Question

Punctul, b c d va rog calcuati pe lung

Answer 1

b)
(2√5+√2)²-(2√5-√2)²+(2√5+√2)(2√5-√2)=
=(2√5+√2-2√5+√2)(2√5+√2+2√5-√2)+(2√5)²-(√2)²=
=2√2×4√5+20-2=
=8√10+18=
=2(4√10+9)

c)
(2/5-2√3)(2/5+2√3)+(-2/5-2√3)(2/5+2√3)=
=(2/5)²-(2√3)²-(2/5+2√3)²=
=4/25-12-4/25-8√3/5-12=
=-24-8√3/5=
=-8(3-√3/5)

d)(0,7+√11)²-(0,7-√11)(-0,7-√11)+(√11-0,7)(0,7+√11)=
=0,49+1,4√11+11+(0,7-√11)(0,7+√11)+(√11-0,7)(√11+0,7)=
=11,49+1,4√11+0,49-11+11-0,49=
=11,49+1,4√11+11,49-11,49=
=1,4√11+11,49