Question

Va rog sa-mi rezolvati si mie macar jumatate din acele calcule a.(2+√3)|2
b.(1+√2)|2
c.(√5+2)|2
d.(2√3+1)|2
e.(√3-2)|2
f.(2-√2)|2
g.(3-√3)|2
h.(3√2-1)|2
i.(√3+5)(5-√3)

Answer 1

a) (2+√3)² = 2²+2*2*√3+√3² = 4+4√3+3 = 7+4√3
b) (1+√2)² = 1²+2*1*√2+√2² = 1+2√2+2 = 3+2√2
c) (√5+2)² = √5²+2*√5*2+2² = 5+4√5+4 = 9+4√5
d) (2√3+1)² = (2√3)²+2*2√3*1+1² = 12+4√3+1 = 13+4√3 
e) (√3-2)² = √3²-2*√3*2+2² = 3-4√3+4 = 7-4√3
f) (2-√2)² = 2²-2*√2*2+√2² = 4-4√2+2 = 2-4√2
g) (3-√3)² = 3²-2*3*√3+√3² = 9-6√3+3 = 12-6√3
h) (3√2-1)² =(3√2)²-2*1*3√2+1² = 18-6√2+1 = 17-6√2 
i) (√3+5)(5-√3) = 3√5-3+25-5√3 = 3√5+22-5√3