Question

a=4$\sqrt{3}$-2$sqrt{6} }$
b=7$sqrt{6} } }$+$\sqrt{3}$
a+b=..........
a-b=.......
Va rog sa-mi explicati cum se face...

Answer 1

sqrt 6 inseamna radical din 6

a= 4sqrt3 - 2 sqrt 6
b= 7sqrt6 + sqrt 3

radicalul il consideram ca un x si un y adica radical din 3 e x si radical din 6 e y.
acum tot ce e in fata lui x se aduna si la fel si la y si obtinem

a+b =  (4sqrt3 - 2 sqrt 6)+(7sqrt6 + sqrt 3) ==
==(4+1) sqrt3 + (7-2) sqrt6 = 5 sqrt3+ 5 sqrt6
a-b=   (4sqrt3 - 2 sqrt 6)-(7sqrt6 + sqrt 3) ==
== (4-1) sqrt3 + (-2 -7) sqrt6 =  3 sqrt3 - 9 sqrt6

Answer 2

$a=4 \sqrt{3} -2 \sqrt{6} \\ b=7 \sqrt{6} + \sqrt{3} \\ a+b=4 \sqrt{3} -2 \sqrt{6} +7 \sqrt{6} + \sqrt{3} =5 \sqrt{3} +5 \sqrt{6} \\ a-b=4 \sqrt{3} -2 \sqrt{6} -(7 \sqrt{6} + \sqrt{3} )=4 \sqrt{3} -2 \sqrt{6} -7 \sqrt{6} - \sqrt{3} = \\ =3 \sqrt{3} -9 \sqrt{6} \\ a+b=5 \sqrt{3} +5 \sqrt{6} ~;~a-b=3 \sqrt{3} -9 \sqrt{6}$