Question

Dau coroana si 15 puncte:

(√3+√2)²=
(√5-√3)²=
(√3-2)²=
(√3-1)x(√3+1)=
(√10-√3)x(√10+3)=
(√2+1)²=
(2√3-3)²=
(2√2-1)x(2√2+1)=
(aicea este radical lung)√11-4√7=
(aicea este radical lung)√7+2√6=
(aicea este radical lung)√7-4√3=
(aicea este radical lung)√13-2√42=

Answer 1

(√3+√2)²=3+2√6+2=5+2√6
(√5-√3)²=5-2√15+3=8-2√15
(√3-2)²=3-4√3+4=7-4√3
(√3-1)(√3+1)=3-1=2          avem formula (a-b)(a+b)=a²-b²
(√10-√3)(√10+√3)=10-3=7
(√2+1)²=2+2√2+1=3+2√2
(2√3-3)²=12-12√3+9=21-12√3
(2√2-1)(2√2+1)=8-1=7
√11-4√7=√(2-√7)²=2-√7
√7+2√6=√(1+√6)²=1+√6
√7-4√3=√(2-√3)²=2-√3
√13-2√42=√(√7-√6)²=√7-√6