sper ca te-am ajutat!!